Figure 2.1 Biologists count spawning adult sockeye salmon to obtain a stock-recruitment relationship (see Figure 2.43).

images

Preview

“Mathematics, in one view, is the science of infinity.”

Philip Davis and Reuben Hersh,
The Mathematical Experience, Birkhäuser
Boston, 1981, p. 152.

Calculus, one of the great intellectual achievements of humankind, came to fruition through the work of Sir Isaac Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716). It consists of two parts, differential calculus and integral calculus, both of which hinge on the concept of a limit in which the behavior of a function is described as its argument approaches a selected limiting value. In this chapter, we first discuss some of the basic properties of a limit and the associated concept of continuity. Using limits, we then introduce the main concept in differential calculus: the derivative. This concept is one of the fundamental ideas in mathematics and a cornerstone of modern scientific thought. It allows us to come to grips with the phenomenon of change in the value of variables over increasingly small intervals of time or space (or any other independent variable—variables representing such things as the velocity of a stooping falcon or the density of a population of fish). By finding the derivative of functions, we calculate the slopes of tangent lines to these functions and examine rates of change defined by these functions. Applications considered in this chapter include enzyme kinetics, biodiversity, foraging of hummingbirds, wolf predation, and population dynamics of the sockeye salmon depicted in Figure 2.1.

2.1 Rates of Change and Tangent Lines

One of the fundamental concepts in calculus is that of the limit. Intuitively, “taking a limit” corresponds to investigating the value of a function as you get closer and closer to a specified point without actually reaching that point. To “motivate limits,” we begin by showing how they arise when we consider rates of change and the tangent lines to graphs of functions.

Rates of change

Rates of change describe how quantities change with respect to a variable such as time or body mass. For instance, in countries where overpopulation is an issue, projections are constantly made about the population growth rate. For a patient receiving drug treatment, physicians perform experiments to estimate the clearance rate of the drug (i.e., the rate at which the drug leaves the blood stream). To calculate a rate of change, we first select an interval of interest and then compute the average rate of change over that interval using the equation in the box below. To calculate the rate of change at a point of interest requires taking limits, as discussed after the next example.

Average Rate of Change

The average rate of change of f over the interval [a, b] is

images

Example 1 Mexico population growth

The (estimated) population size of Mexico (in millions) in the early 1980s is reported in the following table.

images

From this table, we see that N(t) denotes the population size in millions t years after 1980.

  1. Compute the population's average rate of change (i.e., the average population growth rate) for the interval of time [3, 5]. Identify the units for this average rate of change and interpret this rate of change.
  2. Compute the average population growth rate for the interval [1, 3]. How does this compare to your answer in a? What does this imply about the population growth?
  3. Compute the average population growth rates for the intervals [0, 5], [0, 4], [0, 3], [0, 2], and [0, 1]. Discuss the trend of these growth rates.

Solution

  1. The average growth rate for N(t) over [3, 5] is given by

    images

    The units of this growth rate are millions per year. Hence, between the years 1983 and 1985, the population increases on average by 1.915 million individuals per year.

  2. The average growth rate for N(t) over [1, 3] is given by

    images

    The population is growing on average of approximately 1.8 million per year. Thus, the average growth rate from 1981 to 1983 is less than the average growth rate from 1983 to 1985. The population growth rate in Mexico appears to be increasing over the time period 1983 to 1985.

  3. Computing the average growth rates over the requested time intervals yields
    Time interval Average growth rate
    [0, 5] 1.84
    [0, 4] 1.82
    [0, 3] 1.80
    [0, 2] 1.78
    [0, 1] 1.75

    The average population growth rate is decreasing over the smaller time intervals and appears to converge to a value close to, but a little below, 1.75 million per year.

The instantaneous rate of change of f(x) at x = a (if it exists) is defined to be the limiting value of the average rate of change of f on smaller and smaller intervals starting at x = a. For instance, in Example 1, we would estimate the instantaneous rate of change of the population in Mexico in 1980 to be 1.75 million per year. In other words, the population is growing at a rate of 1.75 million per year in 1980. More precisely, we have the following definition.

Instantaneous Rate of Change

The instantaneous rate of change of f at x = a is given by

images

where the symbol images is interpreted as taking b arbitrarily close but not equal to a.

How to calculate such limits is one of the challenges of differential calculus, a problem that we tackle further in Sections 2.2 and 2.3.

Example 2 Rate of change of carbon dioxide

In Example 4 in Section 1.1, we initially approximated the concentration of CO2 in parts per million at the Mauna Loa Observatory of Hawaii with the linear function

images

where t is measured in months after April 1974. We then refined our approximation with the function

images

Using the definition introduced in this section, estimate the instantaneous rate of change of the functions L(t) and F(t) at t = 3.

Solution To estimate the instantaneous rate of change of L at t = 3, we can look at the average rate of change over the interval [3, b] for values of b progressively closer to 3. From our definition for the average rate of change, we have

images

Since L is a linear function of b, the average rate of change of L is independent of b, as the above algebra reveals, and equals the slope of L(t) for all b ≠ 3. Hence, we find that the instantaneous rate of change of L(t) at t = 3 is 0.1225 ppm/month.

On the other hand, the function F(b) is nonlinear in the variable t. To estimate the instantaneous rate of change of F(t) at t = 3, we look at the average rate of change

images

over intervals [3, b] for values of b values progressively closer to 3, but satisfying b > 3. Carrying out a sequence of such calculations yields the following table:

images

This table suggests the instantaneous rate of change of F(t) is a little less than −1.45 ppm/month.

You might ask why there is a difference in the signs between the answers for L(t) and F(t). Recall that L(t) only described the linear trend of the CO2 data that was increasing. On the other hand, F(t) captured the seasonal fluctuations of the CO2 levels, which are sometimes increasing and sometimes decreasing. Turning back to Figure 1.5 in Section 1.1, we see that, indeed, in the third month of the data, the level of CO2 was decreasing.

This example indicates that a linear fit to an oscillating function may provide a reasonable estimate of rates of change averaged over several oscillations; however, it is a poor estimate of the instantaneous rate of change because that depends on the particular stage of each oscillation.

Velocity

Now we consider concepts central to the history of calculus, concepts that motivated much of the work of Newton: the velocity (considered here) and the acceleration (considered in Sections 3.6 and 5.1) of an object moving along a line. For example, the object may be an athlete running along a racing track, or a coffee mug falling to the ground.

Average and Instantaneous Velocity

Let f(t) be the position of an object at time t. The average velocity of an object from time t to time t + h is given by the formula

images

while the instantaneous velocity of an object at time t is given by the formula

images

If we define b = t + h and a = t, then our definitions of average velocity and instantaneous velocity are equivalent to the average rate of change of the object's position and instantaneous rate of change of the object's position.

If you have ever watched a track meet, you may have wondered how fast the winners ran during their races. In the next example, we find out exactly how fast the world's best 100-meter sprinters are and how their velocities change during the race.

images

Figure 2.2 Usian Bolt is widely regarded as the fastest person ever, earning him the nickname “Lightning Bolt.”

Example 3 The fastest humans

Among the fastest 100-meter races ever run are the performances of Ben Johnson at the 1988 Seoul Olympics and Usain Bolt at the 2008 Beijing Olympics. The reaction times at the start of the race and the times taken to run each 10-meter split (the first split is the time to reach 10 meters minus the reaction time) are given in Table 2.1. Ben Johnson was disqualified three days after the event when the steroid Stanozolol was found in his urine.

  1. From these data, estimate Usain Bolt's velocities at the end of each of the 10-meter splits during the course of the race. (For calculation of Ben Johnson's velocities, see Problem 51 in Problem Set 2.1.)

    Table 2.1 Reaction times to start running (RT) and times (in seconds) take to run each of the 10-meter splits in the 1998 and 2008 Olympic 100-meter races

    images

  2. From the data calculated in part a, discuss how Usain Bolt's velocities changed through the course of the race.

Solution

  1. From Table 2.1 and the definition of average velocity given above, Usain Bolt's average velocities in meters per second (m/s) at the end of each split can be shown to be

    images

  2. From Figure 2.3, we see that Bolt's velocity keeps increasing until the end of the sixth 10-meter split. His velocity then remains constant for the next two 10-meter splits but declines noticeably over the final 10-meter split.

    images

    Figure 2.3 Usain Bolt's average velocities at the end of all the 10-meter splits are plotted here with values at each split joined by the solid blue line.

We have all seen objects fall for different periods of time, whether it be a coffee mug falling off a table or a peregrine falcon stooping from high altitudes. The next example uses a basic physical principle (developed in greater detail in Section 5.1) to determine how the distance traveled by a falling object changes in time.

Example 4 Instantaneous velocity of a falling object

Consider a coffee mug falling off a 32-foot ladder. Ignoring air resistance, the height of this coffee mug from the ground at time t seconds can be shown to be

images

Determine what time the mug is halfway to the ground and its instantaneous velocity at this time.

Solution The coffee mug is halfway to the ground when the height above the ground is 16 feet. Therefore, we need to solve 16 = f(t) = 32 − 16t2. Equivalently, t2 = 1, which has solutions t = −1, +1. The only solution that is relevant physically is t = 1 seconds.

To find the instantaneous velocity of f(t) at t = 1, we need the displacement over the time interval t = 1 to t = 1 + h. This displacement is given by

images

Therefore,

images

Letting the elapsed time h go to zero yields an instantaneous velocity of −32 −16(0) = −32 feet/second. This velocity is approximately −22 miles per hour! The negative sign on the velocity corresponds to the fact that the mug is falling downward.

Tangent lines

A linear function is a function with constant slope, which raises this question: What is the slope of a nonlinear function at a point? To answer this question, we need to solve the tangent problem, whose solution resides in the following words of Leibniz quoted by David Berlinski (Infinite Ascent: A Short History of Mathematics (2008) Random House, NY):

We have only to keep in mind that to find a tangent means to draw the line that connects two points of the curve at an infinitely small distance

As illustrated in Figure 2.4 and stated by Leibniz, the tangent line of y = f(x) at a point (a, f(a)) (in blue) can be approximated by secant lines (in red) passing through the points (a, f(a)) and (a + h, f(a + h)) as h ≠ 0 gets closer and closer to 0. The slope of this secant line is given by

images

images

Figure 2.4 Approximating the tangent line with a secant line.

By letting h get arbitrarily close to 0, we get the

images

where the symbol images can be interpreted as taking h > 0 arbitrarily close to, but not equal to 0. This limit may or may not exist, so not every curve will have a tangent line at every point.

Example 5 Approximating a tangent line

Approximate the tangent line of y = ln x at the point x = 1 using the method of secants with decreasing values of h = 1, 0.5, 0.1, 0.01, and 0.001.

Solution As a first approximation to the tangent line, for the case h = 1, we consider the secant line passing through the point (1, ln 1) = (1, 0) and (2, ln 2). The slope of this secant line is given by

images

Using the point slope formula for a line, the equation of the secant line is

images

Graphing y = ln x and (ln 2)(x − 1) yields the plots depicted in Figure 2.5.

images

Figure 2.5 Plots of the functions y = ln x and y = (ln 2)(x − 1).

To obtain a better approximation, we now move closer to (1, 0) by letting h = 0.5. This secant line passes through (1, 0) and (1.5, ln 1.5). The slope of this secant line is

images

and the secant line is

images

This secant looks like a better approximation to a tangent line, as illustrated in Figure 2.6.

images

Figure 2.6 Plots of the functions y = ln x, y = (ln 2)(x − 1), and y = (2 ln 1.5)(x − 1).

To obtain better approximations, repeat the exercise of finding the slope of the secant line passing through (1, 0) and (1 + h, ln(1 + h)) for h = 0.1, 0.01, and 0.001 to obtain the values reported to three decimal places (3 dp) in the following table:

h Slope of secant line (to 3 dp)
1 0.693
0.5 0.811
0.1 0.953
0.01 0.995
0.001 1.000

This table suggests that as h gets closer to 0, the slope of the corresponding secant line approaches 1. Hence, it seems reasonable to approximate the tangent line by y = x − 1, as shown in Figure 2.7.

images

Figure 2.7 Plots of the functions y = ln x, y = 2(ln 1.5)(x − 1) and y = x − 1.

We will see later that this is actually the tangent line.

Sometimes it is possible to algebraically determine the slope of the tangent line.

Example 6 Tangent line for a parabola

Find the equation of the tangent line passing through the point (1, 1) on the curve y = x2.

Solution To find the tangent line, we first need its slope. The slope of the secant line passing through the point (1, 1) and (1 + h, (1 + h)2) is given by

images

The slope of the tangent line is images. Since 2 + h gets arbitrarily close to 2 as h gets close to 0, the slope of the tangent line is 2. Using the point-slope formula, we find the equation of the tangent line:

images

Plotting this line against y = x2 in Figure 2.8 shows that the tangent line just “kisses” the parabola at (1, 1), touching it in a single point in this case.

images

Figure 2.8 Graph of the tangent to y = x2 at (1, 1).

PROBLEM SET 2.1

Level 1 DRILL PROBLEMS

Find the average rate of change for the functions in Problems 1 to 6 on the specified intervals.

1. f(x) = 4 − 3x on [−3, 2]

2. f(x) = 5 on [−3, 3]

3. f(x) = 3x2 on [1, 3]

4. f(x) = −2x2 + x + 4 on [1, 4]

5. f(x) = images on [4, 9]

6. f(x) = images on [1, 5]

Approximate the instantaneous rate of change for the functions in Problems 7 to 12 at the indicated point. These are the same functions as those given in Problems 1 to 6.

7. f(x) = 4 − 3x at x = −3

8. f(x) = 5 at x = 3

9. f(x) = 3x2 at x = 1

10. f(x) = −2x2 + x + 4 at x = 4

11. f(x) = images at x = 4

12. f(x) = images at x = 1

Find the instantaneous velocity of a cup falling from the specified height h feet and time t second in Problems 13 to 16. Use the fact that the height of the cup as a function of time t is given by f(t) = h − 16t2 feet.

13. h = 64 feet; t = 1 seconds

14. h = 64 feet; t = 2 seconds

15. h = 4 feet; t = 1/2 seconds

16. h = 1,600 feet; t = 10 seconds

Trace the curves in Problems 17 to 22 onto your own paper and draw the secant line passing through P and Q. Next, imagine h → 0 and draw the tangent line at P, assuming that Q moves along the curve to the point P. Finally, estimate the slope of the curve at P using the slope of the tangent line you have drawn.

images

images

images

Estimate the tangent line for y = f(x) in Problems 23 to 36 by approximating secant slopes to estimate the limiting value. Graph both the function and the tangent line on the same plot.

23. f(x) = x2x + 1 at x = −1

24. f(x) = 4 − x2 at x = 0

25. f(x) = sin images at x = 0.5

26. f(x) = cos images at x = 0.5

27. f(x) = tan images at x = 0.5

28. f(x) = sin images at x = 1

29. f(x) = images at x = 3

30. f(x) = ex at x = 0

31. f(x) = ln x at x = 9

32. f(x) = tan x at x = 0

33. f(x) = tan images at x = 0.95

34. f(x) = ln images at x = 1

35. f(x) = ln images at x = 0.1

36. f(x) = ex at x = 0

Algebraically determine the tangent line for y = f(x) at the point specified in Problems 37 to 42. Graph both y = f(x) and the tangent line on the same plot.

37. f(x) = 3x − 7 at x = 3

38. f(x) = x2 at x = −1

39. f(x) = 3x2 at x = −2

40. f(x) = x3 at x = 1

41. images at x = 9. Hint: Multiply by images

42. images at x = 5

Level 2 APPLIED AND THEORY PROBLEMS

Find the average rate of change of the given functions over the specified intervals in Problems 43 to 46. Be sure to specify units and briefly state the meaning of the average rate of change.

43. P(t) = 8.3 × 1.33t is the number (in millions) of people living in the United States t decades after 1815 over intervals [0, 2] and [2, 4].

44. L(x) = 20.15 x2/3 is the number of kilograms lifted by Olympic Gold Medal weightlifters weighing x kilograms over intervals [56, 75] and [100, 110].

45. The height H(t) of beer froth after t seconds over intervals [0, 30] and [60, 90]. Use the data in Table 2.2.

Table 2.2 Height of beer froth as a function of time after pouring

Time t (seconds) Froth height H (centimeters)
0 17.0
15 16.1
30 14.9
45 14.0
60 13.2
75 12.5
90 11.9
105 11.2
120 10.7

Data Source: Arnd Leike, “Demonstration of the Exponential Decay Law Using Beer Froth,” European Journal Physics 23 (2002): 21–26.

46. The height f(x), in feet, of the tide at time x hours, where the graph of y = f(x) is provided in Figure 2.9, over intervals [0.25, 0.50] and [1, 2]

images

Figure 2.9 Tidal height.

Approximate the instantaneous rates of change of the given functions at the points specified in Problems 47 to 50. These are the same functions as those in Problems 43 to 46.

47. P(t) = 8.3(1.33)t is the number (in millions) of people living in the United States t decades after 1815 at the points t = 0 and t = 2.

48. L(x) = 20.15x2/3 is the number of kilograms lifted by Olympic Gold Medal weightlifters weighing x kilograms at the points x = 56 and x = 100.

49. Use the data in Table 2.2 to estimate the instantaneous rate of change of height H(t) of beer froth after 0 and 60 seconds respectively.

50. The height f(x), in feet, of the tide at time x hours where the graph of y = f(x) is in given in Figure 2.9 at the points x = 0.25 and x = 1.

