Composition of Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A store takes $15 off a jacket, then adds 8% sales tax. Another store adds the tax first and then takes $15 off. Each store applies two functions, one after the other, but the totals differ by $1.20. Chaining functions is called composition, and the order you chain them in matters.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $g(x) = x^2 - 1$ , what is $g(3)$ ?

$8$
$5$
$4$
$9$

Show the answer

Replace $x$ by $3$ : $3^2 - 1 = 8$ .

Correct: $8$ :
$5$ :
$4$ :
$9$ :

2. How is the graph of $y = \sqrt{x - 2}$ related to the graph of $y = \sqrt{x}$ ?

Shifted 2 units right
Shifted 2 units left
Shifted 2 units down
Stretched by a factor of 2

Show the answer

Subtracting $2$ from the input moves the graph $2$ units right.

Correct: Shifted 2 units right:
Shifted 2 units left:
Shifted 2 units down:
Stretched by a factor of 2:

Part 4 · See it

See it first

Graphs of f(x) = square root of x and f(g(x)) = square root of (x - 2): the composite is f moved 2 units right, so (4, 2) moves to (6, 2). — With $g(x) = x - 2$ inside, $f(g(x)) = \sqrt{x - 2}$ : each point of $f(x) = \sqrt{x}$ moves $2$ units right, and the domain becomes $x \ge 2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Start with an input $x$ the inner function $g$ turns it into $g(x)$
Use $g(x)$ as the input of the outer function $f$ the output is $f(g(x))$ , also written $(f \circ g)(x)$
Change the order to $g(f(x))$ the output usually changes: composition is not commutative
An input $x$ gives a $g(x)$ that $f$ cannot takethat $x$ is left out of the domain of $f(g(x))$

Part 6 · Key ideas

Key ideas

$f(g(x))$ means: find $g(x)$ first, then put that output into $f$ . Work from the inside out and write the inner value down.
Order matters. $f(g(x))$ and $g(f(x))$ are usually different functions.
The domain of $f(g(x))$ : inputs $x$ in the domain of $g$ whose outputs $g(x)$ are in the domain of $f$ .
The identity function $I(x) = x$ changes nothing: $f(I(x)) = I(f(x)) = f(x)$ , as adding $0$ or multiplying by $1$ .
Composites work in every form: read a table or graph twice, or substitute one formula into the other.

Part 7 · Misconception

A common mistake

The wrong idea: To find $f(g(2))$ , find $f(2)$ first and then put it into $g$ .

What actually happens: Work from the inside out: $g(2)$ first, then use that output as the input of $f$ . Show the inner value: $g(2) = 0$ , so $f(g(2)) = f(0) = 4$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ and $g$

Selected values of the functions f and g
$x$	-1	0	1	2	3	4
$f(x)$	2	4	1	3	2	-1
$g(x)$	3	2	4	0	1	-1

1. Using the table, what is the value of $f(g(2))$ ?

$4$
$1$
$0$
$3$

Show the answer

Inside first: $g(2) = 0$ . Then $f(g(2)) = f(0) = 4$ .

Correct: $4$ : Right: the inner value is $g(2) = 0$ , and $f(0) = 4$ .
$1$ : This is $g(f(2)) = g(3) = 1$ : the functions were used in the reverse order.
$0$ : This is only the inner value $g(2) = 0$ ; it still has to go into $f$ .
$3$ : This is $f(2) = 3$ ; the input $2$ must go through $g$ first.

2. Using the table, find $g(f(-1))$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Inside first: $f(-1) = 2$ . Then $g(f(-1)) = g(2) = 0$ .

Answer: 0

3. Using the table, for which values of $x$ is $f(g(x)) = 2$ ? Select every value that works.

$x = -1$
$x = 0$
$x = 3$
$x = 4$

Show the answer

$f$ gives $2$ for two inputs, $-1$ and $3$ . So $g(x)$ must be $-1$ or $3$ : $g(4) = -1$ and $g(-1) = 3$ . Both $x = -1$ and $x = 4$ work.

