Topic 2.7B

Composition of Functions

In 2.7A we evaluated compositions one number at a time: first the inside value, then the outside function. That works, but if you need $f(g(x))$ for many inputs, it is better to build one formula that does both steps at once. This note builds those formulas, finds where they are defined, and runs the process in reverse by breaking a complicated function into simpler pieces.

9 MIN READ6 IDEAS33 PROBLEMS10 flashcards

Practice 33 problems Next: Chapter 2.8 Inverse Functions

Building a Formula for f(g(x))

CONCEPT

Substitute the Whole Inside Function

To find $f(g(x))$ , take the formula for $f$ and replace every $x$ with the entire expression $g(x)$ , in parentheses.

Then simplify. The result is a single function that does $g$ and then $f$ .

Worked example

With $f(x) = x^{2} - 3$ and $g(x) = 2x + 1$ from 2.7A, find formulas for $f(g(x))$ and $g(f(x))$ .

01
$f(g(x))$ : replace $x$ in $f$ with $(2x + 1)$ :
$f\big(g(x)\big) = (2x + 1)^{2} - 3 = 4x^{2} + 4x - 2$
02
$g(f(x))$ : replace $x$ in $g$ with ( $x^{2}$ − 3):
$g\big(f(x)\big) = 2\,(x^{2} - 3) + 1 = 2x^{2} - 5$
03
Check against 2.7A: $4(2)^{2} + 4(2) - 2 = 22 = f(g(2))$ ✓, and $2(2)^{2} - 5 = 3 = g(f(2))$ ✓. Two different formulas, as expected, because order matters.
Remember from 2.4: Nova's weekly model was exactly this. With $T(w) = w/4$ and $N(t) = 256(1.5)^{t}$ , $N(T(w)) = 256(1.5)^{w/4}$ .

COMMON MISTAKE

“ $f(g(x)) = x^{2} - 3$ with $2x + 1$ stuck on: $2x + 1^{2} - 3$ .”

The whole inside function must go in parentheses before squaring: $(2x + 1)^{2}$ , not $2x + 1^{2}$ . Without parentheses only the 1 gets squared. Check with $x = 2$ : the wrong version gives 2, but $f(g(2))$ is 22.

The Domain of a Composition

CONCEPT

Two Conditions

$x$ is in the domain of $f(g(x))$ exactly when $(1) x$ is in the domain of $g$ , AND (2) the value $g(x)$ is in the domain of $f$ .

Remember from 2.7A: $g(f(5))$ was undefined because $f(5) = -1$ was outside g's domain. The same thing happens with formulas.

Worked example

Example 1. $f(x) = \sqrt{x}$ and $g(x) = 5 - x$ . Find the domains of $f(g(x))$ and $g(f(x))$ .

01
$f(g(x))$ = √( $5 - x$ ). The square root needs $5 - x \ge 0$ , so $x \le 5$ . (For example, $x = 5$ gives 0 and $x = 1$ gives 2, but $x = 6$ gives $\sqrt{-1}$ , not real.)
02
$g(f(x)) = 5 - \sqrt{x}$ . Here the inside function $\sqrt{x}$ needs $x \ge 0$ first. After that, 5 − (anything) is fine. Domain: $x \ge 0$ .
03
Same two functions, opposite orders, different domains: $x \le 5$ versus $x \ge 0$ .

Worked example

Example 2. $f(x) = 1/(x - 3)$ and $g(x) = x^{2} - 1$ . Find $f(g(x))$ and its domain.