51. From the data presented in Table 2.1 in Example 3, calculate Ben Johnson's velocities at the end of each of the 10-meter splits during the course of his race and discuss how his velocity changes over the course of the race.

52. The population of a particular bacterial colony was determined to be given by the function

images

thousand individuals t hours after observation began. Find the rate at which the colony was growing after exactly five hours.

53. The biomass of a particular bush growing in a field is given by the function B(t) = 1 + 10t1/2 kg, where t is years. Find the rate at which the bush is increasing in biomass after exactly sixteen years.

54. An environmental study of a certain suburban community suggests that t years from now, the average level of CO2 in the air can be modeled by the formula

images

parts per million.

  1. By how much will the CO2 level change in the first year?
  2. By how much will the CO2 level change over the next (second) year?
  3. At what rate will the CO2 level be changing with respect to time exactly one year from now?

2.2 Limits

In defining the instantaneous rate of change and the slope of a tangent line, we use the notation for a limit in Section 2.1. The concept of a limit is one of the foundations of both differential and integral calculus. Thus, we devote this section to exploring the concept of limits in the context of functions before proceeding to the calculus itself.

Introduction to Limits

We begin our study of limits with the following mathematically “informal” definition.

Limit (Informal)

Let f be a function. The notation

images

is read as “the limit of f(x) as x approaches a is L” and means that the functional values f(x) can be made arbitrarily close to L by requiring that x be sufficiently close to, but not equal to, a.

Note that sometimes f(x) may not be defined at x = a but may be defined for all x as close as you like to the value of a.

This definition is made mathematically precise at the end of this section. At several other places in this book we favor informal over formal definitions, using the following statement of the historian E. T. Bell as our justification (see Men of Mathematics, New York: Simon & Schuster, 1937, p. 98):

To the early developers of the calculus the notions of variables and limits were intuitive; to us they are extremely subtle concepts hedged about with thickets of semi-metaphysical mysteries concerning the nature of numbers....

Our goal is to provide definitions that make the concepts usable without getting caught up in the technicalities of formal definitions.

Example 1 Finding limits

Find the following limits using the informal definition provided above.

  1. Graph the function y = x2 and find images x2 for x > 2.
  2. Numerically evaluate the function y = images.
  3. Use the informal definition of a limit to find images e−1/x2.

Solution

  1. Graph y = x2. Choose several values of x (getting closer to 2, but not equal to 2) and then corresponding y values. As illustrated in Figure 2.10, we can see that as x gets closer to 2 from above, the value of the function gets closer to 4.

    In fact, if we zoom in around the point x = 2, we obtain the plot illustrated in Figure 2.11.

    These graphs (correctly) suggest that images x2 = 4. This limit corresponds to simply evaluating x2 at x = 2. This is the idea of continuity, which is described in the next section.

    images

    Figure 2.10Illustration of the process of finding images x2.

    images

    Figure 2.11 Illustration of the process of finding images x2 when zooming in around the point x = 2.

  2. The function f(x) = images is not defined at x = 2 (division by zero), so we cannot simply evaluate this function at x = 2 as suggested at the end of part a. Instead, we can only consider values near (but not equal to) 2. Since x2 − 4 = (x −2)(x + 2), we have

    images

    Evaluating f at x values near 2 yields the following table:

    images

    This table suggests that images = 0.25, which corresponds to evaluating images at x = 2.

  3. The function f(x) = e−1/x2 is not defined at x = 0. However, if x is close to zero, then −images is very large and negative. Consequently, if x is close to zero, e−1/x2 is close to zero. This suggests that images e−1/x2 = 0. We can reinforce this conclusion by looking at the graph of f, as shown in Figure 2.12.

images

Figure 2.12 Graph of the function e−1/x2.

The existence of a limit images f(x) = L can be interpreted in terms of choosing the appropriate window for viewing a function.

Example 2 Choosing the correct viewing window

For the following limits images f(x) = L, determine how close x needs to be to a to ensure that f(x) is within 0.01 and 0.00001 of L. In each case plot the function in the appropriate window to illustrate your findings.

images

Solution

  1. To have 2x − 1 within 0.01 of 7, we need

    images

    Plotting y = 2x − 1 in the window [3.995, 4.005] × [6.99, 7.01], yields the graph illustrated in Figure 2.13. Notice that the graph of the function plotted in Figure 2.13 just fits in this window!

    images

    Figure 2.13 Plot of the function y = 2x − 1 in the window [3.995, 4.005] × [6.99, 7.01].

    To have 2 x − 1 within 0.00001 of 7, we need

    images

    Plotting y = 2x − 1 in the window [3.999995, 4.000005] × [6.99999, 7.00001] yields Figure 2.14. Again, notice that the graph of the function plotted in Figure 2.14 just fits in this window.

    images

    Figure 2.14 Plot of the function y = 2x − 1 in the window [3.999995, 4.000005] × [6.99999, 7.00001].

  2. To ensure that e−1/x2 is within 0.01 of 0, we need

    images

    Since the left-hand inequality is always true, we can ignore it. Also, since the natural logarithm is an increasing function, we have

    images

    images

    Figure 2.15 Plot of the function y = e−1/x2 in the window [−0.46599, 0.46599] × [−0.01, 0.01].

    Thus, if we plot y = e−1/x2 in the window [−0.46599, 0.46599] × [−0.01, 0.01], we obtain the graph illustrated in Figure 2.15. In this case, the upper half of the function just fits the window.

    images

    Since the left-hand inequality is always true, we can ignore it. Also, since the natural logarithm is an increasing function, we have

    images

    images

    Figure 2.16 Plot of the function y = e−1/x2 in the window [−0.294718, 0.294718] × [−0.00001, 0.00001].

    Thus, if we plot e−1/x2 in the window [−0.294718, 0.294718] × [−0.00001, 0.00001] we obtain the graph illustrated in Figure 2.16, whose upper half also just fits the window.

The statement images f(x) = L can fail in two ways. First, images f(x) exists but does not equal L. Second, the images f(x) does not exist. In other words, no matter what L we choose, the statement images f(x) = L is false. You typically encounter the first failure when someone is testing you on this material or when someone has defined a function such that f(a) ≠ images f(x). The second failure is more interesting.

Example 3 Limit failures

Determine whether the following limits exist.

images

Solution

  1. To understand images, we might begin by evaluating cos images at smaller and smaller x values. But we need to be careful!
    • If we evaluate at x = 1, x = 0.1, x = 0.01, x = 0.001,..., we obtain cos 2π = 1, cos 20π = 1, cos 200π = 1,.... This suggests images cos images = 1.
    • If we evaluate at x = 2, x = 2/3, x = 2/5..., we obtain cos π = −1, cos 3π = −1, cos 5π = −1,.... This suggests that images cos images = −1.

    Both of these statements cannot be true simultaneously. Lets consider the statement images cos images = 1; this requires that cos images can be made arbitrarily close to 1 for all x sufficiently close (but not equal to) 0. However, there are x's arbitrarily close but not equal to 0 (namely, x = 2/3, 2/5, 2/7,···) such that cos images = −1, which is 2 units away from 1. Hence, images cos images ≠ 1. This argument can be refined to show that images cos imagesL for any choice of L. Therefore, the limit does not exist.

    images

    Figure 2.17 Graph of f (x) = cos images.

    Graphing this function in Figure 2.17 illustrates the dramatic nature of this nonexisting limit.

  2. To understand images x cos images, we begin by noticing that cosine takes on values between −1 and 1. Hence, for x ≠ 0, −1 ≤ cos images ≤ 1 and thus

    images

    for all x ≠ 0. Therefore, by choosing x sufficiently close to 0 but not equal to 0, we can make |x| as close to 0 as we want, so that x cos images becomes arbitrarily close to 0. Therefore,

    images

    images

    Figure 2.18 Graph of f (x) = cos images.

    as the graph of y = x cos images in Figure 2.18 illustrates. In the next section, we will see that this example is an application of the squeeze theorem.

We occasionally rely on technology to compute limits. Although technology almost always steers us in the right direction, cases exist where it drives us to incorrect conclusions.

Example 4 A computational dilemma

Consider the function

images

  1. Use technology to evaluate f(x) at x = ±0.1, ±0.01, ±0.001, ±0.0001. Based on these evaluations, formulate a conclusion about images f(x).
  2. Use technology to evaluate f(x) at x = ±10−5, ±10−6, ±10−7, ±10−8 ±10−9. Based on these evaluations, formulate a conclusion about images f(x). Compare your results to those of part a.

Solution

  1. We begin with a table of values.
    x f(x)
    ±0.1 0.498756
    ±0.01 0.499988
    ±0.001 0.500000
    ±0.0001 0.500000

    This table suggests that the limit is 0.5. Moreover, plotting this function in Figure 2.19 over the interval −1 ≤ x ≤ 1 reaffirms this conclusion.

    images

    Figure 2.19 Graph of numerical solution to f(x) = images over the range [−1, 1].

  2. Next, we evaluate f for even smaller values of x.
    x f(x)
    ±10−5 0.500000
    ±10−6 0.500044
    ±10−7 0.488498
    ±10−8 0.000000
    ±10−9 0.000000

    It appears that f is approaching the value 0, not 0.5. Plotting the graph of y = f(x) over this smaller range of x values yields Figure 2.20, which suggests that the limiting value is 0—very strange indeed!

    images

    Figure 2.20 Graph of numerical solution to f(x) = images over the range [−10−7, 10−7].

    So we have a dilemma. Should the answer be 0 or 0.5? Later, we will develop more reliable methods that will show this limit is 0.5. Hence, when you use technology, always be aware that technology may mislead you.

One-sided limits

The definition of the limit of f(x) as x approaches a requires that f(x) approach the same value independent of whether x approaches a from the right or the left. In this sense, images f(x) is a “two-sided” limit. One-sided limits, on the other hand, only require that f(x) approach a value as x approaches a from the left or the right.

One-sided Limits (Informal)

Right-hand limit: We write

images

if we can make f(x) as close to L as we please by choosing x sufficiently close to a and to the right of a.

Left-hand limit: We write

images

if we can make f(x) as close to L as we please by choosing x sufficiently close to a and to the left of a.

A two-sided limit cannot exist if the corresponding one-sided limits are different. Conversely, it can be shown that if the two one-sided limits of a given function f as xa and xa+ both exist and are equal, then the two-sided limit, images f(x), exists. These observations are so important that we restate them as follows:

Matching Limits

Let f be a function. Then

images

Example 5 Finding one-sided limits

Consider the function f(x) = images. Find the right-hand limit as x → 3+ and the left-hand limit as x → 3; discuss whether images f(x) exists.

Solution Since f(x) = images = 1 whenever x > 3, we have

images

Since f(x) = images = −1 whenever x < 3, we have

images

images

Figure 2.21 Graph of f(x) = images.

A graph of this function is given in Figure 2.21.

Since the right-hand and left-hand limits are not the same, images f(x) does not exist.

Example 6 The floor function

The floor function, sometimes called the step-function, is the function that returns the largest integer less than or equal to x. The function is typically denoted by images. For instance, images.

  1. Graph the y= images over the interval [−π, π].
  2. Determine at what values of a, images does not exist.

Solution

  1. The graph of y= images is shown in Figure 2.22. Notice that the closed circles include the end point and the open circles exclude the end point.
  2. From Figure 2.22, we see that the limit will not exist at integer values. That is, at the points a = −3, −2, −1, 0, 1, 2, 3, we have

    images

    Therefore, by the matching limits property, the limit does not exist at a = −3, −2, −1, 0, 1, 2, 3.

    images

    Figure 2.22 Graph of images on images.

One-sided limits are particularly useful when considering piece-wise defined functions (see Section 1.1 for a definition).

In the next example, the piece-wise defined function describes the feeding behavior of the copepod Calanus pacificus, illustrated in Figure 2.23. Planktonic copepods are small crustaceans found in the sea. These organisms play an important role in global ecology as they are a major food source for small fish, whales, and sea birds. It is believed that they form the largest animal biomass on Earth. Given their importance, scientists are interested in understanding how their feeding rate depends on availability of resources. In a classic ecology paper, C. S. Holling classified feeding rates into three types. (See his article, “The Functional Response of Invertebrate Predators to Prey Density,” Memoirs of the Entomological Society of Canada 48 (1966): 1–86.) The first type, so-called type I, assumes that organisms consume at a rate proportional to the amount of food available until they achieve a maximal feeding rate. The type II feeding rate was studied in Example 6 of Section 1.6.

images

Figure 2.23 The planktonic copepod Calanus pacificus as seen under an electron microscope. Copepods such as this are believed by some scientists to form the largest animal biomass on Earth.

Example 7 Type I functional response

In the 1970s, B. W. Frost, a scientist at the Department of Oceanography at the University of Washington, measured feeding rates of the planktonic copepod Calanus pacificus in the lab. In one of his experiments, C. pacificus were offered different concentrations of the diatom species Coscinodiscus anstii. Frost found that C. pacificus reached its maximal feeding rate of 1,250 cells/hour when the concentration of C. anstii was approximately 200 cells/milliliter (see Figure 2.24). If you assume that the feeding rate is proportional to the concentration x of C. anstii until they achieve their maximal feeding rate, then the feeding rate as a function of x is of the form

images

where a > 0 is a proportionality constant.

images

Figure 2.24 Feeding rate I (cells/hour) of a copepod as a function of the density of the diatoms (cells/milliliter) upon which it feeds.

Data Source: B. W. Frost, 1972. “Effects of Size and Concentration of Food Particles on the Feeding Behavior of the Marine Planktonic Copepod Calanus pacificus.” Limnology and Oceanography 17(6): 805–815.

images

Solution

  1. Since f(x) = 1,250 for all x > 200, we find

    images

    On the other hand, f(x) = ax for all x ≤ 200. Hence, as x increases to 200, f(x) approaches 200a and

    images

  2. By the matching limit property, images f(x) exists if and only if the left- and right-hand limits are equal. Therefore, we need to ensure that 1,250 = 200a or a = 6.25. In which case, images f(x) = 1,250. The graph of this function, along with the data as plotted in Figure 2.24, illustrates that by choosing a = 6.25, the linear function and constant function are pasted together in such a way that their values agree at x = 200.

Limits: A formal perspective (optional)

This section can be omitted by those not going on to major in mathematics at the undergraduate level. Our informal definition of the limit provides valuable intuition that allows you to develop a working knowledge of this fundamental concept. For theoretical work, however, the intuitive definition will not suffice, because it gives no precise, quantifiable meaning to the terms “arbitrarily close to L” and “sufficiently close to a.” In the nineteenth century, leading mathematicians, including Augustin-Louis Cauchy (1789–1857) and Karl Weierstrass (1815–1897), sought to put calculus on a sound logical foundation by giving precise definitions for the foundational ideas of calculus. The following definition, derived from the work of Cauchy and Weierstrass, gives precision to the definition of a limit.

Limit (Formal Definition)

Let f be a real-valued function.

images

if for every images > 0 there is some δ > 0 such that |f(x) − L|≤ images whenever 0 < |xa| < δ.

images

Figure 2.25 The epsilon-delta definition of limit.

Behind the formal language is a fairly straightforward idea. Given any images > 0 specifying a desired degree of proximity to L, a number δ > 0 is found that determines how close x must be to a to ensure that f(x) is within images of L. This is shown in Figure 2.25.

Because the Greek letters images (epsilon) and δ (delta) are traditionally used in this context, the formal definition of limit is sometimes called the epsilon-delta definition of the limit. The goal of this subsection is to show how it can be used rigorously to establish a variety of results.

One can view this definition as setting up an adversarial relationship between two individuals. One person shouts out a value of images > 0. The opponent has to come up with a δ > 0 such that f(x) is within images of L whenever x is within δ of a. This relationship is illustrated in Figure 2.26.

images

Figure 2.26 Formal definition of limit: images f(x) = L.

Note that f(a) does not need to equal the value of L in order for images f(x) = f(a). If images f(x) = f(a), then the epsilon-delta argument confirms that f(x) is continuous at x = a. On the other hand, if images f(x) ≠ f(a), then the epsilon-delta argument confirms that f(x) is not continuous at x = a.

Notice that whenever x is within δ units of a (but not equal to a), the point (x, f(x)) on the graph of f must lie in the rectangle (shaded region) formed by the intersection of the horizontal band of width 2images centered at L and the vertical band of width 2δ centered at a. The smaller the images interval around the proposed limit L, generally the smaller the δ interval will need to be for f(x) to lie in the images interval. If such a δ can be found no matter how small images is, then L must be the limit. The following examples illustrate epsilon-delta proofs, one in which the limit exists and one in which it does not.

Example 8 An epsilon-delta proof of a limit statement

Show that images (4x − 3) = 5.

Solution The object is to prove that the limit is 5. We have

images

For a given images > 0 choose δ = images. Then

images

images

Figure 2.27 Illustration of the epsilon-delta proof of a limit statement.

This process is illustrated for images = 2 in Figure 2.27.

Example 9 Limit of a constant

Use an epsilon-delta proof to show that the limit of a constant is a constant; that is, show images c = c.

Solution Let f(x) = c

images

Thus, for any given images > 0, pick any δ > 0. Then, if |xa| < δ, we have

images

Example 10 An epsilon-delta proof to show that a limit does not exist

Show that images images does not exist.

Solution Let f(x) = images and L be any number. Suppose that images f(x) = L. Consider the graph of f plotted in Figure 2.28.

images

Figure 2.28 Illustration of the epsilon-delta proof that the limit images f(x) = L exists.