Correct: $x = -1$ : $g(-1) = 3$ and $f(3) = 2$ , so $f(g(-1)) = 2$ .
$x = 0$ : $g(0) = 2$ and $f(2) = 3$ , so $f(g(0)) = 3$ .
$x = 3$ : $f(3) = 2$ , but the input $3$ goes into $g$ first: $f(g(3)) = f(1) = 1$ .
Correct: $x = 4$ : $g(4) = -1$ and $f(-1) = 2$ , so $f(g(4)) = 2$ .

Graph

The graphs of $f$ and $g$

The line is

f(x) = 2x - 1

and the parabola is

g(x) = (x - 1)^2 - 2

.

4. Use the graphs to find $g(f(1))$ .

$-2$
$-5$
$1$
$-1$

Show the answer

Read $f(1) = 1$ on the line, then $g(1) = -2$ at the vertex of the parabola. So $g(f(1)) = -2$ .

Correct: $-2$ : Inside first: $f(1) = 1$ , then $g(1) = -2$ .
$-5$ : This is $f(g(1)) = f(-2) = -5$ , the reverse order.
$1$ : This is the inner value $f(1) = 1$ ; it still goes into $g$ .
$-1$ : This adds $f(1) + g(1) = 1 + (-2)$ ; a composite feeds one output into the other function.

5. Use the graphs to find $f(g(3))$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Read $g(3) = 2$ on the parabola, then $f(2) = 3$ on the line. So $f(g(3)) = 3$ .

Answer: 3

6. Let $f(x) = \sqrt{x}$ and $g(x) = 6 - 2x$ . What is the domain of $f(g(x))$ ?

$x \le 3$
$x \ge 3$
$x \ge 0$
$x \le 6$

Show the answer

$f(g(x)) = \sqrt{6 - 2x}$ needs $6 - 2x \ge 0$ , so $2x \le 6$ and $x \le 3$ .

Correct: $x \le 3$ : The inner output $6 - 2x$ must be in the domain of $f$ : $6 - 2x \ge 0$ gives $x \le 3$ .
$x \ge 3$ : Dividing $-2x \ge -6$ by $-2$ flips the inequality: $x \le 3$ .
$x \ge 0$ : That is the domain of $f$ alone. The input of $f$ is $g(x)$ , so $g(x) \ge 0$ is the condition.
$x \le 6$ : The $2$ in front of $x$ was dropped: $6 - 2x \ge 0$ gives $x \le 3$ .

7. The function $h(x) = (3x + 1)^4$ can be written as $h(x) = f(g(x))$ . Which pair of functions works?

$f(x) = x^4$ and $g(x) = 3x + 1$
$f(x) = 3x + 1$ and $g(x) = x^4$
$f(x) = (3x)^4$ and $g(x) = x + 1$
$f(x) = x^4 + 1$ and $g(x) = 3x$

Show the answer

The inner step builds $3x + 1$ , the outer step raises it to the fourth power: $f(g(x)) = (3x + 1)^4$ .

Correct: $f(x) = x^4$ and $g(x) = 3x + 1$ : $f(g(x)) = (3x + 1)^4$ : $g$ does the first step and $f$ the last.
$f(x) = 3x + 1$ and $g(x) = x^4$ : Reversed: $f(g(x)) = 3x^4 + 1$ .
$f(x) = (3x)^4$ and $g(x) = x + 1$ : $f(g(x))$ $= (3(x + 1))^4$ $= (3x + 3)^4$ .
$f(x) = x^4 + 1$ and $g(x) = 3x$ : $f(g(x)) = (3x)^4 + 1$ : the $+1$ happens after the power.

Part 9 · Summary

Summary

A composite function chains two functions:

f(g(x))

puts the output of

g

into

f

. Evaluate from the inside out and show the inner value. The order matters, the domain is the inputs whose inner outputs

f

can take, and the identity function

I(x) = x

leaves any function unchanged. You can compose from tables, graphs or formulas, and take a function apart into an inner and an outer step.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff and gg

The graphs of ff and gg

Summary

What comes next

Connections

Values of $f$ and $g$

The graphs of $f$ and $g$