Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
340
Note: Problems 15.33–15.34 refer to the function j(x) = 3(x + 4) – 1.
15.34 Verify that j(x) and the function j
–1
(x) generated in Problem 15.33 are
inverses.
Verify that j(j
–1
(x)) = j
–1
(j(x)) = x.
Because j(j
–1
(x)) = j
–1
(j(x)) = x, j(x) and j
–1
(x) are inverse functions.
Note: Problems 15.35–15.36 refer to the function f(x) = (x + 2)
2
.
15.35 Why does no inverse function exist for f(x)? Explain how to alter the domain
of the function such that an inverse does exist.
The graph of f(x), illustrated in Figure 15-6, fails the horizontal line test, so no
inverse function exists.
Figure 15-6: Any horizontal line above the x-axis intersects the graph of f(x) twice.
The domain
is the collection
of numbers you’re
allowed to plug into
a function. You can
plug any real number x
into the function
f(x) = (x + 2)
2
, so the
domain of f(x) is all
real numbers. Look at
Problems 16.9–16.20 to
practice identifying
the domain of a
function.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
341
However, if you limit the domain of the function, an inverse does exist. For
instance, if you consider only the portion of the function for which x ≥ –2,
the corresponding graph (illustrated in Figure 15-7) does pass the horizontal
line test, and therefore an inverse exists.
Figure 15-7: The portion of the graph of f(x) for which x ≥ –2 increases monotonically
and thus has an inverse.
Note: Problems 15.35–15.36 refer to the function f(x) = (x + 2)
2
.
15.36 Identify the inverse of the modified function identified in Problem 15.35.
According to Problem 15.35, the function f(x) = (x + 2)
2
has an inverse when
x ≥ –2. Rewrite f(x) as y.
y = (x + 2)
2
Follow the procedure outlined in Problem 15.33 to calculate the inverse:
Reverse x and y, solve the resulting equation for y, and then replace y with f
–1
(x).
Therefore, the inverse function is either or .
Notice that the graph of f(x) passes through point (0,4) in Figure 15-7.
Therefore, the graph of f
–1
(x) must pass through the point (4,0). Substitute x =
4 into both inverse function candidates and determine which results in f(4) = 0.
Because when , that is the inverse function.
The
function also has an
inverse for x ≤ –2.
When you
take the root
of both sides of
an equation and
the index of the root
is even, multiply the
side of the equation
that DOESN’T contain
the variable you’re
isolating by “±1.”
Reverse
x and y in a
point on the graph of
f(x) to get a point on
the graph of f
-1
(x).
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
342
15.37 Identify the inverse function of .
Rewrite g(x) as y to get . Reverse x and y in the equation and
solve for y.
Either or is the inverse function
of g(x). To determine which is correct, evaluate g(x) for an x-value (such as
x = 1) to produce a coordinate pair.
Because , it must follow that . Substitute
into both potential inverse functions to determine which produces the correct
value.
Because when , that is the inverse
function.
Normally,
x = 0 is the best
number to test the
function and nd its
inverse. That would
make g(0) = –4 in this
problem. Unfortunately,
BOTH possible inverse
functions equal 0 when
you plug –4 into them,
so that’s why you
have to go with
something else
(like x = 1).
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