It would seem that no matter what value images > 0 is chosen, it would be impossible to find a corresponding δ > 0. Indeed, suppose that

images

Then

images

and

images

This inequality holds whenever

images

and hence will be violated whenever

images

Thus, no matter what value of images > 0 we choose, we cannot find a δ > 0 such that |1/xL| < images for all 0 < |x − 0| < δ. Since this statement is true for any L we chose, it follows that the limit does not exist.

PROBLEM SET 2.2

Level 1 DRILL PROBLEMS

Given the functions f and g defined by the graphs in Figure 2.29, find the limits in Problems 1 to 4.

images

Figure 2.29 Graphs of the functions f and g.

images

Given the functions defined by the graphs in Figure 2.30, find the limits, if they exist, in Problems 5 to 8. If the limits do not exist, discuss why.

images

Figure 2.30 Graphs of the functions F and G.

images

Describe each figure in Problems 9 to 12 with a one-sided limit statement. For example, for Problem 9, the answer is images f(x) = 2.

images

Approximate the limits by filling in the appropriate values in the tables in Problems 13 to 15 using a one-sided statement.

images

Determine the limits in Problems 16 to 24. If the limit exists, explain how you found the limit. If the limit does not exist, explain why.

images

25. Consider the function

images

whose graph is given Figure 2.31. Does images f(x) exist? If so, what is it? If not, why not?

images

Figure 2.31 Graph of the function f(x) = images.

26. Consider the function

images

whose graph is given Figure 2.32. Does images f(x) exist? If so, what is it? If not, why not?

images

Figure 2.32 Graph of the function f(x) = images.

27. Consider the statement images (4 + x) = 5. How close does x need to be to 1 to ensure that 4 + x is within the given distance of 5?

images

28. Consider the statement images x2 = 4. How close does x need to be to 2 to ensure that x2 is within the given distance of 4?

images

29. Consider the statement images images = 0. How close does x need to be to 0 to ensure that images is within the given distance of 0?

images

30. Consider the statement images ln x = 1. How close does x need to be to e to ensure that ln x = 1 is within the given distance of 1?

images

31. Consider the function

images

  1. Use technology to graph y = f(x) over the interval [−2, 2].
  2. Use technology to graph y = f(x) over the interval [−0.1, 0.1]. Based on your graph, guess the value of images f(x).
  3. Use technology to graph y = f(x) over the interval [−10−7, 10−7]. Based on your graph, guess the value of images f(x).
  4. Most technologies can keep track of sixteen or fewer digits. In light of this observation, discuss what might be happening in parts b and c.

32. Consider the function f(x)) = images.

  1. Use technology to graph y = f(x) over the interval [−2, 2].
  2. Use technology to graph y = f(x) over the interval [−0.1, 0.1]. Based on your graph, guess the value of images f(x).
  3. Use technology to graph y = f(x) over the interval [−10−7, 10−7]. Based on your graph, guess the value of images f(x).
  4. Most technologies can keep track of sixteen or fewer digits. In light of this observation, discuss what might be happening in parts b and c.

In Problems 33 to 38, prove the limit exists using the formal definition of the limit.

images

images

Level 2 APPLIED AND THEORY PROBLEMS

39. The federal income tax rates for singles in 2010 is shown in Table 2.3.

Table 2.3 Schedule X—Single

images

Express the income tax f(x) for an individual in 2010 with adjusted income x dollars as a piecewise defined function.

  1. Graph y = f(x) over the interval [0, 500,000].
  2. Determine at what values of a, images f(x) does not exist.

40. In 2011, the U. S. postal rates were 44 cents for the first ounce or fraction of an ounce, and 17 cents for each additional ounce or fraction of an ounce up to 3.5 ounces. Let p represent the total amount of postage (in cents) for a letter weighing x ounces.

  1. Graph y = p(x) over the interval [0, 3.5] ounces.
  2. Determine at what values of a, images f(x) does not exist.

41. A wildlife ecologist who studied the rate at which wolves kill moose in Yellowstone National Park found that when moose were plentiful, wolves killed moose at the rate of one moose per wolf every twenty-five days. (Note this doesn't mean that wolves only eat every twenty-five days, because they hunt in packs and share kills.) However, when the density of moose drops below x = 3 per km2, then the rate at which wolves kill moose is proportional to the density. Construct a type I functional response f(x) (see Example 7) such that f(x) has a limit at x = 3.

42. A student looking at Figure 2.24 decided that the following function might provide a better fit to the data:

images

Find values for the parameters a and b that ensure f(x) has limits at x = 150 and x = 300.

2.3 Limit Laws and Continuity

Having defined limits, we are ready to develop some tools to verify their existence and to compute them more readily. In some cases, taking the limit of a function reduces to evaluating the function at the limit point, and in other cases we cannot find the limit by evaluation. In this section, we find when evaluation is acceptable and when it is not.

Properties of limits

With a definition of a limit in hand, it is important to understand how the definition acts under functional arithmetic. For instance, if images f(x) = L and images g(x) = M, then f(x) and g(x) can be made arbitrarily close to L and M, respectively, for all x sufficiently close but not equal to a. Hence, f(x)g(x) must be arbitrarily close to LM for all x sufficiently close but not equal to a. Therefore, it is reasonable to conjecture that the limit of the product f · g is the product LM of the limits. Indeed, this is true and can be proved using the formal definition of the limit. In fact, limits satisfy all the arithmetic properties that you would think they should, as summarized in the following limit laws.

Limit Laws

Let f and g be functions such that images f(x) = L and images g(x) = M. Then

images

Example 1 Using limit laws

Using the limit laws, find the following limits. You may assume that images x = 4 and images 1 = 1.

images

Solution

images

The preceding example illustrates that applying the product and sum limit laws repeatedly allows us to quickly compute limits of polynomials and rational functions as x approaches a by evaluating them at the value a, provided a is in the domain.

Limits of Polynomials and Rational Functions

Let f be either a polynomial or a rational function. If a is in the domain of f, then

images

Proof. We have previously shown that images c = c (the limit of a constant is a constant) and images x = a. By applying the limit law for products repeatedly, we have images xn = an for n = 1, 2, 3,.... Let p(x) = b0 + b1x + b2x2 + ... bnxn be a polyno mial. Then

images

Thus, we have shown images p(x) = p(a) for any polynomial. You will be asked to prove the result for a rational function in problem 34 in Problem Set 2.3.

Example 2 Finding limits algebraically

Find the limits and show each step of your derivation.

images

Solution

  1. Since 2x4 − 5x3 + 2x2 − 5 is a polynomial, it is sufficient to evaluate the polynomial at x = 2:

    images

  2. Since images is a rational function and x = 2 is in the domain, it is sufficient to evaluate the rational function at x = 2:

    images

  3. Since x = −2 is not in the domain, we cannot simply evaluate the function at x = −2 to determine the limit. However, we can factor and then evaluate at x = −2:

    images

In Example 3b of Section 2.2, we computed a limit by “squeezing” a function for an unknown limit between functions whose limits we understand. The squeeze theorem provides a general approach for computing limits in this manner. This theorem, stated below, is sometimes called the sandwich theorem or pinching theorem.

Theorem 2.1 Squeeze theorem

Let f, g, and h be functions such that

images

Example 3 Using the squeeze theorem

Find images

Solution Since −1 ≤ sin images ≤ 1 for x ≠ 5, we have

images

for x ≠ 5. Therefore,

images

for all x > −1 and x ≠ 5. Define

images

Since f and h are rational functions with x = 5 in their domain,

images

Since f(x) ≤ g(x) ≤ h(x) for x near 5, the squeeze theorem implies

images

Continuity at a point

Example 2 illustrated how easy it is to find limits images f(x) when f is a polynomial or rational function and x = a is in the domain of f. One simply evaluates f at x = a. When one is able to evaluate a limit so easily, a function is continuous at x = a. The idea of continuity corresponds to the intuitive notion of a curve “without breaks or jumps”.

Continuity at a Point

A function f is continuous at the point x = a if f is defined at x = a and images f(x) = f(a).

A function f is discontinuous at the point x = a if it is not continuous at the point x = a.

Example 4 Checking continuity

Test the continuity of each of the following functions at x = 0 and x = 1. If the function is not continuous at the point, explain. Discuss whether the function can be redefined at points of discontinuity to make it continuous.

  1. The function f is defined by the graph y = f(x):

    images

  2. g(x) = images if x ≠ 1, g(x) = 6 if x = 1

Solution

  1. Since f(x) approaches 1 from both sides of x = 0, we see images f(x) = 1. However, as f(0) = 2, we see that

    images

    and f is not continuous at x = 0. However, this discontinuity can be fixed by redefining f(0) to be 1.

    At x = 1, we see

    images

    Thus the limit does not exist, so f is not continuous at x = 1. Since the left- and right-hand side limits do not agree, there is no way redefine f at x = 1 to repair this discontinuity.

  2. At x = 0 we use the limit law for quotients

    images

    and g(0) = 3, so g is continuous at x = 0. At x = 1 we see g(1) = 6. We cannot use the limit of a quotient law because of division by zero at x = 1. However, if we factor the numerator and then take the limit, we get

    images

    Since images g(x) ≠ g(1), g is not continuous at x = 1. However, this discontinuity is reparable by redefining g(1) = 4.

In general, continuity of a function f(x) at x = a can fail in three ways. First, the limit images f(x) = L may be well defined, but either f(a) is not defined or f(a) ≠ L. When this occurs, f has a point discontinuity at x = a. These discontinuities can be repaired by redefining f at x = a by f(a) = L. For example, in Example 4a, we repaired a point discontinuity at x = 0. The second type of discontinuity is a jump discontinuity where images f(x) = L and images f(x) = M with LM. The left and right limits don't agree, there is no way to repair the discontinuity. Example 4a illustrates this type of discontinuity at x = 1. The final type of discontinuity is an essential discontinuity, which occurs when one or both of the one-sided limits, images f(x) or images f(x), do not exist or are infinite. In Example 3a and Figure 2.17 in Section 2.2, we showed that f(x) = x cos images exhibits an essential discontinuity at x = 0.

With the concept of continuity, we are able to get a limit law for a composition of functions.

Composition Limit Law

Let f and g be functions such that images f(x) = L and g is continuous at x = L. Then

images

The next examples illustrate why images g(x) existing does not suffice for images g[f(x)] to exist.

Example 5 Compositional limits

Consider the functions f and g, whose graphs are shown on the left in blue and red, respectively.

images

  1. Find images f(x) and images g(x).
  2. Does images g[f(x)] exist? If it doesn't exist, determine whether you can redefine one of the functions at a point to ensure the limit exists.

Solution

  1. In the graph f(x) approaches 0 as x approaches 1. Hence, images f(x) = 0. Since g(x) = 0.5 for all x ≠ 0, images g(x) = 0.5.
  2. Consider the right-sided limit, images g[f(x)]. Since f(x) < 0 for x > 1 and g(x)= 0.5 for x ≠ 0, we have g[f(x)] = 0.5 for all x > 1. Therefore, images g[f(x)] = 0.5.

    Now, consider the left-sided limit, images g[f(x)]. Since f(x) = 0 for x in [−1, 1], g[f(x)] = 1 for x in [−1, 1]. Therefore, images g[f(x)] = 1.

    Since the right-sided and left-sided limits are not equal, the limit images g[f(x)] does not exist. However, if we redefine g(0) = 0.5. Then,

    images

    and the limit does exist. Notice that by redefining g to be 0.5 at x = 0, we made g continuous at x = 0 and, consequently, resulted in the limit of the composition becoming well defined.

Using the limit laws, we can derive some laws of continuity.

Continuity Laws

Let f and g be functions that are continuous at x = a. Then

Sums f + g is continuous at x = a

Differences fg is continuous at x = a

Products f · g is continuous at x = a

Quotients f/g is continuous at a provided that g(a) ≠ 0

Compositiong images f is continuous at x = a, provided g is continuous at x = f(a)

Proof. We will illustrate the proof of the continuity property for products. All other parts follow in a similar manner. Assume that f and g are continuous at a. Then images f(x) = f(a) and images g(x) = g(a). Hence

images

Therefore, fg is continuous at x = a.

Since we have shown that images f(x) = f(a) for polynomial and rational functions at points in their domain, these functions are continuous at all points on their domain. As it turns out, this statement holds for all elementary functions.

Theorem 2.2 Continuity of elementary functions theorem

Let f be a polynomial, a rational function, a trigonometric function, a power function, an exponential function, or a logarithmic function. Then f is continuous at all points in its domain.

Armed with the tools of continuity, we can readily calculate many limits.

Example 6 Quick limits

Use the results of this section to find the given limits, and justify each step of your derivation.

images

Solution

images

images

Combining continuity theorems with the limit laws, we can compute limits that we could not otherwise find.

Example 7 Technology vanquished

Recall in Example 4 of Section 2.2, we used technology to study the limit

images

and this study was inconclusive. Now find this limit using algebra and the results of this section.

Solution To work with the expression f(x) = images, we need to simplify it. One way to simplify is to multiply the numerator and denominator by images.

images

We now turn to evaluating the limit.

images

Notice that this value of images corresponds to our initial guess of 0.5 in Example 4 of Section 2.2, based on using technology with x = 10−5 but not when using technology with x ≤ 10−8.

Intermediate Value Theorem

The function f is said to be continuous on the open interval (a, b) if it is continuous for each number (i.e., at each point) in this interval. Note that the end points are not part of open intervals. If f is also continuous from the right at a, we say it is continuous on the half-open interval[a, b). Similarly, f is continuous on the half-open interval (a, b] if it is continuous at each number between a and b and is continuous from the left at the end point b. Finally, f is continuous on the closed interval [a, b] if it is continuous at each number between a and b and is both continuous from the right at a and continuous from the left at b.

Example 8 Intervals of continuity

For the following functions, determine on which intervals the function is continuous.

images

Solution

  1. Since images is a rational function, it is continuous on its domain, that is, when-ever its denominator is nonzero. Since 1 − x2 = 0 if and only if x = ±1, images is continuous on the open intervals (−∞, −1), (−1, 1), and (1, ∞).
  2. Since images equals 1 for all x > −3 and equals −1 for all x < −3, images is continuous on the open intervals (−∞, −3) and (3, ∞).
  3. Since tan x = images is a quotient of the elementary functions sin x and cos x, it is continuous at all points where cos x ≠ 0. Therefore, tan x is continuous on all intervals of the form (π/2 + kπ, 3π/2 + kπ) where k is an integer.

The graphs of functions that are continuous on an interval cannot have any breaks or gaps as shown by the following theorem.

Theorem 2.3 Intermediate value theorem

Let f be continuous on the closed interval [a, b]. If L lies strictly between f(a) and f(b), then there exists at least one number c in the open interval (a, b) such that f(c) = L.

This theorem says that if f is a continuous function on some closed interval [a, b], then f(x) must take on all values between f(a) and f(b). The intermediate value theorem is extremely useful in ensuring that we can solve certain nonlinear equations. Consider the following example.

Example 9 Proving the existence of roots

Use the intermediate value theorem to prove that there exists a solution to

images

Use technology to estimate one of the solutions.

images

Solution Let f(x) = x5x2 + 1. Since f is a polynomial, it is continuous at all points on the real number line. To use the intermediate value theorem, we need to find an interval [a, b] such that f(a) and f(b) have opposite signs (since 0 is a value between a positive number and a negative one). A little experimentation reveals that f(−1) = −1 < 0 and f(1) = 1 > 0. Hence, there must be a c in (−1, 1) such that f(x) = 0. Using technology, we see that there is a solution around x = −0.8.

While the intermediate value theorem allows us to prove when a function has a root, sometimes we need to solve for these roots numerically. One numerical approach is the bisection method.

The Bisection Method

Let f(x) be a continuous function on [a, b] where f(a) and f(b) have opposite signs. By the intermediate value theorem, there is a root in [a, b]. To find the root,

Step 1. Calculate f images. If f images = 0, then you have found a root and you are done.

Step 2. If f images has the same sign as f(a), then by the intermediate value theorem, the root lies between f images and f(b). In this case, rename the interval images as [a, b] and repeat the process.

Step 3. If f images has the same sign as f(b), then by the intermediate value theorem, the root lies between f(a) and f images. In this case, rename the interval images as [a, b] and repeat the process.

Step 4. Keep repeating until the width ba of the interval is smaller than the desired accuracy for the root.

Example 10 Bisection method for solving equations

Given a plot of the polynomial

images

and a calculator use the bisection method to find the largest root of this equation correct to two decimal places (two decimal places).

Solution We see from a plot of this polynomial in Figure 2.33 that it has five roots, one each, respectively, on the integer intervals [−3, −2], [−2, −1], [−1, 0], [1, 2], and [2, 3]. Table 2.4 presents the calculations for the bisection method for the interval [2, 3], which contains the largest root. After ten iterations of the method, Table 2.4 implies that the root lies in the interval [2.5615, 2.5621]. Thus, to two decimal places, the root is x = 2.56. If we wanted to find the root to more than two decimal places, we could keep going, as illustrated in Table 2.4, until the desired accuracy is obtained.

images

Figure 2.33 Graph of f(x) = x5 − 10x3 + 21x + 4.

Table 2.4 Bisection method for finding roots of a nonlinear function

images

Equations that take the form of finding the roots of polynomials, such as the problem in the previous example of finding the largest root of a fifth-order polynomial, are known as algebraic equations. Equations that involve exponential or trigonometric functions, such as x sin x − 1/2 = 0 or images − π = 0, are known as transcendental equations and generally have no analytical solutions but need to be solved numerically.