01
Substitute:
$f\big(g(x)\big) = \frac{1}{(x^{2} - 1) - 3} = \frac{1}{x^{2} - 4}, \qquad x \ne 2,\; x \ne -2$
02
$f$ can't take the input 3, so exclude every $x$ with $g(x) = 3$ : $x^{2} - 1 = 3$ gives $x^{2} = 4$ , so $x = 2$ or $x = -2$ . Remember from 2.1 and 2.5A: an even power has two roots, and here BOTH must be excluded.
03
Domain: all real $x$ except 2 and −2. Check a value that works: $f(g(3)) = 1/(9 - 4) = 1/5$ .
A hidden restriction. Let $f(x) = x^{2}$ and $g(x) = \sqrt{x}$ . Then $f(g(x)) = (\sqrt{x})^{2}$ , which simplifies to $x$ . But the simplified formula hides a restriction: $g$ only accepts $x \ge 0$ , so $f(g(x))$ is defined only for $x \ge 0$ .

The formulas look the same after simplifying, but the composition keeps the domain of the inside function.

COMMON MISTAKE

“ $(\sqrt{x})^{2} = x$ , so the domain of $f(g(x))$ is all real numbers.”

Find the domain BEFORE simplifying. The inside function $\sqrt{x}$ already throws out every negative input, and simplifying afterward can't bring them back. (Compare √( $x^{2}$ ): at $x = -3$ it gives 3, not −3, so it is not $x$ either.)

Quick check

Let $f(x) = \sqrt{x}$ and $g(x) = 5 - x$ . What is the domain of $f(g(x))$ ?

Decomposing a Function

Sometimes you need to go backwards: given one function, find an inside function $g$ and an outside function $f$ so that $h(x) = f(g(x))$ . Ask yourself: what is calculated FIRST, and what is done to it?

h(x)	inside g(x)	outside f(x)
$(3x - 1)^{4}$	3x − 1	$x^{4}$
$2^{x^{2}}$	$x^{2}$	$2^{x}$
√( $x^{2}$ + 9)	$x^{2}$ + 9	√x
$256(1.5)^{w/4}$	w/4	$256(1.5)^{w}$

There is usually more than one correct decomposition. For $(3x - 1)^{4}$ you could also use $g(x) = (3x - 1)^{2}$ and $f(x) = x^{2}$ . What doesn't help is the trivial split $f(x) = x$ , $g(x) = h(x)$ : it is correct, but it says nothing.

Worked example

Example 3. Decompose $h(x) = \sqrt{2x + 5}$ and $k(x) = 5^{x^{2} - 1}$ .

01
Ask: if I plug in a number, what do I calculate FIRST? For $h$ at $x = 2$ : first $2(2) + 5 = 9$ , then $\sqrt{9} = 3$ . So the inside is $g(x) = 2x + 5$ and the outside is $f(x) = \sqrt{x}$ .
02
For $k$ at $x = 2$ : first $2^{2} - 1 = 3$ , then $5^{3} = 125$ . Inside $g(x) = x^{2} - 1$ , outside $f(x) = 5^{x}$ .
03
Check by composing back: $f(g(x))$ = √( $2x + 5$ ) ✓ and $5^{x^{2} - 1}$ ✓.

Quick check

Write $h(x) = (x^2 - 9)^4$ as $f(g(x))$ with $g$ the inner function.

Transformations Are Compositions

CONCEPT

Composing With a Linear Function

A change to the OUTPUT happens after $f$ , so it is a linear function on the outside: $2f(x) - 3 = L(f(x))$ with $L(x) = 2x - 3$ .

A change to the INPUT happens before $f$ , so it is a linear function on the inside: $f(x + 2) = f(S(x))$ with $S(x) = x + 2$ , and $f(3x) = f(M(x))$ with $M(x) = 3x$ .

The identity function $I(x) = x$ changes nothing: $f(I(x)) = I(f(x)) = f(x)$ .

Remember from 2.4: “If Nova had launched 2 months earlier” was $N(t + 2) = N(S(t))$ , an input change, and it equaled $576(1.5)^{t}$ , which is an output change (a vertical stretch). The same function can be seen as either composition.

In 2.8 you will meet pairs of functions that undo each other, so that $f(g(x)) = x$ : composition with the result being the identity.

Worked example

Example 4. Let $f(x) = x^{2}$ . Write $g(x) = 3f(x - 2) + 1$ as a composition, and find its formula.

