Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
343
Piecewise-Defined Functions
Function rules that change based on the x-input
15.38 Given the piecewise-defined function f(x) below, evaluate f(–2), f(0), and f(6).
A piecewise-defined function consists of two or more individual functions (in
this case g(x), h(x), and j(x)), each of which represents values of the function
for specific values of x. Here, f(x) = g(x) when x < 0. For instance, when x = –2,
f(–2) = g(–2). When x = 0, f(x) = h(x); therefore, f(0) = h(0). Finally, f(x) = j(x)
when x > 0, so f(6) = j(6).
Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) below.
15.39 Evaluate k(–3).
Note the three domain intervals defined for k(x): x ≤ –3, –3 < x < 1, and x ≥ 1.
The only domain interval that contains x = –3 is x ≤ –3. Therefore, k(x) = 3x + 4
when x = –3.
Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) below.
15.40 Evaluate k(1).
When x ≥ 1, k(x) = 5. Because 1 ≥ 1, k(1) = 5.
f(x) changes
based on INPUT
values. The little
statements right of
the functions (x ‹ 0,
x = 0, and x › 0) tell
you which function
to plug x into.
–3 is less than OR
equal to –3; 3 is not
less than itself, but it is
equal to itself, and only
one of those two cases
has to be true.
No matter
what x-value you
plug in that’s greater
than or equal to 1,
the output will
be 5.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
344
Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) here.
15.41 Graph k(x).
Use the techniques outlined in Chapter 5, “Graphing Linear Equations in Two
Variables,” to construct the graphs of y = 3x + 4, , and y = 5. Figure 15-8
illustrates the graphs as dotted lines.
Figure 15-8: The graphs of the individual functions that comprise k(x). They are
drawn as dotted lines because only specific pieces of each graph (to be
identified in later steps) contribute to the graph of k(x).
Darken the portions of each graph that represent portions of k(x). In other
words, darken the interval of y = 3x + 4 for which x ≤ –3, the portion of the
graph left of the vertical line x = –3. The inequality symbol in the statement
x ≤ 3 indicates that a closed point should be used at the endpoint x = –3, as
illustrated in Figure 15-9.
Similarly, darken the portion of y = 5 for which x ≥ 1 and indicate a closed
endpoint. Finally, darken the portion of between x = –3 and x = 1. The
corresponding inequality statement (–3 < x < 1) indicates that the interval has
open endpoints. The graphs are illustrated in Figure 15-9.
Remember, ≤
and ≥ mean solid
dots on the graph,
and < and > mean
hollow dots.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
345
Figure 15-9: The darkened interval of each graph represents a portion of the graph
of k(x).
To graph k(x), draw the darkened portions of each linear graph in Figure 15-9
on a single coordinate plane, as illustrated by Figure 15-10.
Figure 15-10: The graph of k(x).
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
346
Note: Problems 15.42–15.43 refer to the relation f(x) defined below.
15.42 Evaluate f(–1).
When x ≤ 2, f(x) = 3x
2
– 6x + 5.
Note: Problems 15.42–15.43 refer to the relation f(x) defined below.
15.43 For what value of a is f(x) is a function?
Substituting any real number x into f(x) produces a single output except when
x = 2. The domain restrictions in this function are x ≤ 2 and x ≥ 2; the value
x = 2 satisfies both conditions, so x = 2 can either be substituted into 3x
2
– 6x + 5
or . Therefore, both expressions must be equal when x = 2 to ensure
that f(x) is a function.
Substitute x = 2 into the equation and solve for a.
If f(x) is
a function,
each input can
only have one
output. That means
plugging x = 2 into f(x)
has to produce ONE
real number output.
Unfortunately, two
different rules apply
to x = 2, so both of
them have to equal
the same thing
when you plug in
x = 2.
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