Example 11 Limiting global warming

According to an article in the New Scientist,* recent research suggests that stabilizing carbon dioxide concentrations in the atmosphere at 450 parts per million (ppm) could limit global warming to 2°C. In Section 1.2, we modeled carbon dioxide concentrations in the atmosphere with the function (which we now present to higher precision to make more transparent the numerical details of the convergence process)

images

where x is months after April 1974. Use the bisection method to find the first time that the model predicts carbon dioxide levels of 450 ppm. Get a prediction that is accurate to 2 decimal places.

Solution Solving f(x) = 450 is equivalent to solving g(x) = 0 where g(x) = f(x) − 450. Using technology to plot g(x), we see from the left panel in Figure 2.34 that g(x) first equals zero around t = 970 months.

images

Figure 2.34 Plots of the function g(x) = f(x) − 450 on relatively large and small intervals containing x = 970 ppm, where f(x) is defined in Example 11.

There appear to be several zeros in the interval [900, 1000]. Zooming into the interval [960, 980], we get the right panel in Figure 2.34. Since the first zero appears to be between 965 and 972, we can set a = 965 and b = 972 and apply the bisection method, which yields the following table of values (where all the values have all been rounded to 4 decimal places throughout the calculations):

images

Hence, the model predicts that carbon dioxide concentrations will reach 450 ppm in 970.32 months, which is 80 years and 10 months, after April 1974. In other words, in February 2055.

PROBLEM SET 2.3

Level 1 DRILL PROBLEMS

Determine the limits images f(x), images f(x) and images f(x) in Problems 1 to 6. If they do not exist, discuss why.

images

3. f(x) = x/|x| with a = 0.

4. f(x) = x2/|x| with a = 0.

5. f(x) defined by the graph with a = 1.

images

6. f(x) defined by the graph with a = 0.

images

Find the limits in Problems 7 to 14. Justify each step with limit laws and the appropriate results from this section.

images

The graph of a function f is shown in Problems 15 to 18. Determine at what points f is not continuous and whether f can be redefined at these points to make it continuous. Explain briefly.

images

Each function in Problems 19 to 22 is defined for all x > 0, except at x = 2. In each case, find the value that should be assigned to f(2), if any, to guarantee that f will be continuous at 2. Explain briefly.

images

In Problems 23 to 28, use the intermediate value theorem to prove the following equations have at least one solution.

images

29. Use the bisection method explicated in Example 10 to find the following roots of the polynomial

images

to an accuracy of 3 decimal places.

  1. smallest root
  2. second smallest root
  3. largest negative root
  4. smallest positive root

In Problems 30 to 33, use the squeeze theorem, limit laws, and continuity of elementary functions to find the limit.

images

Level 2 APPLIED AND THEORY PROBLEMS

34. Prove that if p and q are polynomial functions with q(a) ≠ 0, then

images

35. Use limit laws to prove if f and g are continuous at x = a, then f + g is continuous at x = a.

36. Use limit laws to prove if f and g are continuous at x = a, then fg is continuous at x = a.

37. Use limit laws to prove if f and g are continuous at x = a and g(a) ≠ 0, then images is continuous at x = a.

38. Use limit laws to prove if f is continuous at x = a and g is continuous at x = f(a), then g images f is continuous at x = a.

39. Why does the cubic equation x3 + ax2 + bx + c = 0 have at least one root for any values of a, b, and c?

40. For any constants a, b, why does the equation xn = a + b cos(x) always have at least one real root when n is an odd integer but not necessarily when n is an even integer?

41. Consider an organism that can move freely between two spatial locations. In one location, call it patch 1, the number of progeny produced per individual is

images

where N is the population size. In the other location, call it patch 2, the number of progeny produced per individual is always 5. Assume that all individuals in the population move to the patch that allows them to produce the greatest number of progeny. Let g(N) represent the number of progeny produced per individual for such a population.

  1. Write an explicit expression for g(N) such that g(N) = f(N) whenever N < c for some constant c > 0 and g(N) = 5 whenever Nc.
  2. Determine the value of c that ensures g(N) is continuous for N ≥ 0.

42. As discussed in Example 11, recent research suggests that stabilizing carbon dioxide concentrations in the atmosphere at 450 parts per million (ppm) could limit global warming to 2°C. Use the bisection method and the carbon dioxide concentration model

images

where x is months after December 1973, to find the second time that this model predicts carbon dioxide levels of 450 ppm. Get a prediction that is accurate to 2 decimal places.

43. Use the bisection method to find the last time that the model in Problem 42 predicts carbon dioxide levels of 450 ppm. Get a prediction that is accurate to 2 decimal places.

44. Scientists believe that it will be extremely difficult to rein in carbon emissions enough to stabilize the atmospheric CO2 concentration at 450 parts per million, as discussed in Example 11, and think that even 550 ppm will be a challenge. Use the bisection method to find the first time that the model in Problem 42 predicts carbon dioxide levels of 550 ppm. Get a prediction that is accurate to 2 decimal places.

45. Fisheries scientists often use data to establish a stock-recruitment relationship of the general form y = f(x), where x is the number of adult fish participating in the spawning process (i.e., the laying and fertilizing of eggs) that occurs on a seasonal basis each year and y is the number of young fish recruited to the fishery as a result of hatching from the eggs and surviving through to the life stage at which they become part of the fishery (i.e., available for harvesting). Two fisheries scientists found that the following stock-recruitment function provides a good fit to data pertaining to the southeast Alaska pink salmon fishery:

images

Use the bisection method to find the spawning stock level x that is expected to recruit 10,000 individuals to the fishery. (Hint: For the value of y in question, find the root of the equation yf(x) = 0.) For more information on the research study, see T. J. Quinn and R. B. Deriso, Quantitative Fish Dynamics (New York: Oxford University Press, 1977).

46. Use the bisection method to find the spawning stock level x that is expected to recruit 20,000 individuals to the fishery modeled by the stock-recruitment function given in Problem 45.

47. Use the bisection method to find the spawning stock level x that is expected to recruit 5,000 individuals to the fishery modeled by the stock-recruitment function given in Problem 45.

2.4 Asymptotes and Infinity

In Chapter 1, we introduced the notion of infinity and represented it with the symbols ∞ and −∞. This symbol was used by the Romans to represent the number 1,000 (a big number to them). It was not until 1650, however, that it was first used by John Wallis (1616–1703) to represent an uncountably large number. From childhood many of us come to think of infinity as endlessness, which in a sense it is since infinity is a not a number. To mathematicians, however, infinity is a much more complicated idea than simply endlessness. As the famous mathematician David Hilbert (1862–1943) said, “The infinite! No other question has ever moved so profoundly the spirit of man” (in J. R. Newman, ed., The World of Mathematics, New York: Simon & Schuster, 1956, 1593).

In this section, we tackle limits involving the infinite in two ways. First, we determine under what conditions functions approach a limiting value as their argument becomes arbitrarily positive or arbitrarily negative. Second, we study functions whose value becomes arbitrarily large as their arguments approach a finite value where the function is not well defined.

Horizontal asymptotes

To understand the behavior of functions as their argument becomes very positive or negative (i.e., further from the origin in either direction), we introduce horizontal asymptotes.

Horizontal Asymptotes (Informal Definition)

Let f be a function. We write

images

if f(x) can be made arbitrarily close to L for all x sufficiently large. We write

images

if f(x) can be made arbitrarily close to L for all x sufficiently negative. Whenever one of these limits occurs, we say that f(x) has a horizontal asymptote at y = L.

Example 1 Finding horizontal asymptotes

Find the following limits involving a given function f. In each, indicate how positive or negative x needs to be to ensure that f(x) is within one ten-millionth of the limiting value L.

images

Solution To help visualize the solutions, plots of these three functions are given in Figure 2.35 over domains of x that illustrate the asymptotic behavior of these functions.

  1. For x sufficiently large, images is arbitrarily close to 0. Hence

    images

    images

    Figure 2.35 Plot of functions. Left panel: f(x) images; right panel: f(x) = ex; bottom panel: f(x) = images for 0 ≤ x ≤ 10.

    To say that f(x) is within one ten-millionth of the limiting value, L, is to say that

    images

    For positive x, we need

    images

    Thus, if x is greater than 10,000,000, then f(x) will be within one ten-millionth of x = 2.

  2. For x sufficiently negative, ex is arbitrarily small. Hence, we would expect that

    images

    To see how negative x needs to be to ensure that ex is less than one ten-millionth, we have

    images

    Thus, if x is less than −16.2, then f(x) will be within one ten-millionth of zero.

  3. To find the limiting value of this function, we can divide the numerator and denominator of f(x) by x:

    images

    In order to be within one ten-millionth of the limiting value of 2, we need

    images

    Thus, if x is greater than about 4,000,000, then f(x) will be within one tenmillionth of two.

Understanding the asymptotic behavior of a function can help us graph and interpret it. The next example involves dose–response curves that arise in many kinds of biological experiments. For example, the x axis of a dose–response curve may represent concentration of a drug or hormone delivered at various “doses.” The y axis represents the response, which could be many things, depending on the experiment. For example, the response might be the activity of an enzyme, accumulation of an intracellular messenger, voltage drop across a cell membrane, secretion of a hormone, increase in heart rate, or contraction of a muscle. In the next example, dose is the amount of a histamine (measured in millimoles) administered to a patient, and response is the percentage of patients exhibiting above-normal temperatures.

Example 2 Dose–response curves

The percentage of patients exhibiting an above-normal temperature response to a specified dose of a histamine is given by the function

images

where x is the natural logarithm of the dose in millimoles (mmol or mM).

  1. Find the horizontal asymptotes of R(x).
  2. Show that R(x) is increasing and sketch y = R(x).
  3. Calculate how large x needs to be to ensure that it is within 1% of its asymptotic value.

Solution

  1. To find the horizontal asymptotes, we find

    images

    and

    images

    Thus, the horizontal asymptotes are y = 100 and y = 0.

  2. To show R(x) is increasing, notice that

    images

    Since 1 + e−5−x is a decreasing function, R(x) is an increasing function of x. The y-intercept of R(x) is R(0) = images ≈ 100. Thus, the graph of the function R(x) has the form illustrated in Figure 2.36. Notice that this curve fits the data fairly well.

    images

    Figure 2.36 Percentage of patients responding to a dose of histamine. The x axis corresponds to the natural logarithm of the doses in millimoles.

    Data Source: K. A. Skau, “Teaching Pharmacodynamics: An Introductory Module on Learning Dose–Response Relationships,” American Journal of Pharmaceutical Education 68 (2004), article 73.

  3. To find when R(x) is within 1% of 100, that is when it reaches the value 99, we first note that R(x) < 100 for all x. Hence, we only need to solve

    images

Limits at infinity can be very nonintuitive, as the following example shows.

Example 3 A difference of infinities

Find

images

Solution To deal with this limit, it is useful to multiply and divide by the conjugate images.

images

Since images gets arbitrarily large as x gets arbitrarily large,

images

This asymptote is somewhat surprising, as one might first guess that images is much bigger than x for large x and therefore the limit is arbitrarily large. However, our calculations show that this initial guess is wrong.

Vertical asymptotes

Many functions, such as rational functions, logarithms, and certain power functions, are not defined at isolated values. As the argument of the function gets close to these isolated values, the function may become arbitrarily large or arbitrarily negative and exhibit a vertical asymptote.

Vertical Asymptotes (Informal Definition)

Let f be a function. We write

images

if f(x) can be made arbitrarily large for all x sufficiently close to a and to the left of a. We write

images

if f(x) can be made arbitrarily large for all x sufficiently close to a and to the right of a. We write

images

if f(x) can be made arbitrarily negative for all x sufficiently close to a and to the left of a. We write

images

if f(x) can be made arbitrarily negative for all x sufficiently close to a and to the right of a.

Whenever any one of these limits occurs, we say that f(x) has a vertical asymptote at x = a.

Example 4 Finding vertical asymptotes

Find images f(x) and images f(x) for the given functions, then sketch the graph of y = f(x) near x = a.

images

images

Figure 2.37 Graph of the function images on [−1, 1].

Solution

  1. images f(x) = −∞ since for x < 0 sufficiently close to 0, images is arbitrarily negative. images f(x) = ∞ since for x > 0 sufficiently near 0, images is arbitrarily large. Since y = images is decreasing for all x ≠0, the graph of y = images near x = 0 is as shown in Figure 2.37.
  2. images = ∞ since for x < 2 and sufficiently close to 2, images is arbitrarily large.

    images = ∞ since for x > 2 and sufficiently close to 2, images is arbitrarily large.* The graph of y = images close to the vertical asymptote y = 2 is illustrated in Figure 2.38.

    images

    Figure 2.38 Graph of the function images on [0, 4].

  3. images = ∞ since for x < images and arbitrarily close to images, sin x is arbitrarily close to 1 and cos x is positive and arbitrarily close to 0, so the quotient of sine and cosine is arbitrarily large.

    images = −∞ since for x > images and arbitrarily close to images, sin x is arbitrarily close to 1 and cos x is negative and arbitrarily close to 0, so the quotient of sine and cosine is arbitrarily negative. The graph of y = tan x close to the vertical asymptote x = π/2 is illustrated in Figure 2.39.

    images

    Figure 2.39 Graph of the function tan x on [−1, 4].

Combining the information about horizontal and vertical asymptotes can provide a relatively complete sense of the graph of a function, as illustrated in the next example. This example, described by Francois Messier (“Ungulate Population Models with Predation: A Case Study with the North American Moose.” Ecology 75 (1994): 478–488), examines wolf-moose interactions over a broad spectrum of moose densities throughout North America. One of his primary objectives was to determine how the predation rate of moose by wolves depends on moose density. Messier found that a Michaelis-Menton function provides a good fit to moose-killing rates by wolves as a function of moose density. Recall that we encountered the Michaelis-Menton function in Example 6 of Section 1.5 for modeling bacterial nutrient-uptake rates.

Example 5 Wolves eating moose

The rate f(x) at which wolves kill moose, as a function of moose density x (numbers per square kilometer), can be described by the function

images

Here, we examine the shape of the function f(x) over the biologically relevant range of values x ≥ 0, as well as the biologically irrelevant range x < 0.

  1. Find all horizontal and vertical asymptotes for y = f(x). Discuss the biological meaning of one of the horizontal asymptotes.
  2. Sketch its graph y = f(x) for all x and discuss the biological meaning of the graph for nonnegative x.
  3. Relate the graph to the following statement of Sir Winston Churchill (1874–1965) (quoted in H. Eves, Return to Mathematical Circles, Boston: Prindle, Weber & Schmidt, 1988):

    “I had a feeling once about Mathematics—that I saw it all. Depth beyond depth was revealed to me—the Byss and Abyss. I saw—as one might see the transit of Venus or even the Lord Mayor's Show—a quantity passing through infinity and changing its sign from plus to minus. I saw exactly why it happened and why the tergiversation was inevitable but it was after dinner and I let it go.”

Solution

  1. First, let us find the horizontal asymptotes.

    images

    Thus, f(x) has a horizontal asymptote y ≈ 3.36, which f(x) approaches as x approaches ∞; this means that when the moose density is very large, the wolf killing rate stabilizes around 3.36 moose per wolf per 100 days. Similarly (without the corresponding biological meaning) for x < 0, we obtain

    images

    Next, let us find the vertical asymptotes. Since f(x) is not defined at x = −0.46, there is a possible vertical asymptote at x = −0.46. We have images f(x) = ∞ because when x < −0.46, but close to −0.46, x is negative and 0.46 + x is arbitrarily small and also negative. Alternatively, we have images f(x) = −∞ because when x > −0.46, but close to −0.46, x is negative and 0.46 + x is arbitrarily small and positive.

  2. We begin by drawing the asymptotes: y = 3.36 and x = −0.46. The y-intercept is found at x = 0 as f(0) = 0. Our observation that images for x ≠ 0 implies that this function is increasing for all x ≠ 0. Using this information, we draw the graph shown in Figure 2.40.

    Looking at the nonnegative portion of this graph, we see that the killing rate is zero at x = 0 (i.e., moose cannot be killed if they are not around) and that this rate increases with increasing moose density. The rate, however, saturates at approximately x = 3.36 moose per wolf per 100 days. Biologists refer to this saturation rate as the killing rate when moose are “not limiting.”

    images

    Figure 2.40 Graph of the function images

  3. As viewed from left to right, the function passes from positive infinity to negative infinity as it passes through the value x = −0.46, which is Churchill's “quantity passing through infinity and changing its sign from plus to minus.” Perhaps Churchill saw a wolf after his dinner and that's why he let it go.

Infinite limits at infinity

As x gets larger and larger without bound, the value of f might also get larger and larger without bound. In such a case, it is natural to say that f(x) approaches infinity as x approaches infinity.

Infinity at Infinity (Informal Definition)

Let f be a function. We write

images

if f(x) can be made arbitrarily large for all x sufficiently large. We write

images

if f(x) can be made arbitrarily large for all x sufficiently negative. We write

images

if f(x) can be made arbitrarily negative for all x sufficiently large. We write

images

if f(x) can be made arbitrarily negative for all x sufficiently negative.

Example 6 Limits to infinity

Find the following limits.

images

Solution

  1. For large x the number x2 can be made arbitrarily large for all sufficiently large x, so we say images x2 = ∞.
  2. It is tempting to use a limit law here and write

    images

    However, this is incorrect! Limit laws do not apply to infinite limits. Indeed, ∞ − ∞ is not a meaningful statement as ∞ is not a real number. Luckily, we can deal with this by noticing that for large x, xx2 = x(1 − x) is the product of two numbers such that for large x, one of these numbers is large and positive and the other can be made arbitrarily negative. Thus, for sufficiently large x, x(1 − x) is arbitrarily negative. Hence, images (xx2) = −∞.

