Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
330
Note: Problems 15.15–15.17 refer to the following functions: and
.
15.17 Evaluate .
Evaluate r(1) and p(1) and then calculate the quotient.
Simplify the expression.
Composition of Functions
Plug one function into another
Note: Problems 15.18–15.19 refer to the functions f(x) = x
2
– 1 and g(x) = 3x.
15.18 Evaluate f(5).
Substitute x = 5 into f(x).
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
331
Note: Problems 15.18–15.19 refer to the functions f(x) = x
2
– 1 and g(x) = 3x.
15.19 Express f(g(x)) in terms of x.
Problem 15.18 requires you to substitute x = 5 into f(x). Here, you substitute g(x)
into f(x).
Recall that g(x) = 3x.
Note: Problems 15.20–15.21 refer to the functions and .
15.20 Express in terms of x.
The expression is equivalent to j(h(x)). Replace the x in 7x
2
– 3 with
h(x).
Substitute into the function.
Note: Problems 15.20–15.21 refer to the functions and .
15.21 Evaluate .
Note that . To evaluate the composite function, begin by
evaluating j(–2).
“Composing”
functions
means plugging
functions into each
other. In Problem 15.19,
you plug g(x) into f(x).
That can be written
f(g(x)), which is read
“f of g of x”, or
, which is
real “f circle
g of x.”
Keep the
letters in the
same order—j
comes rst in
and j(h(x)).
Composition of functions
is not necessarily
commutative, so there’s
no guarantee that
j(h(x)) and h(j(x))
are equal.
Squaring
a square root
cancels it out, leaving
only the radicand
x + 7 behind.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
332
Substitute j(–2) = 25 into h(j(–2)).
h(j(–2)) = h(25)
Evaluate h(25).
Simplify the radical expression.
Therefore, .
Note: Problems 15.22–15.24 refer to the following functions: f(x) = x
2
– 4x + 3, ,
h(x) = 2x – 1.
15.22 Express in terms of x and evaluate .
Note that ; to write the function in terms of x, substitute
g(x) into f(x).
Evaluate f(g(6)).
Note: Problems 15.22–15.24 refer to the following functions: f(x) = x
2
– 4x + 3, ,
h(x) = 2x – 1.
15.23 Evaluate f(g(h(5))).
Begin by evaluating h(5).
h(5) = 2(5) – 1 = 10 – 1 = 9
Substitute h(5) = 9 into the expression: f(g(h(5))) = f(g(9)). Evaluate g(9).
h(x) is plugged
into g(x), which is
plugged into f(x). Start
with the innermost
function (the function
inside the most
parentheses) and
work your way out.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
333
Substitute g(9) = 3 into the expression: f(g(9)) = f(3). Evaluate f(3).
Therefore, f(g(h(5))) = 0.
Note: Problems 15.22–15.24 refer to the following functions: f(x) = x
2
– 4x + 3, ,
h(x) = 2x – 1.
15.24 Express f(h(g(x))) in terms of x. Assume x > 0.
Begin by substituting g(x) into h(x).
Substitute into f(h(g(x))).
15.25 If r(s(x)) = (x
3
– 1)
2
and r(x) = x
2
, then express s(r(x)) in terms of x.
The function r(s(x)) is defined the expression x
3
– 1 squared, and r(x) = x
2
, so
s(x) = x
3
– 1.
You don’t
need to put
this x in absolute
values because the
problem says to assume
x is positive (x > 0).
Otherwise,
.
In the
function r(s(x)),
s(x) is plugged into
r(x). You know that
r(x) = x
2
, so anything
you plug into r(x) gets
squared. To gure
out s(x), ask yourself
this question: “What
expression, when
squared, is equal
to (x
3
– 1)
2
?”. The
answer is s(x) =
x
3
– 1.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
334
15.26 If and , evaluate g(g(7)).
Substitute g(x) into f(x) to generate f(g(x)).
If and , then .
Solve the equation for g(x) by squaring both sides of the equation to
eliminate the radicals.
Rather than expressing g(x) as a fraction containing a sum, express it as a sum
of fractions.
To evaluate g(g(7)), begin by substituting x = 7 into g(x).
Therefore, .
This is true
because of the
transitive property,
which says that two
things equal to the
same thing are equal
to each other.
and
mean the
same thing:
.
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