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Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
326
Note: Problems 15.7–15.8 refer to the relation s(x) = 5 ± x.
15.8 Is s(x) a function? Why or why not?
According to Problem 15.7, s(–1) = 4 or s(–1) = 6. For a relation to be a
function, each value substituted into the relation must produce only one result.
In this case, substituting x = –1 into s(x) produces two results (4 and 6), so s(x)
is not a function.
Operations on Functions
+, –,
˙
, and ÷ functions
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x 4, g(x) = 3x + 2, and
h(x) = 2x
2
– 13x + 20.
15.9 Express (f + g)(x) in terms of x.
The function (f + g)(x) represents the sum f(x) + g(x).
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x 4, g(x) = 3x + 2, and
h(x) = 2x
2
– 13x + 20.
15.10 Evaluate (f + 4gh)( –2).
The function (f + 4gh)(x) represents the sum f(x) + 4g(x) – h(x). Evaluate
f(x), g(x), and h(x) for x = –2.
Use the values of f(–2), g(–2), and h(–2) to calculate (f + 4gh)( –2).
This is
NOT the
distributive
property—youre not
distributing x through
(f + g) to get f(x) + g(x).
It’s just notation:
(f + g)(x) = f(x) + g(x),
(f – g)(x) = f(x) – g(x),
(fg)(x) = f(x)
˙
g(x), and
.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
327
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x 4, g(x) = 3x + 2, and
h(x) = 2x
2
– 13x + 20.
15.11 Evaluate (fg)(6).
The function (fg)(x) is deﬁned as the product f(x) · g(x). Therefore,
(fg)(6) = f(6) · g(6). Evaluate f(x) and g(x) for x = 6.
Use the values of f(6) and g(6) to calculate (fg)(6).
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and
h(x) = 2x
2
13x + 20.
15.12 Express (fg)(x) in terms of x and evaluate that expression for x = 6 to verify the
solution to Problem 15.11.
The function (fg)(x) is deﬁned as the product f(x) · g(x). Calculate the product.
Evaluate the expression for x = 6.
To calculate
(fg)(6), you can
either plug 6 into
f(x) and g(x) and then
multiply those results (like
Problem 15.11) or multiply
f(x) and g(x) rst and
THEN plug in x = 6 (like
Problem 15.12).
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
328
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x 4, g(x) = 3x + 2, and
h(x) = 2x
2
– 13x + 20.
15.13 Express in terms of x.
The function is deﬁned as the quotient of h(x) and f(x).
Reduce the fraction to lowest terms by factoring the numerator.
Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and
h(x) = 2x
2
– 13x + 20.
15.14 Evaluate .
Substitute x = –1 into g(x) and h(x).
Calculate the quotient of h(1) and g(–1).
Chapters
20 and 21
go into a lot
rational expressions
(fractions that contain
polynomials), including
how to simplify
them.
The original
denominator of
the fraction was
x – 4, so when x = 4 the
denominator is 4 – 4 = 0.
Dividing by 0 is not
allowed. Sure you can
simplify the fraction to
get 2x – 5, but
is STILL undened
at x = 4.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
329
Note: Problems 15.15–15.17 refer to the following functions: and
.
15.15 Evaluate (p + r)(4).
Calculate the sum p(4) + r(4).
Therefore, .
Note: Problems 15.15–15.17 refer to the following functions: and
.
15.16 Express (pr)(x) in terms of x and evaluate (pr)(–2).
Multiply functions p(x) and r(x).
Evaluate (pr)(–2) by substituting x = –2 into (pr)(x).
You cant
because they’re
they dont have
the same numbers
under the