  3. Again, it is tempting to use a limit law to conclude the limit is images. This is meaningless. However, if we divide the numerator and denominator by x, we find (for x ≠ 0)

    images

    Since images + 10 approaches 0 + 10 = 10 as x approaches ∞, we find images for x sufficiently large. Therefore

    images

Part b of the previous example and Example 3 illustrate the subtlety of taking the limits of the form images f(x) − g(x) when images f(x) = ∞ and images g(x) = ∞. In the previous example with f(x) = x and g(x) = x2, the limit of the difference was −∞. In Example 3 with f(x) = images and g(x) = x, the limit of the difference was 0. The following example illustrates that any limiting value is possible.

Example 7 Another limit of an infinite difference

Find images images where is a constant.

Solution Multiplying and dividing by the conjugate images yields

images

Since images and images go to zero as x gets very positive, we have

images

We have shown that the limit of a difference of terms can take on any value!

Example 8 Unabated population growth

At the beginning of Section 1.4, we modeled population growth in the United States with the function

images

where t represents the number of decades after 1815.

  1. Find images N(t).
  2. Determine how large t has to be to ensure that N(t) is greater than 300,000,000. Discuss how your answer relates to the current U.S. population size.

Solution

  1. Since 8.3(1.33)t gets arbitrarily large for large t, we have that images 8.33(1.33)t = ∞.
  2. We want N(t) ≥ 300,000,000. Solving for t in this inequality yields

    images

    Therefore, the model predicts that t = 61 decades after 1815, in other words in the year 2425, there will be approximately N(t) = 300 million people in the United States. Given that the population size in January 2007 was over 300 million, we can see that the model from the 1800s considerably underestimated the future growth of the U.S. population.

PROBLEM SET 2.4

Level 1 DRILL PROBLEMS

In Problems 1 to 24, find the specified limits.

images

images

images

For images in Problems 25 to 30, determine how close x > a needs to be to a to ensure that f(x) ≥ 1,000,000.

images

For images f(x) = L in Problems 31 to 34, determine how negative x needs to be to ensure that |f(x) − L| ≤ 0.05.

images

For the limit images f(x) = ∞ in Problems 35 to 38, determine how large x needs to be to ensure that f(x) > 1,000,000.

images

Level 2 APPLIED AND THEORY PROBLEMS

39. In Example 8, we showed that images N(t)= ∞ whereN(t) = 8.3(1.33)t represents U.S. population size in millions t decades after 1815. To see that N(t) can get arbitrarily large for x sufficiently large, do the following:

  1. Determine how large t needs to be to ensure that N (t) ≥ 500,000,000.
  2. Determine how large t needs to be to ensure that N(t) ≥ 1,000,000,000.

40. In Example 5 of Section 1.4, we modeled the height of beer froth with the function H(t) = 17(0.99588)tcm where t is measured in seconds.

  1. Determine L such that images H(t) = L.
  2. Determine how large t needs to be to ensure that H(t) is within 0.1 of L.
  3. Determine how large t needs to be to ensure that H(t) is within 0.01 of L.

41. In Example 6 of Section 1.5, we modeled the uptake rate of glucose by bacterial populations with the function f(x) = images mg per hour where x is measured in milligrams per liter.

  1. Find the horizontal and vertical asymptotes of f(x). Interpret the horizontal asymptote(s).
  2. Graph f(x) for all values of x.

42. In Example 5, we examined how the predation rate of wolves depended on moose density. Messier also studied how wolf densities in North America depend on moose densities. He found that the following function provides a good fit to the data:

images

where x is number of moose per square kilometer.

  1. Find the horizontal and vertical asymptotes of f(x). Interpret the horizontal asymptotes.
  2. Graph f(x) for all values of x > 0.

43. In Problem 42, you were asked to find L such that images f(x) = L.

  1. Determine how large x needs to be to ensure that f(x) is within 0.1 of L.
  2. Determine how large x needs to be to ensure that f(x) is within 0.01 of L.

44. The von Bertalanffy growth curve is used to describe how the size L(usually in terms of length) of an animal changes with time. The curve is given by

images

where t measures time after birth and a, b, and t0 are positive parameters. We will derive this curve in Chapter 6. To better understand the meaning of the parameters t0 and b, carry out these steps.

  1. Evaluate L(t0). What does this imply about the meaning of t0?
  2. Find images L(t). What does this limit say about the biological meaning of a?
  3. Graph L(t) and discuss how an organism grows according to this curve.

45. At the beginning of the twentieth century, several notable biologists, including G. F. Gause and T. Carlson, studied the population dynamics of yeast. For example, Carlson grew yeast under constant environmental conditions in a flask; he regularly monitored their population densities.* In Chapter 6, we will show that the following function describes the growth of the population:

images

where N(t) is the population density and t is time in hours. Find images N(t) and discuss the meaning of this limit.

46. The following equation is used to calculate the average firing rate f of a neuron (in spikes per second) as a function of the concentration x of neurotransmitters perfusing its synapses.

images

Find the horizontal asymptote; then find the values of x such that f(x) is within 0.5% of its asymptotic values.

47. The following equation is used to calculate the average firing rate f of a neuron (in spikes per second) as a function of the concentration x of neurotransmitters perfusing its synapses.

images

Find the values of x such that f(x) is within 0.5% of its asymptotic values.

48. Compare the solutions obtained to Problems 46 and 47 and decide which of these represents a tighter on-off switch of the neuron, that is, from firing at its maximum rate to being inactive. What do you conclude in terms of which of the parameters a, b, and c in the function

images

controls the narrowness of the range of x over which on-off switching occurs? Note that this function is called the logistic function and will be encountered in many different examples in upcoming chapters.

2.5 Sequential Limits

In Section 1.7, we considered sequences a1, a2,... of real numbers, which can be used to model drug concentrations, population dynamics, and population genetics. In some cases, these sequences converged to a limiting value as n got very large. In this section, we study the limits of sequences, their relationship to continuity, and a convergence theorem, as well as how these concepts can be used to understand the asymptotic behavior of difference equations. While limits of functions form the basis of differentiation as we shall soon see, limits of sequences form the basis of integration (as we will discuss in Chapter 5).

Sequential limits and continuity

For sequences, there is only one type of limit to consider: the sequential limit, defined as the limiting value of an as n → ∞.

Sequential Limits (Informal Definition)

Let a1, a2, a3,... be a sequence. We write

images

provided that we can make an arbitrarily close to L for all n sufficiently large. In this case, we say the sequence converges to L.

We write

images

provided that we can make an arbitrarily large for all n sufficiently large.

We write

images

provided that we can make an arbitrarily negative for all n sufficiently large.

Example 1 Finding sequential limits

In each of the following, if it exists, calculate images a(n) where

images

Solution

  1. Since 2n gets arbitrarily large as n gets arbitrarily large, images 2n = ∞.
  2. Since images approaches zero as n gets arbitrarily large, images.
  3. Since the numerator and denominator are polynomials in n, we divide the numerator and denominator by the term with largest exponent, namely, n2.

    images

  4. Since cos images alternates between the values 0, 1, and −1, this sequence does not have a limit. There is no unique value that the sequence approaches.
  5. Since images and we can make images arbitrarily close to 0 for n sufficiently large,

    images

    Graphing this sequence, illustrated in the figure below, confirms this convergence to zero.

    images

As in our previous limit definitions, the existence of a sequential limit implies that we can make an as close to L as we like, provided that n is sufficiently large. But how do we verify this statement? What is meant by sufficiently large? The following example illustrates the answer to this question.

Example 2 Finding sufficiently large n

Consider an = images.

  1. Find images an = L.
  2. Determine how large n needs to be to ensure that |anL| < 0.002.

Solution

  1. Dividing the numerator and denominator by n yields

    images

    Hence, L= 1.

  2. We want that

    images

    The number n must be greater than 998.

There is an important relationship between limits of sequences and limits of functions. This relationship is most useful for proving discontinuity of a function.

Theorem 2.4 Sequential continuity theorem

Let f be a function. Then images f(x) = L if and only if images f(an) = L for any sequence satisfying images an = a.

One direction of this theorem is clear. If images an = a, then an can be made arbitrarily close to a for n sufficiently large. Therefore, if images f(x) = L, then f(an) is arbitrarily close to L for n sufficiently large. Hence, if images f(x) = L, then images f(an) = L. If you are feeling sufficiently adventuresome, try proving this direction using formal definitions of limits. To do so, you will have to come up with a formal definition of sequential limits. The other direction of the sequential continuity theorem is more subtle, and the ideas of the proof are beyond the scope of this text.

Example 3 Proving nonexistence of limits

Show that images sin images does not exist.

Solution Let f(x) =sin images. Our goal is to find two sequences an and bn satisfying images an = 0 and images bn = 0, but images f(an) = L and images f(bn) = M with LM. Then, we can apply the sequential continuity theorem to conclude that the limit images f(x) does not exist. For this example, we let an = images. Then,

images

We now find the limits of f(an) and f(bn):

images

and

images

Since images f(an) ≠ images f(bn), it follows from the sequential continuity theorem that images sin images does not exist because it cannot be equal to both 0 and 1 at the same time.

Asymptotic behavior of difference equations

When we introduced sequences in Section 1.7, we considered a special class of sequences that arises through a difference equation

images

where a1 is specified and f is a function. In some instances, we can actually find explicit expressions for the sequence defined by the difference equation and take the limit.

Example 4 Finding the limit of a sequence

Find an explicit expression for the sequences defined by the following difference equations and find the limit as n becomes large.

images

Solution

  1. We have a1 = 0.1, a2 = 0.1a1 = 0.12, and a3 = 0.1a2 = 0.13. Hence, we can see inductively that an = 0.1n. Since an gets arbitrarily small as n gets sufficiently large, we obtain images an = 0.
  2. We have,

    images

    To find this limit, consider the logarithm of this sequence, that is,

    images

    Clearly, images ln an = 0. Thus, by the continuity of ex and the sequential continuity theorem, we get that

    images

In part b of Example 4, we saw that sometimes it is useful to find images an by finding images f(an) for an appropriate choice of a continuous one-to-one function f.

Example 5 Lethal recessives revisited

In Example 5 of Section 1.7, we modeled the proportion xn of a lethal recessive allele in a population at time n with the difference equation:

images

Keeping with the notation xn rather than an, assume that the initial proportion of the lethal allele is x1 = 0.5.

  1. Verify that xn = images satisfies the difference equation.
  2. Determine images xn. Discuss the implication for the proportion of the lethal recessive allele in the long term.
  3. Determine how large n needs to be to ensure that xn ≤ 0.1.
  4. Determine how large n needs to be to ensure that xn ≤ 0.01. Discuss the implications.

Solution

  1. First we verify that the formula holds for n = 1, namely, x1 = images = 0.5. To verify that xn = images satisfies the difference equation for any n > 1, we proceed inductively and substitute our expression for xn into both sides of the difference equation and show that they are equal. Evaluating the proposed solution on the left-hand side of the difference equation images gives us

    images

    Evaluating the proposed solution on the left-hand side of the difference equation equation give us

    images

    Hence, we have shown that the formula xn = images satisfies the difference equation xn+1 = images for all n.

  2. Since images gets arbitrarily small as n gets arbitrarily large, images xn = 0. Hence, in the long term, we expect the lethal recessive genes to vanish from the population.
  3. We want

    images

    Hence, after nine generations the proportion of lethal recessives is less than 0.1.

  4. We want

    images

    Hence, after ninety-nine generations the proportion of lethal recessives is less than 0.01. These calculations suggest that initially the proportion of lethal recessives decreases rapidly, but further decreases in the proportion occur more and more slowly.

Returning to our an notation, recall from Section 1.7, that a point a is an equilibrium of a difference equation an+1 = f(an) if f(a) = a. In Example 4a and Example 5, the only solution to the equation f(a) = a is a = 0, and the sequences generated by these difference equations converge to this equilibrium. In Example 4b, the equation images = a has two solutions: a = 0 and a = 1, and the sequence we examined converged to the latter rather than former equilibrium.

To see why this convergence to equilibria occurs, consider a sequence an that satisfies the difference equation

images

where f is a continuous function. Assume that images an = a. By the sequential continuity theorem, we have

images

Hence, the limiting value a is an equilibrium for this difference equation.

Limits of Difference Equations

Let f be a continuous function and an be a sequence that satisfies

images

If images an = a, then f(a) = a. In other words, a is an equilibrium.

Example 6 To converge or not to converge

Find the equilibria of the following difference equations and use technology to determine whether the specified sequence converges to one of the equilibria.

images

Solution

  1. To find the equilibria, we solve

    images

    Hence, if the sequences determined by this difference equation have well-defined limits, then these limits are either images. Computing the first twenty terms of the difference equation with a1 = 1 and plotting yields the following graph.

    images

    It appears that the sequence is converging to the positive equilibrium.

  2. To find the equilibrium, we solve

    images

    Computing and plotting the first twenty terms of the difference equation with a1 = 0.1 and plotting yields the following graph.

    images

    It appears that the sequence is converging to images.

  3. To find the equilibrium, we solve

    images

    Computing and plotting the first one hundred terms of the difference equation with a1 = 0.1 and plotting yields the following graph.

    images

    It appears that the sequence does not converge; rather, it seems to eventually oscillate between four values.

One of the most important models in population biology is the discrete logistic model:

images

where the parameter r > 0 is called the intrinsic rate of growth and K > 0 is called the environmental carrying capacity.

Example 7 Dynamics of the discrete logistic model

  1. Find the equilibrium solutions associated with the discrete logistic equation. What do you observe about the roles of the parameters r and K in determining this equilibrium?
  2. Calculate the first twenty points of the sequence an+1 = an + 0.3an(1 − an) with a1 = 0.1.
  3. Repeat part b with a1 = 1.5.
  4. Calculate the first twenty points of the sequence an+1 = an + 1.9an(1 − an) with a1 = 0.6.
  5. Calculate the first twenty points of the sequence an+1 = an + 2.2an(1 − an) with a1 = 0.6.

Solution

  1. The equilibria are solutions to the equation

    images

    From this it is clear that the value of r does not influence the value of the equilibria, one of which is equal to K.

  2. The values in this sequence are plotted in the left-hand side of Figure 2.41. The sequence appears to be converging to the positive equilibrium K = 1.

    images

    Figure 2.41 Plots of the sequences an+1 = an + 0.3an(1 − an) from starting values a1 = 0.1 (left panel) and a1 = 1.5 (right panel).

  3. The values in this sequence are plotted in the right-hand side of Figure 2.41. The sequence appears to be converging to the equilibrium K = 1.
  4. The values in this sequence are plotted in the left-hand side of Figure 2.42. The sequence exhibits dampened oscillations around the equilibrium K = 1. It still appears to be converging to K = 1.

    images

    Figure 2.42 Plots of the sequences an+1 = an + ran(1 − an) from the starting value a1 = 0.6 for the cases r = 1.9 (left panel) and r = 2.2 (right panel).

  5. The values in this sequence are plotted in the right-hand side of Figure 2.42. The sequence oscillates between a lower and higher population density. The oscillations are not dampened and the sequence does not appear to be converging to the equilibrium K = 1.

Examples 6 and 7 illustrate that the existence of equilibria for a difference equation does not ensure the convergence of the sequences generated by it. This raises the question, when do the sequences generated by a difference equation converge to an equilibrium? In general, this is a hard question. The following theorem, stated without proof, provides a criterion that ensures convergence of solutions of a difference equation. Later, when we have covered the basics of derivatives of functions, we will present another criterion that ensures convergence of a sequence to an equilibrium. We make two definitions before stating this theorem. We say that a sequence is increasing (respectively, decreasing) if a1a2a3 ≤ ... (respectively, a1a2a3...).

Theorem 2.5 Monotone convergence theorem

Let f be a continuous, increasing function on an interval I = [a, b] such that the image of f lies in I (i.e., if x is in I, then f(x) is a value that also lies within I). If a1, a2, a3,..., is a sequence that satisfies an+1 = f(an), then the sequence is either increasing or decreasing. Moreover, images an = a in I exists and satisfies f(a) = a.

The next example illustrates the application of the monotone convergence theorem to the Beverton-Holt stock-recruitment model. This model has been used extensively by fisheries scientists to formulate models of fish, such as Pacific salmon, that breed once and then die.Stock-recruitment curves describe how a spawning stock of N individuals in one generation contributes recruits R (i.e., new individuals) to the next generation. An example of a Beverton-Holt function fitted to data obtained for sockeye salmon spawning in Karluk Lake, Alaska, is shown in Figure 2.43.

images

Figure 2.43 Sockeye salmon (Oncorhynchus nerka) on the left; and, on the right, the relationship between recruits and stock for sockeye salmon in Karluk Lake, Alaska.

Data Source: John A. Gulland, Fish Stock Assessment: A Manual of Basic Methods (New York: Wiley, 1983).

Example 8 Beverton-Holt sockeye salmon dynamics

The function plotted in the right panel of Figure 2.43 is given by

images

Note that it is more natural for us to continue using Nn rather then the sequence notation an, since N is the symbol most commonly used to denote population size in the ecology literature.

where N is the stock size (spawners) in a particular generation and R(N) is the number of recruits produced for the next generation. Since the number of recruits determines the size of the stock that will be spawners in the next generation, it follows that the salmon dynamics can be modeled by the difference equation

images

where Nn is the stock size of the nth generation.

