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 + 4g – h)( –2).
The function (f + 4g – h)(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 + 4g – h)( –2).
This is
NOT the
distributive
property—you’re 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
more detail about
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 undened
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).
Simplify the radical expressions.
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 can’t
add these yet
because they’re
not like radicals—
they don’t have
the same numbers
under the
radical symbol.
If you’re multiplying
two radicals with the
same index, you can
multiply the contents of
the radicals inside one
big radical sign with a
matching index.
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