  1. Find the equilibria of this difference equation.
  2. Graph R(N) and y = N.
  3. Apply the monotone convergence theorem to determine what happens to Nn when N1 = 10 and when N1 = 200. Use cobwebbing to illustrate your results.

(Note that it is possible to find an explicit solution of this difference equation. This is explored in Problem Set 2.5.)

It may seem strange for population size to have a fractional value, but not if size is measured as a density—that is, numbers per unit area—or is in units of millions of individuals.

Solution

  1. To find the equilibria, we solve

    images

    Hence, the equilibria are given by 0 and images.

  2. Plotting the two functions yields the following graph.

    images

    The equilibria correspond to the points where the functions intersect.

  3. Since R(N) is increasing on I = [0, ∞) and the image of I under f is I, we can apply the monotone convergence theorem.

    Assume N1 = 10. Since N2 = images ≈ 38.46 ≥ N1, the monotone convergence theorem implies that Nn is increasing. On the other hand, since the graph of R(N) is saturating at 166images, we have 10 ≤ Nn+1 = R(Nn) ≤ 166images for all n ≥ 1. Therefore, by the monotone convergence theorem images Nn must equal the equilibrium 133images. Cobwebbing with N1 = 10 illustrates this convergence in Figure 2.44 (left).

    images

    Figure 2.44 Cobwebbing solutions to salmon model for initial values N1 = 10 (left) and N1 = 200 (right).

    Assume N1 = 200. Since N2 = images ≈ 142.85 ≤ N1, the monotone convergence theorem implies that Nn is a decreasing sequence. On the other hand, 200 ≥ Nn ≥ 133images for all n ≥ 1. Therefore, by the monotone convergence theorem, images Nn must equal the equilibrium 133images (from part a). Cobwebbing with N1 = 200 illustrates this convergence in Figure 2.44 (right).

In Example 5 of Section 1.7, we developed a population genetics model under the assumption that individuals could carry and pass on a recessive lethal allele. In formulating this model, we assumed that two alleles, A and a, exist and determine three possible genotypes: AA, Aa, and aa. If the allele a is the recessive lethal, then this implies that genotype aa is the least viable since, by definition, aa individuals all die before they are able to reproduce. In the next example, we assume that all three genotypes are viable, but that the so-called heterozygous genotype Aa is the least viable. Extreme instances of heterozygous inviability arise when different genotypes can mate but produce infertile offspring, such as mules, which are produced when horses mate with donkeys.

Example 9 Disruptive selection

If two homozygous genotypes AA and aa produce equal numbers of progeny that exceed the number of progeny produced by the heterozygous Aa genotype, then the proportion xn of allele a at time n can be shown to be modeled by xn+1 = f(xn), where the graph of f(x) has the S-shaped curve depicted in Figure 2.45. Curves of this shape are said to depict disruptive selection.

images

Figure 2.45 Disruptive selection.

Using cobwebbing methods, determine from this graph what happens to xn in the long term for the two cases given in parts a and b below.

  1. x1 = 0.6
  2. x1 = 0.4
  3. As reported in a 1972 Science article, Foster and others experimentally examined changes in two chromosomal proportions in Drosophila melanogaster. Data from a set of experiments is graphed in Figure 2.46.

images

Figure 2.46 Time series for disruptive selection in chromosomal proportions for a species of fruit fly.

Source: G. G. Foster, M. J. Whitten, T. Prout, and R. Gill, “Chromosome Rearrangements for the Control of Insect Pests,” Science 176: 875–880. Reprinted with permission.

These experimentally determined graphs show how the proportion of an allele changes over generations for initial conditions that through a small amount of random variation lead to different population levels at time 1 on the x-axis. Discuss whether these experiments are consistent with the model predictions.

Solution

  1. Since the graph of f is increasing, we can apply the monotone convergence theorem. Since f(0.6) > 0.6, the sequence xn is increasing if x1 = 0.6. Since xn ≤ 1 for all n and f(1) = 1 is the only equilibrium greater than 0.6, xn converges to 1 as n increases. In other words, the proportion of a alleles approaches one. Cobwebbing reaffirms this prediction, as shown in the following graph.

    images

  2. Since the graph of f is increasing, we can apply the monotone convergence theorem. Since f(0.4) < 0.4, the sequence xn is decreasing if x1 = 0.4. Since xn ≥ 0 for all n and f(0) = 0 is the only equilibrium less than 0.4, xn converges to 0 as n increases. In other words, the proportion of a alleles approaches zero. Cobwebbing reaffirms this prediction, as shown in this graph.

    images

  3. The experiments of Foster and others are consistent with the model predictions, if we look at the proportion of allele a at the start of the second generation (i.e., at 1 rather than 0 on the horizontal axis). Specifically, within the limits of a small amount of random variation, the data show that if the proportion of allele a is greater or less than one-half, then the proportion of allele a, respectively, approaches one or goes extinct.

Fibonacci famously posed the following problem (reworded here) over 800 years ago. Suppose a newly born pair of rabbits, one male and one female, are put in an enclosed, but very large, field. Further, suppose all rabbits are able to mate at the age of one month and that the impregnated female gives birth to a male-female pair one month later. Ignoring questions relating to inbreeding, assuming that rabbits never die and there is always enough food for them to eat, what is the asymptotic annual rate of increase of the rabbits in the enclosed field? This is the problem we address in the next example.

Example 10 Fibonacci and the growth rate of rabbits

Fibonacci's rabbit model (see images, Problem 38 of Section 1.7) has the form

images

where Rn is the number of rabbit pairs in month n.

  1. What is the asymptotic behavior of Rn as n → ∞?
  2. What is the annual rate of increase in the population? Calculate the population produced by one pair at the end of the first year (i.e., calculate R12when R0 = 1).

Solution

  1. To find the asymptotic behavior of Rn, we can transform the model into a familiar sequence model problem by dividing both sides of the above equation by Rn−1 to obtain

    images

    and then define an = Rn/Rn−1. In this case an, the ratio of the number of rabbits in month n to those in month n − 1, satisfies the equation

    images

    The equilibrium solution to this equation is

    Table 2.5 Fibonacci rabbit growth

    Month Number of pairs
    0 1
    1 1
    2 2
    3 3
    4 5
    5 8
    6 13
    7 21
    8 34
    9 55
    10 89
    11 144
    12 233

    images

    Only the positive solution a ≈ 1.6180 applies here, and it can be shown that the sequence converges to this value as the number of months increases (see Problem 44 in Problem Set 2.5). Thus, although Rn increases without bound, the ratio an = Rn/Rn−1 approaches a constant.

  2. To get the annual rate of increase we need to calculate (1.618)12 ≈ 322, a really stunning rate of growth. In Table 2.5, we list the first 13 terms and note that the rate of increase over the first 12 iterations is 233 rather than 322 because the equilibrium value for the rate of increase applies asymptotically rather than initially.

PROBLEM SET 2.5

Level 1 DRILL PROBLEMS

Determine whether the sequential limits in Problems 1 to 8 exist. If they exist, find the limit. If they do not exist, explain briefly why.

images

Consider the sequences defined in Problems 9 to 14.

  1. Find images an.
  2. Determine how large n needs to be to ensure that |anL| < 0.001.

images

images

All the sequences in Problems 15 to 18 satisfy images an = ∞. Determine how large n has to be to ensure that an ≥ 1,000,000.

15. an = 2n

16. an = n2

17. an = 2n − 10,000

18. an = images

Find the sequences determined by the difference equation xn+1 = f(xn) with the initial condition x1 specified in Problems 19 to 24. Determine images an. Justify your answer.

19. f(x) = x + 2 with x1 = 0

20. f(x) = images with x1 = 27

21. f(x) = images with x1 = 100

22. f(x) = x2 with x1 = 1.00001

23. f(x) = x2 with x1 = 0.99999

24. f(x) = 4x2 with x1 = 1

Find the equilibrium of the difference equations in Problems 25 to 28 and use technology to determine which of the specified sequences converge to one of the equilibria.

images

Use the monotone convergence theorem in Problems 29 to 34 to determine the limits of the following specified sequences.

images

Level 2 APPLIED AND THEORY PROBLEMS

35. In Example 5 of Section 2.5 we introduced a model for the frequency of lethal recessive alleles in a population. The model is given by

images

where xn is the frequency of the recessive allele in the population.

  1. If x1 = 0.25, then verify that xn = images satisfies the difference equation.
  2. Find images xn.
  3. Determine how large n needs to be to ensure that xn ≤ 0.1.
  4. Determine how large n needs to be to ensure that xn ≤ 0.001.

36. Let us consider the lethal recessive allele model in greater generality.

  1. Verify that xn = images satisfies the difference equation xn+1 = images for any choice of x1.
  2. For any x1, find images xn.
  3. Assuming that x1 lies in (0, 1), determine how large n needs to be to ensure that xn ≤ 0.01.

37. In Example 5 of Section 1.7, we discussed a population genetics model under the assumption that there were two alleles, A and a, that determined three possible genetic types, AA, Aa, and aa. We assume that genetic type Aa(so called heterozygote) is the least viable. If the genotype aa produces nine times more progeny than genotype AA progeny, then the frequency xn of allele a at time n can be modeled by xn+1 = f(xn) where the graph of f is given by

images

  1. Determine what happens to xn in the long term if x1 = 0.91.
  2. Determine what happens to xn in the long term if x1 = 0.89.
  3. As reported in a 1972 Science article, Foster and others experimentally examined changes in two chromosomal frequencies in Drosophila melanogaster. Data from a set of experiments are graphed next:

    images

    These experimentally determined graphs show how the frequency of an allele changes over generations for different initial conditions. Discuss whether these experiments are consistent with the model predictions.

38. The Beverton-Holt model has been used extensively by fisheries. This model assumes that populations are competing for a single limiting resource and reproduce at discrete moments in time. If we let Nn denote the population abundance in the nth year (or generation), r > 0 the maximal per capita growth rate, and a > 0 as a competition coefficient, then the model is given by

images

with r > 0 and a > 0.

  1. Assume N1 = 1 and find the first four terms of the sequence.
  2. Guess the explicit expression for the sequence and verify that your guess is correct.
  3. Find the equilibria of this difference equation.
  4. By considering the images Nn, determine under what conditions the population goes extinct (i.e., converges to 0) versus under what conditions it is able to persist (i.e., converge to a positive equilibrium N > 0) if it is initially at N1 = 1.

In the Fibonacci rabbit problem laid out in Example 10, suppose only a proportion p of the females that could become pregnant actually do become pregnant each month. (We assume the population starts with a large number of pairs so that when we refer to proportions, we are actually thinking of whole numbers of pairs rather than, say one-third of two pairs, which makes no biological sense.) What is the annual rate of increase in Problems 39 to 43 if p equals the specified value?

39. 3/4

40. 2/3

41. 1/2

42. 1/3

43. 1/4

44. Consider the Fibonacci sequence in Example 10. Show that

images

and hence apply the monotone convergence theorem to find out what happens to the sequences of even elements of an.

45. Use technology to calculate the first twenty points of the sequence an+1 = an + ran(1 − an) with a1 = 0.5 for the cases r = 0.9, r = 1.5 and r = 2.1. How does this fit in with the discussion in the solution to part e of Example 7.

46. Revisiting the sockeye salmon stock-recruitment relationship considered in Example 8, we see from Figure 2.43 that we could just as credibly fit the Ricker function

images

If x is the stock in one generation and y is the stock recruited in the next generation, then this relationship is actually a population model of the form

images

If the population is now at the level x1 = 100 individuals, use technology to generate the number of individuals that you expect in the next ten generations. Deduce the equilibrium value and check this value by using technology or the bisection method to solve the equation x = 3.7xe−0.01x. In Figure 2.47, the solid line is fit to the data using a Ricker functional form that is similar to the Beverton-Holt form considered in Example 8.

images

Figure 2.47 Sockeye salmon stock-recruitment.

Data Source: John A. Gulland, 1983.Fish Stock Assessment: A Manual of Basic Methods, Wiley.

2.6 Derivative at a Point

In this section, we introduce one of the major concepts in calculus, the idea of a derivative. Although functions are fundamental and limits are essential, they simply laid the foundation for the excitement to come. To motivate the idea of the derivative at a point, let us recast Example 1 of Section 2.1 using different notation. If we let f(x) represent the population of Mexico in year x, then the average population between the years 1980 and 1985 can be found by

images

where a = 1980 (the base year) and h = 5 (the duration of the time interval). We also found the average rate of change over smaller and smaller intervals to guess that the instantaneous rate of change of the Mexico population in 1980 would be 1.75 million per year. This idea can be written as

images

In Chapter 1 and Section 2.1, we previewed the notion of the derivative at a point by defining the tangent line for f(x) at a point x = a to be the limit of the slope of the secant lines

images

Slopes of tangent lines and instantaneous rates of changes have the same formula, and it is this limiting process that is the basis for the concept of the derivative.

Derivative at a Point

The derivative of function f at a point x = a, denoted by f′(a), is

images

provided this limit exists. If the limit exists, we say that f is differentiable at x = a.

Example 1 Finding derivatives using the definition

Use the definition of a derivative to find the following derivatives.

  1. f′(3) where f(x) = 1
  2. f′(2) where f(x) = 3x
  3. f′(1) where f(x) = 1 + 3x2

Solution

  1. Let f(x) = 1 and a = 3.

    images

  2. Let f(x) = 3x and a = 2.

    images

  3. Let f(x) = 1 + 3x2 and a = 1.

    images

Example 1 illustrates two facts. First, the derivative of a constant function is zero. Intuitively this makes sense, as a constant function by definition does not change and, consequently, its rate of change should be zero. Second, the derivative of a linear function is the slope of the linear function. Intuitively, this makes sense as the rate at which the function is increasing is given by the slope of the function. More interestingly, Example 1 illustrates that we can explicitly compute the slope (equivalently, the instantaneous rate of change) of a quadratic function.

The derivatives in Example 1 were pretty straightforward to compute, but other derivatives require certain algebraic procedures to compute, as illustrated by the next example.

Example 2 Algebraic steps to find a derivative

Find the following derivatives algebraically.

images

Solution

  1. To find this derivative, we multiply the numerator and denominator of the quotient by the “conjugate” of the original numerator.

    images

    Hence, taking the limit as h goes to 0 yields f′(4) = images.

  2. To find this derivative, we can multiply by the common denominator in the numerator.

    images

    Taking the limit as h goes to 0 yields f′(5)= images.

Slopes of tangent lines

The definition of the derivative was inspired directly by the slope of the tangent line. Using derivatives, we can find the tangent line.

Tangent Line

Let f be a function that is differentiable at the point x = a. The tangent line of f at x = a is the line with slope f′(a) that passes through the point (a, f(a)).

Example 3 Tangent line to a parabola

Find the tangent line to f(x) = 1 + 3x2 at x = 1. Sketch the parabola and the tangent line.

Solution In Example 1 we found that the slope of the tangent line is f′(1) = 6. Since the tangent line passes through (1, f(1)) = (1, 4), we can use the point-slope formula to find the equation of the tangent line:

images

images

Figure 2.48 Graph of the parabola y = 1 + 3x2 with tangent line at (1, 4).

The graph of the parabola, along with the tangent line at (1, 4) is shown in Figure 2.48.

In Example 3, the tangent line intersects the graph of the function in exactly one point. This unique intersection is not typical, as illustrated in the next example.

Example 4 Multiple intersections

Find the tangent line to the cubic f(x) = x3 at x = 1. Sketch the cubic and the tangent line.

Solution We first find the slope of the tangent line:

images

The tangent line passes through (1, f(1)) = (1, 1), so the equation of the tangent line is

images

Figure 2.49 Graph of the cubic y = x3 and its tangent line at (1, 1).

images

Equivalently, y = 3x − 2. The graph of y = x3 and the tangent line at (1, 1) is shown in Figure 2.49.

Instantaneous rates of change

In Section 2.1, we defined

images

to be the average rate of change of f over the interval [a, b]. Taking the limit as b approaches a yields the instantaneous rate of change

images

In the next example, we relate this definition of the instantaneous rate of change to the derivative.

Example 5 Instantaneous rates of change

Show that

images

provided that the limits exist.

Solution Let h = ba. Then b = a + h so that

images

Since b approaching a is equivalent to h approaching 0, we have

images

provided the limits exist. By the definition of a derivative, we get

images

The solution to Example 5 allows us to equate the derivative with an instantaneous rate of change.

Instantaneous Rate of Change as a Derivative

Let f be a function that is differentiable at x = a. The instantaneous rate of change of f at x = a is f′(a).

Example 6 Instantaneous velocity

On a calm day, a seed cone of a coastal California redwood tree drops from one of its high branches 305 feet above the ground. From physics, we know that the distance s in feet an object falls after t seconds, when air resistance is negligible, is given by the formula

images

  1. Find s′(1) and interpret this quantity.
  2. Find the velocity (instantaneous rate of change) of the cone at the moment it hits the ground.

Solution

images

After one second, the cone is falling at a velocity of 32 feet per second (or 32 ft/s).

b. First, we need to find how long it takes the cone to fall to the ground. When the cone hits the ground, it has fallen 305 feet. Hence, we need to solve 305 = s(t) = 16t2, which yields t = images. TO find simages, we use the defi nition of a derivative

images

Hence, at the moment the cone hits the ground it is falling at a velocity of 139.7 ft/s. This is equivalent to 95.3 miles per hour (mi/h). Of course, if the effects of air resistance are taken into account, velocity will be less.

Example 7 Melting Arctic sea ice

One of the important consequences of increasing temperature on Earth is that sea ice is melting at both the North and South poles. In 2012, the extent of Arctic sea ice was approximately 3.61 million square kilometers, the lowest in the past thirty years. Figure 2.50 contains the plot of the average Arctic sea ice extent for the past thirty years. These data can be approximated by a quadratic function of the form

images

where t is years after 1980.

images

Figure 2.50 Sea ice extent as a function of years since 1980.

  1. Find S′(32).
  2. Determine the units of this derivative and discuss their meaning.

Solution

  1. Using the definition of a derivative, we get

    images

  2. The units of S′(32) are millions of square kilometers per year. Since S′(32) = −0.233, we conclude that we are losing 233 thousands of square kilometers of Arctic sea ice per year.

Differentiability and continuity

If a function f(x) is differentiable at a point a, then it is also continuous at a point a, as stated in the following theorem.

Theorem 2.6 Differentiability implies continuity theorem

If f is differentiable at the point x = a, then f is continuous at x = a.

Proof. To prove this theorem, assume that f is differentiable at x = a. Then

images

Therefore, by the limit law for sums,

images

Hence,

images

and thus f is continuous at x = a.

The reverse of this theorem, however, is not true. To fully appreciate differentiability, it is useful to understand examples of where continuity holds but the function is not differentiable.

Example 8 Absolute value functions have nondifferentiable corners

The absolute value function f(x) = |axb|, for positive values of a and b, is defined to be

images

Show that f(x), which is illustrated in Figure 2.51 for the case a = 2 and b= 1, is continuous but not differentiable at x = b/a.

Solution Since images is continuous at x = b/a.

images

Figure 2.51 Graph of the function f(x) = |axb| for the case a = 2 and b = 1.

Now consider the limit from the left in the definition of a derivative applied to the given function:

images

Similarly consider the limit from the right in the definition of a derivative applied to the given function:

images

Since these two limits are not equal, the function is not differentiable at x = a.

A biological model where we have continuity but nondifferentiability at a point is illustrated in the following example.

Example 9 Continuous but not differentiable

Examine the continuity and differentiability of f(x) at x = a for the following two functions.

images

where we recall from Example 7 in Section 2.2 that this function models the feeding rate of planktonic copepods and x is the concentration of planktonic cells per liter.

b. f(x) = x1/3 and a = 0.

Solution

  1. Since

    images

    and

    images

    we see f is continuous at x = 200.

    On the other hand, for h < 0,

    images

    and for h > 0,

    images

    Since the left- and right-hand limits are not equal, the limit does not exist, so f is not differentiable at x = 200. As you can see in Figure 2.52, the function is continuous but is still not differentiable at x = 200.

    images

    Figure 2.52 Feeding rate of planktonic copepods.

  2. Since f(x) = x1/3 is arbitrarily close to 0 for all x sufficiently close to 0, images f(x) = 0. Since f(0) = 0, f is continuous at x = 0. To determine the derivative of x1/3 at x = 0, we need to consider

    images

    images

    Hence, the derivative is not defined, as the limit is not finite. Graphing y = x1/3 reveals that the slope of the tangent line at x = 0 is infinite; that is, it is vertical, as illustrated on the left.

Example 9 illustrates that continuity does not ensure differentiability and that differentiability can fail in at least two ways. The limit of the slopes of the secant lines might not converge or this limit may become infinitely large. While continuity does not imply differentiability, Theorem 2.6 ensures that the opposite is true; that is, differentiability ensures continuity. Hence, differentiability can be viewed as an improvement over continuity.

PROBLEM SET 2.6

Level 1 DRILL PROBLEMS

Using the definition of a derivative, find the derivatives specified in Problems 1 to 10.

1. f′(−2) where f(x) = 3x − 2

2. f′(3) where f(x) = 5 − 2x

3. f′(1) where f(x) = −x2

4. f′ (0) where f (x) = x + x2

images

Find the tangent line at the specified point and graph the tangent line and the corresponding function in Problems 11 to 20. Notice these functions are the same as those given in Problems 1 to 10.

11. f(x) = 3 x − 2

12. f(x) = 5 − 2x at x = 3

13. f(x) = −x2 at x = 1

14. f(x) = x + x2 at x = 0

images

17. f(x) = x3 at x = −1

18. f(x) = x3 + 1 at x = 2

19. f(x) = images at x = 9

20. f(x) = images at x = 5

Determine at which values of x in Problems 21 to 26 that f is not differentiable. Explain briefly.

images

images

25. f(x) = |x − 2|

26. f(x) = 2|x + 1|

27. Let f(x) = images

  1. Sketch the graph of f.
  2. Show that f is continuous, but not differentiable, at x = 1.

28. Give an example of a function that is continuous on (−∞, ∞) but is not differentiable at x = 5.

29. Consider the function defined by

images

Sketch this graph and find all points where the graph is continuous but not differentiable.

30. Consider the function defined by

images

Sketch this graph and find all points where the graph is continuous but not differentiable.

Level 2 APPLIED AND THEORY PROBLEMS

31. A baseball is thrown upward and its height at time t in seconds is given by

images

  1. Find the velocity of the baseball after two seconds.
  2. Find the time at which the baseball hits the ground.
  3. Find the velocity of the baseball when it hits the ground.

32. A ball is thrown directly upward from the edge of a cliff and travels in such a way that t seconds later, its height above the ground at the base of the cliff is

images

feet.

  1. Find the velocity of the ball after two seconds.
  2. When does the ball hit the ground, and what is its impact velocity?
  3. When does the ball have a velocity of zero? What physical interpretation should be given to this time?

33. If the data in Figure 2.53 represents a set of measurements relating enzyme activity to temperature in degrees Celsius, and the quadratic equation

images

provides a good fit to this data, then find A′(50) and discuss its meaning.

images

Figure 2.53 Enzyme activity as a function of temperature.

34. In Example 7 we discussed how the extent of Arctic sea ice is modeled by the function

images

where t is years after 1980. Use the definition of the derivative to compute S′(20). How does this value compare to S′(30) found in Example 7? What does this comparison suggest about the rate at which arctic ice is being lost?

35. In 2010, W. B. Grant published an article about the prevalence of multiple sclerosis in three U.S. communities and the role of vitamin D. The best-fitting quadratic relationship to the data published in this article, relating prevalence of multiple sclerosis (MS) to latitude (exposure to sunlight, hence vitamin D synthesis decreases with latitude) is shown in Figure 2.54. This quadratic function is given by

images

where x is latitude.

images

Figure 2.54 Prevalence of MS (in cases per 100,000) as a function of latitude in the United States.

Data Source: W. B. Grant, “The Prevalence of Multiple Sclerosis in 3 US Communities: The Role of Vitamin D” [letter].Prevention of Chronic Diseases 7 (2010): A89.

  1. Find P′(40).
  2. Find P′(45).
  3. Interpret and compare the numbers that you found in a and b.

36. An environmental study of a certain suburban community suggests that t years from now, the average level of carbon monoxide in the air can be modeled by the formula

images

parts per million.

  1. At what rate will the carbon monoxide level be changing with respect to time one year from now?
  2. By how much will the carbon monoxide level change during the first year?

37. Perelson and colleagues studied the viral load of HIV patients during antiviral drug treatment.* They estimated the viral load of the typical patient to be

images

particles per milliliter on day t after the drug treatment.

  1. Estimate V′(2).
  2. Describe the units of V′(2) and interpret this quantity.

38. Stock-recruitment data and a fitted Beverton-Holt function for sockeye salmon in Karluk Lake, Alaska, were was shown in Figure 2.43. The fitted function was

images

where x is the current stock size and y is the number of recruits for the next year. To determine the number of recruits produced per individual, consider the function

images

  1. Algebraically find g′(10).
  2. Describe the units of g′(10), and discuss the meaning of this quantity.

39. In Example 6 of Section 1.5, we developed the Michaelis-Menton model for the rate at which an organism consumes its resource. For bacterial populations in the ocean, this model was given by

images

where x is the concentration of glucose (micrograms per liter) in the environment. To determine the rate of glucose consumption per microgram of glucose in the environment, consider the function

images

  1. Algebraically compute g′(0) and g′(20).
  2. Describe the meaning of the derivatives that you computed.

40. In Example 5 of Section 2.4, we found that the rate at which wolves kill moose can be modeled by

images

where x is measured in number of moose per square kilometer. To determine the per capita killing rate of moose, consider the function

images

  1. Algebraically compute g′(1) and g′(2).
  2. Describe the meaning of the derivatives that you computed.

2.7 Derivatives as Functions

Our notion f′(a) for the derivative at the point x = a suggests that we can think of f′ as a function. Indeed this is true.

Derivative as a Function

Let f be a function. The derivative of f is defined by

images

for all x for which this limit exists.

Example 1 Finding Derivatives

Find the derivatives f′ of the following functions f.

images

e. Guess the derivative of f(x) = xn for n a whole number.

images

Figure 2.55 The function f(x) = x (blue) and its derivative f′(x) = 1 (red) are plotted on the interval [−2, 2].

Solution

  1. If f(x) = 1, then f′(x) = 0 for every x(see Example 1 of Section 2.6). The derivative of a constant is 0.
  2. Use the definition of the derivative of a function.

    images

    The function and its derivative are illustrated in Figure 2.55.

  3. For f(x) = x2 and a fixed value of x, the definition of a derivative implies

    images

    images

    Figure 2.56 The function f(x) = x2(blue) and its derivative f′(x) = 2x (red).

    The function x2 and its derivative 2x are illustrated in Figure 2.56.

  4. Again, we use the definition. For a fixed number x

    images

    images

    Figure 2.57 The function f(x) = x3 (blue) and its derivative f′(x) = 3x2 (red).

    The function x3 and its derivative 3x2 are illustrated in Figure 2.57.

  5. The above parts suggest that f′(x) = nxn−1 for n a whole number. Indeed, this turns out to be true, as we will show in Chapter 3.

When given numerical data, we can estimate the derivative of the data using the definition of the derivative with the smallest possible h value.

Example 2 Estimating the derivative using a table

In Example 1 of Section 2.1, we considered the population size of Mexico (in millions) in the early 1980s as reported in the following table:

Year Population (millions)
1980 67.38
1981 69.13
1982 70.93
1983 72.77
1984 74.66
1985 76.60

Let P(t) denote the population size t years after 1980. That is, t = 0 corresponds to 1980. For example, we found the average population growth rate in 1980 to be about 1.75 million/year, so we would write this as P′(0) ≈ 1.75.

  1. Estimate P′(t) at t = 0, 1, 2, 3, 4 using h = 1 in the definition of a derivative.
  2. In Problem 28 of Section 1.4, you may have found the population size could be represented by the exponential function P(t) = 67.37(1.026)t. Approximate the derivative with an exponential function and compare it with P(t). What do you notice?

Solution

  1. The estimates of P′(t) are calculated from the data as indicated in the third column in Table 2.6.

    Table 2.6 Estimates of the rate of population growth in Mexico

    images

  2. To approximate P′(t) by an exponential function, we can look at these ratios:

    images

    These ratios are all about the same. In fact, the average is 1.026, which is the ratio for the population function itself! We approximate P′(t) by

    images

    Comparing this function with the function for the population growth, we see that

    images

    Thus

    images

    If we use this formula to calculate the derivative at times t = 0, 1,..., 5 we see in column 4 of Table 2.6 that values obtained are very close to the estimates obtained directly from the data, with only the t = 4 differing by an amount of 0.02. The advantage of having the formula is that we can calculate the derivative at any t value.

The solution to part b of Example 2 suggests that whenever P(t) has the general exponential form P(t) = abt, then the derivative has the same form but differing by some constant; that is, P′(t) = cP(t), where for Example 3 we obtained c = 0.026. This equation is an example of what is known as a differential equation, as it relates a function to its derivative. You will learn more about differential equations in Chapters 6 and 8. In Chapter 3, we will verify that the derivative of an exponential function is a constant multiple of the exponential function.

Notational alternatives

The primed-function notation f′ that we have been using to denote derivatives is but one of several used in various texts. Newton used a “dot” notation, which we will not consider here. The notation that mathematicians prefer was developed by Leibniz. This notation is inspired by the following presentation of the derivative.

Let Δx represent a small change in x. The change of y = f(x) over the interval [x, x + Δx] is given by

images

The average rate of change of y = f(x) over the interval [x, x + Δx] is given by

images

Hence, the derivative of f at x is

images

Leibniz represented this limit as

images

where in some sense dy corresponds to an “infinitesimal” change in y and dx represents an “infinitesimal” change in x. Commonly used variations in notation that you will find in this and other calculus texts include

images

To indicate the derivative at the point x = a using Leibniz notation, we use the cumbersome expression

images

Example 3 Using alternative derivative notations

Find the following derivatives.

images

Solution

  1. In Example 1, we found that the derivative of x3 is 3x2. Since images is the derivative of y = x3 evaluated at x = −1, we have

    images

  2. In Example 1, we guessed that the derivative of xn is nxn−1; in which case, for n = 5 we have

    images

The next example draws upon research undertaken by ecologist Nathan Sanders and colleagues, who assessed the number of local ant species along an elevational gradient, Kyle Canyon, in the Spring Mountains of Nevada to obtain a measure called species richness. These data, illustrated in Figure 2.58, are plotted in terms of number of species of ants as a function of elevation (in kilometers).

Example 4 Ant biodiversity

Ecologists, noting that the number of ant species declines at both low (close to sea level) and high (at the tops of mountains) levels, fitted a parabola to the data plotted in Figure 2.58. The fit they obtained is given by

images

where x is elevation measures in kilometers.

  1. Find images.
  2. Identify the units of images and interpret images.

images

Figure 2.58 Number of species of ants as a function of elevation.

Data Source: N. Sanders, J. Moss, and D. Wagner, “Patterns of Ant Species Richness along Elevational Gradients in an Arid Ecosystem,” Global Ecology and Biogeography, 12 (2003): 93–102.

Solution

images

b. The units of images are species per kilometer elevation. images represents the rate of change of species richness with respect to elevation. For elevations less than 24.9/15.4 ≈ 1.6 kilometers, images > 0. Consequently, for elevations of less than 1.6 kilometers, an ant-loving entomologist would encounter more species of ants by hiking higher up. However, for elevations greater than 1.6 kilometers, an ant-loving entomologist would encounter more species of ants by walking downward.

Mean Value Theorem

To understand what the derivative tells us about the shape of a function, we need the mean value theorem (MVT). The proof of this theorem is given as a series of challenging exercises in Problem set 2.7.

Theorem 2.7 Mean value theorem

Let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists c in (a, b) such that

images

images

Figure 2.59 Mean value theorem in action. The slope of the tangent line at x = c equals the slope of the secant line from x = a to x = b.

Notice that the right-hand side of this equation is the average rate of change of f over the interval [a, b]. Hence, the MVT states that for a differentiable function on an interval [a, b], there is a point in the interval where the instantaneous rate of change equals the average rate of change. Alternatively, we can think of the MVT in geometric terms. Recall that the right-hand side of the MVT equation is the slope of the secant line passing through the points (a, f(a)) and (b, f(b)). Hence, the MVT asserts that there is a point in the interval such that the slope of the tangent line at this point equals the slope of the secant line. A graphical representation of this interpretation is given in Figure 2.59.

Example 5 Mean value theorem in action

Determine whether the MVT applies for the following functions f on the specified intervals [a, b]. If the MVT applies, then find c in (a, b) such that the statement of the MVT holds.

  1. f(x) = x2 on the interval [0, 2]
  2. f(x) = |x| on the interval [−1, 1]

Solution

  1. Recall that f′(x) = 2x for all x. Hence, f is differentiable on the interval [0, 2]. Consequently, the MVT applies and we should be able to find the desired value c. The average rate of change of f on [0, 2] is given by

    images

    images

    Solving f′(x) = 2x = 2 yields x = 1. Hence, the instantaneous rate of change at x = 1 equals the average rate of change over the interval [0, 2]. The plot on the left with y = x2 in red, the tangent line in blue, and the black line connecting (0, f(0)) to (2, f(2)) illustrates our calculations.

  2. We need to find the derivative of f(x)= |x|. Since f(x) = x for x > 0, we get

    images

    whenever every h is sufficiently small but not equal to zero. Hence, f′(x) = 1 for x > 0. On the other hand, since f(x) = −x for x < 0, we get

    images

    whenever h is sufficiently small but not equal to zero. Hence, f′(x) = −1 for x < 0.

    What happens at the point x = 0? Our calculations imply that images = −1. Since the one-sidedd limits do not agree, f is not differentiable at x = 0 and the MVT need not apply. In fact, since the average rate of change over the interval [−1, 1] equals images = 0, there is no instantaneous rate of change of f that equals the average rate of change.

Example 6 Foraging for food

Hummingbirds (Figure 2.60) are small birds that weigh as little as three grams and have an energetically demanding lifestyle. With their wings beating at rates of eighty to a hundred beats per second, the hummingbird can lose 10% to 20% of its body weight in one to two hours. To survive, hummingbirds require relatively large amounts of nectar from flowers. Therefore, they spend much of the day flying between patches of flowers extracting nectar. As a hummingbird extracts nectar in a patch, its energetic gains E(t) in calories increase with time t (in seconds). Figure 2.61 shows a hypothetical graph of energetic gains E(t), in calories, in one patch.

images

Figure 2.60 Ruby-throated hummingbird (Archilochus colubris).

images

Figure 2.61 Energy gains over time.

  1. Approximate the average rate of energy intake over the interval [0, 60].
  2. Use the geometric interpretation of the mean value theorem to estimate the time when the instantaneous rate of energy intake equals the average rate of energy intake.

Solution

  1. Since E(0) = 0 and E(60) ≈ 1000, we obtain

    images

    calories per second.

  2. Graphing the line connecting the points (0, 0) and (60, 1000) yields the secant line whose slope equals the average rate of change:

    images

    The slope of this line is approximately 16.7. To estimate the time at which E′(t) = images, we can place a straightedge on top of the red line segment and slowly slide it upward keeping it parallel to the red segment. If we slide it upward until the straightedge is tangent to the curve y = E(t), then we obtain the following graph:

    images

    The blue segment shows the location of the tangent line at t ≈ 20 seconds. Hence, the instantaneous rate of energy intake rate equals the average energy intake rate at t ≈ 20.

It is worth noting from Figure 2.61 that E′(t) is above the average rate of energy intake for t < 20 and below the average for t > 20. Hence, as we explore in more detail in Chapter 4, the hummingbird may consider leaving the patch after 20 seconds.

Derivatives and graphs

Using the mean value theorem, we can prove the following two facts about the relationship of the sign of the derivative f′ to the graph of y = f(x).

Increasing-Decreasing

Let f be a function that is differentiable on the interval (a, b). If f′(x) > 0 for all x in (a, b), then f is increasing on (a, b). If f′(x) < 0 for all x in (a, b), then f is decreasing on (a, b).

To prove these properties, assume that f′ > 0 on (a, b). Take any two points x2 > x1 in the interval (a, b). By the mean value theorem, there exists a point c in the interval [x1, x2] such that

images

Since f′(c) > 0, we have

images

Since x2x1 > 0, we have f(x2) − f(x1) > 0. Equivalently, f(x2) > f(x1). Therefore, f is increasing on the interval [a, b]. The case of f′ < 0 on [a, b] can be proved similarly, and it appears as an exercise in Problem Set 2.7.

Example 7 Identifying signs of f

Let the graph of y = f(x) be given by Figure 2.62. For the interval [−3, 2], determine where the derivative of f is positive and where the derivative of f is negative.

Solution Since the graph is increasing on the intervals (−3, −2) and (0, 1), f′ > 0 on these intervals. Since the graph is decreasing on the intervals (−2, 0) and (1, 3), f′ < 0 on these intervals.

For a function y = f(x), a turning point is an x value where the function switches from increasing to decreasing, or vice versa. More precisely, if f is continuous on (a, b), then c in (a, b) is a turning point provided that either (i) f is increasing on (a, c) and decreasing on (c, b), or (ii) f is decreasing on (a, c) and increasing on (c, b). When f is differentiable on (a, b), turning points correspond to where the derivative f′ changes sign. For example, if f′(x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a turning point as the function switches from increasing to decreasing at x = c.

images

Figure 2.62 Graph of y = f(x).

Example 8 Mix and match

Match the graphs of y = f(x)

images

with the graph of their derivatives y = f′(x)

images

Solution

  1. Looking at graph (a), we see three turning points at approximately −0.6, 0, and 0.6. Turning points corresponds to x values where the derivative of function equals zero. Since only graph (iii) of the graphs (i)–(iii) intersects the x axis in three points, the derivative graph for (a) must be (iii).
  2. The turning points for graph (b) are at approximately −0.4 and 0.4; the graph of (i) shows the derivative to be 0 at those points. Hence, graph (i) must be the derivative of the graph of (b).
  3. There are no turning points on graph (c), so the derivative graph should not cross the x axis. Therefore, the derivative graph is (ii).

Example 9 Reconstructing f from f

Let the graph of y = f′(x) be given by the graph in Figure 2.63. Sketch a possible graph for y = f(x).

images

Figure 2.63 Graph of a derivative.

Solution We can sketch the graph of f by looking at the intervals for which the graph of f′(x) is positive or negative, as shown in Figure 2.64. On intervals where f′(x) is positive, we sketch a curve that is increasing, and on intervals where f′(x) is negative, we sketch a curve that is decreasing.

images

images

Figure 2.64 Construction of the graph of a function f given its derivative f′.

A possible graph of y = f(x) is shown on the left.

PROBLEM SET 2.7

Level 1 DRILL PROBLEMS

Use the definition of a derivative to find f′(x) for the functions in Problems 1 to 8.

images

Use the derivatives found in Problems 1 to 8 to find the values requested in Problems 9 to 16.

images

Find at what point the slope of the instantaneous rate of change equals the average rate of change over the specified intervals in Problems 17 to 22. Also, provide a sketch that illustrates this relationship.

17. f(x) = 8 over the interval [−5, 5]

18. f(x) = 3x − 2 over the interval [3, 4]

19. f(x) = −x2 over the interval [−1, 1]

20. f(x) = x + x2 over the interval [0, 1]

21. f(x) = images over the interval [1, 2]

22. f(x) = images over the interval [1, 4].

Mix and match the graphs in Problems 23 to 28 with the graphs labeled (A) to (F), which are the derivative graphs.

images

images

For each of the functions given in Problems 29 to 34, find intervals for which f is increasing and intervals for which f is decreasing.

29. f(x) = x2x + 1

30. f(x) = 5 − x2

31. f(x) = x3 + x

32. f(x) = 8 − x3

33. Let f be the function for which the graph of the derivative y = f′(x) = g(x) is given by

images

34. Let f be the function for which the graph of the derivative y = f′(x) = g(x) is given by

images

35. For the graph y = g(x) given in Problem 33, estimate all values of c in [−3, 0] such that

images

36. For the graph y = g(x) given in Problem 34, estimate all values of c in [−2, 2] such that

images

In each of Problems 37 to 40, the graph of a function f′ is given. Draw a possible graph of f.

images

Level 2 APPLIED AND THEORY PROBLEMS

41. Let f be differentiable on the interval (a, b). Use the mean value theorem to prove if f′ < 0 on [a, b], then f is decreasing on (a, b).

42. Rolle's theorem: Let f be differentiable on (a, b) and continuous on [a, b]. Assume f(a) = f(b) = 0. Without using the mean value theorem, argue that there exists a c in (a, b) such that f′(c) = 0.

43. Use Rolle's theorem to prove the mean value theorem.

44. A baseball is hit upward and its height at time t in seconds is given by

images

  1. Find the velocity of the baseball after t seconds.
  2. Find the time at which the velocity of the ball is 0.
  3. Find the height of the ball at which the velocity is 0.

45. To study the response of nerve fibers to a stimulus, a biologist models the sensitivity, S, of a particular group of fibers by the function

images

where t is the number of days since the excitation began.

  1. Over what time period is sensitivity increasing? When is it decreasing?
  2. Graph S′(t).

46. During the time period 1905–1940, hunters virtually wiped out all large predators on the Kaibab Plateau near the Grand Canyon in northern Arizona. The data for the deer population, P, over this period of time are as follows:

images

  1. Estimate P′(t) for 1905 ≤ t ≤ 1939.
  2. Graph and interpret P′(t).

47. In 1913, Carlson studied a growing culture of yeast (see Problem 45 in Problem Set 2.4 and Section 6.1). The table of population densities N(t) at one-hour intervals is shown here:

images

  1. Estimate N′(t) for 0 ≤ t ≤ 17.
  2. Graph N′(t) and briefly interpret this graph.

48. Our ruby-throated hummingbird has entered another patch of flowers, and the energy she is getting as a function of time in the patch is plotted below.

images

  1. Find the average energy intake rate.
  2. Estimate at what time the instantaneous energy intake rate equals the average intake rate.

49. Two radar patrol cars are located at fixed positions 6 miles apart on a long, straight road where the speed limit is 65 miles per hour. A sports car passes the first patrol car traveling at 60 miles per hour; then 5 minutes later, it passes the second patrol car going 65 miles per hour. Analyze this situation to show that at some time between the two clockings, the sports car exceeded the speed limit. Hint: Use the MVT.

50. Suppose two race cars begin at the same time and finish at the same time. Analyze this situation to show that at some point in the race they had the same speed.

CHAPTER 2 REVIEW QUESTIONS

  1. Find at what point the instantaneous rate of change of the function f(x) = x3x equals the average rate of change over the interval [−1, 2]. Provide a sketch that illustrates this relationship.
  2. In the 2012 Summer Olympics in London, swimmer Ye Shiwen swam the last 50 meters of her 400-meter individual medley final quicker than the winner of the men's race, Ryan Lochte. The 50-meter split times for both swimmers are shown in the following table:

    images

    1. Find the average velocity of both swimmers in the last split of the race.
    2. Find the average velocity of both swimmers over the entire race.
  3. Use the definition of derivative to find f′(x) for f(x) = images.
  4. Find the average rate of change of f(x) = x2 − 2x + 1 on [1, 3] and the instantaneous rate of change at x = 1.
  5. Consider f(x) = 9 − x2 and g(x) = ln x.
    1. Graph y = f(x) and y = g(x) on the same coordinate axes.
    2. Plot the point P(2, ln 2) on the graph of g. Graphically, estimate the position of the line tangent to g at the point P.
    3. Plot the point Q(2, 5) on the graph of f. Algebraically find the line tangent to f at the point Q. Use the equation of this tangent line to show that it “kisses” the graph of f at the point Q.
  6. Find images with the suggested methods.
    1. Graphically
    2. Using a table of values
    3. By algebraic simplification
    4. By using the informal definition of limit
    5. Using technology
  7. Let f(x) = images

    Find the requested limits.

    images

  8. Evaluate the sequential limits, if they exist.

    images

  9. The graph of a function f′ is given. Draw a possible graph of f.

    images

  10. An environmental study of a certain suburban community suggests that t years from now, the average level of carbon dioxide in the air can be modeled by the formula

    images

    parts per million.

    1. At what rate will the CO2 level be changing with respect to time one year from now?
    2. By how much will the CO2 level change in the first year?
    3. By how much will the CO2 level change over the next (second) year?
  11. Consider the function y = images
    1. Find the horizontal asymptote(s) of this function.
    2. Find the right-hand side and left-hand side limits of this function at all of its vertical asymptotes.
  12. The canopy height (in meters) of a tropical elephant grass (Pennisetum purpureum) is modeled by

    images

    where t is the number of days after mowing.

    1. Sketch the graph of h(t).
    2. Sketch the graph of h′(t).
    3. Approximately when was the canopy height growing most rapidly? Least rapidly?
  13. The concentration C(t) of a drug in a patient's blood stream is given by

    images

    1. Estimate C′(t) for t = 0, 0.1,... 0.9.
    2. Sketch C′(t) and interpret it.
  14. Suppose that systolic blood pressure of a patient t years old is modeled by

    images

    for 0 ≤ t ≤ 65, where P(t) is measured in millimeters of mercury.

    1. Sketch the graph of y = P(t).
    2. Using the graph in part a, sketch the graph of y = P′(t).
  15. Find the limit images and determine how positive x needs to be to ensure that images is within one millionth of the limiting value L.
  16. Whales have difficulty finding mates in the vast oceans of the world when their population numbers drop below a critical value. Thus, a model of the growth of whale populations from one whale generation to the next is going to be relatively more robust at intermediate whale densities than at low densities when finding mates is a problem, or at high densities when competition for food is a problem. The form of a hypothetical function f in the difference equation

    images

    that reflects the above properties is illustrated in Figure 2.65, where an is the density of the whales in generation n(units are whales per 1000 sq km).

    images

    Figure 2.65 A function, f, modeling the growth of a hypothetical whale population.

    Determine images an when a1 = 55. Justify your answer.

  17. Give a proof that the function f(x) = xex − 0.1 has at least one positive root.
  18. Sketch a graph of a function y = f(x) on the interval [−2, 2] such that f(−2) = f(0) = f(2) = 0, images > 0 on [−2, −1) and (1, 2], and images < 0 on (−1, 1).
  19. The average population growth rate of Mexico from 1981 to 1983 was 1.82 million per year and from 1983 to 1985 was 1.915 million per year. Assume the population size N(t) as a function of time and N′(t) are continuous. Prove that at some point in time between 1981 and 1985, the instantaneous rate of population growth was 1.85 million per year.
  20. Consider the graph of f shown in Figure 2.66. On the interval [−3, 4] find the points of discontinuity as well as places where the derivative does not exist. Explain your reasoning.

    images

    Figure 2.66 A function, f, on the interval [−3, 4].

GROUP PROJECTS

Seeing a project through on your own, or working in a small group to complete a project, teaches important skills. The following projects provide opportunities to develop such skills.

Project 2A A simple model of gene selection

One of the simplest problems in population genetics is to consider what happens to a particular version of a gene, where each version is referred to as an allele, that is being selected for or against because it confers some advantage or disadvantage to carriers of that allele. Examples of disadvantageous or deleterious alleles are those associated with genetic diseases such as sickle cell anemia, hemophilia, and Tay-Sachs disease. Most of our genes come in pairs of alleles, and if one allele in the pair is deleterious, then the effect of that allele may often be partially or fully masked by the other allele in the pair. If a person has a double dose of the deleterious allele, the disease is expressed in its severest form. If a person has a single dose of the deleterious allele (i.e., one normal and one deleterious allele) then, depending on the disease, a milder version of the disease may be expressed (partial masking), or the individual is completely healthy (full masking). In the latter case, the individual is said to be a carrier for the disease (e.g., hemophilia).

On the other hand, alleles may confer a strong advantage to an organism that carries them. For example, if an insect carries an allele of a particular gene that allows it to detoxify a pesticide or a virus carries an allele that allows it to neutralize an otherwise effective drug, then we say that these pests and pathogens carry alleles of genes that confer resistance to the chemicals that would otherwise control or kill them.

Sometimes individuals who carry two different alleles of a particular gene are better off than individuals who have two copies of the same allele, regardless of which allele it is Biologist call this condition heterozygous superiority, and it is associated with the phenomenon called hybrid vigor. For example, an individual human is going to be better at fighting disease if he or she has two different alleles at an immune system gene responsible for the production of antibodies that protect against invading pathogens, such as the influenza virus.

Population geneticists have devised a simple model that allows them to assess what happens to such alleles. The form of this model is pn+1 = f(pn), where pn represents the proportion of the allele in question in the population in the nth generation: If pn = 1, then every individual in the population has a double dose of the allele in question. If pn = 0, then no one has even a single dose of this allele. If pn = 0.5, some individuals do not have the allele, some have a single dose, and some a double dose of the allele, but the proportion in the population of this allele is 1/2.

In this simple allele proportion model, the specific form of f(p) is

images

where a ≥ 0 and b ≥ 0 are constants that determine whether the allele in question confers an overall advantage or disadvantage or is associated with heterozygote superiority.

In this project, investigate the value of the equilibria that arise for various combinations of a and b, paying particular attention to whether a or b is greater than or less than 1. Interpret the various cases in terms of the limiting values of several sequences of proportions pn that start at different values p1 satisfying 0 < p1 < 1. Also, describe how these cases correspond to classification of the alleles as advantageous, deleterious, or associated with heterozygote superiority. Find specific cases in the literature, or by searching the Web, to illustrate these three phenomena.

Project 2B Fibonacci rabbit growth when death is included

In the rabbit population growth process proposed by Fibonacci (Example 10 of Section 2.5), the assumption was that all the rabbits live forever. As an alternative, let's assume that only a proportion s of rabbits alive each month survive to the next month (independent of how old they are or what gender they are) and, of those that survive, only a proportion p of the females from the month before produce a litter that always consists of r male-female pairs. Investigate the growth of this population by carrying out the following tasks.

  1. Derive an equation for the rate at which the proportion of pairs increases from month to month as a function of the three population parameters 0 < s ≤ 1, 0 < p ≤ 1, and r a positive integer. Note that getting the correct equation can be a little tricky, so use a diagram similar to the one outlined in Chapter 1: see the Fibonacci images, that is, Problem 38 in Section 1.7. In particular, starting out with a suitable number of new-born pairs, draw diagrams for the cases (s, p, r) = (1/2, 1, 1), (1, 1/2, 1), and (1, 1, 2) and use these to construct a general expression for an in terms of an−1 and an−2 that contains the three parameters in question.
  2. What must hold true for the parameters s, r, and p to ensure the population size remains constant for all time for any initial population size? For the case r = 1, express s as a function of p such that the population is neither growing nor declining. Hence, in the square of the positive quadrant of the p-s plane defined by 0 ≤ p ≤ 1 and 0 ≤ s ≤ 1, shade all points where the rabbit population is growing and all points where it is declining. Hint: Write a difference equation for xn = an/an−1.
  3. Repeat exercise b for the case r = 2 and make a general statement about how the two shaded areas change as r increases.

* Catherine Brahic, “Carbon Emissions Rising Faster than Ever,” New Scientist.com news service 17 (November 10, 2006): 29.

* If you want to support this statement with a technical argument, suppose you want images ≥ 1,000,000. Taking the square root of both sides and cross-multiplying yields images ≥ |x − 2|. Hence, f(x) ≥ 1,000,000 whenever 0 < |x − 2| < images.

* T. Carlson, “Uber Geschwindigkeit und Größe der Hefevermehrung in Würze,” Biochem Z57 (1913): 313–334.

* A. S. Perelson, A. U. Neumann, M. Markowitz, J. M. Leonard, D. D. Ho, “HIV-1 Dynamics in Vivo: Virion Clearance Rate, Infected Cell Lifespan, and Viral Generation Time,” Science 271 (1996): 1582–1586; and A. S. Perelson and P. W. Nelson “Mathematical Analysis of HIV-1 Dynamics in Vivo,” SIAM Review 41 (1999): 3–44.

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