01
Inside (acts first on x): $S(x) = x - 2$ , a shift right by 2.
02
Outside (acts on the output of f): $L(x) = 3x + 1$ , a vertical stretch by 3 and a shift up 1. So $g(x) = L(f(S(x)))$ : three functions in a row.
03
Formula: $g(x) = 3(x - 2)^{2} + 1 = 3 x^{2} - 12x + 13$ . Check: $g(2) = 1$ (the vertex moved to $(2, 1)$ ) and $g(3) = 3(1) + 1 = 4$ .

Worked example

Try it yourself. (a) $f(x) = x + 4$ , $g(x) = x^{2}$ . Find $f(g(x))$ and $g(f(x))$ , and check both at $x = 1$ . (b) $f(x) = \sqrt{x}$ , $g(x) = x - 7$ . Find $f(g(x))$ and its domain. (c) Decompose $h(x) = (x + 5)^{3}$ .

Answers: (a) $f(g(x)) = x^{2} + 4$ and $g(f(x)) = (x + 4)^{2} = x^{2} + 8x + 16$ . At $x = 1$ : 5 and 25. (b) $\sqrt{x - 7}$ , domain $x \ge 7$ . (c) $g(x) = x + 5$ , $f(x) = x^{3}$ .

Practice

Worked example

P1. $f(x) = 3x$ and $g(x) = x - 4$ . Find $f(g(x))$ and $g(f(x))$ .

Answer: $f(g(x)) = 3(x - 4) = 3x - 12$ . $g(f(x)) = 3x - 4$ .

Worked example

P2. $f(x) = 1/x$ and $g(x) = x - 6$ . Find $f(g(x))$ and its domain.

Answer: $f(g(x)) = 1/(x - 6)$ . $f$ can't take 0, so exclude $x$ with $x - 6 = 0$ : all real $x$ except 6.

Worked example

P3. Decompose $h(x) = e^{2x + 1}$ into $f(g(x))$ .

Answer: $g(x) = 2x + 1$ (inside) and $f(x) = e^{x}$ (outside).

Worked example

P4. $f(x) = \sqrt{x}$ and $g(x) = x^{2} - 4$ . Find $f(g(x))$ and its domain.

Answer: $f(g(x))$ = √( $x^{2}$ − 4). Need $x^{2} - 4 \ge 0$ . The boundary $x^{2} = 4$ has two roots, $x = \pm 2$ , so the domain is $x \le -2$ or $x \ge 2$ . ( $x = 1$ fails: $\sqrt{-3}$ is not real; $x = 3$ and $x = -3$ both give $\sqrt{5}$ .)

Common slips

“ $f(g(x)) = x^{2} - 3$ with $2x + 1$ stuck on: $2x + 1^{2} - 3$ .”

The whole inside function must go in parentheses before squaring: $(2x + 1)^{2}$ , not $2x + 1^{2}$ . Without parentheses only the 1 gets squared. Check with $x = 2$ : the wrong version gives 2, but $f(g(2))$ is 22.
“ $(\sqrt{x})^{2} = x$ , so the domain of $f(g(x))$ is all real numbers.”

Find the domain before simplifying. The inside function $\sqrt{x}$ already throws out every negative input, and simplifying afterward can't bring them back. (Compare √( $x^{2}$ ): at $x = -3$ it gives 3, not −3, so it is not $x$ either.)

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10 cards · Composition and inverses

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Recap card

6 lines to re-read the night before.

01
To build $f(g(x))$ , substitute the whole expression $g(x)$ , in parentheses, for $x$ in $f$ . Then simplify.
02
Domain of $f(g(x))$ : $x$ must be in the domain of $g$ And $g(x)$ must be in the domain of $f$ . Find it before simplifying.
03
When excluding values, solve $g(x)$ = (bad input) completely; an even power gives two values to exclude.
04
Decomposing: the inside function is what is computed first. Several decompositions can be correct.
05
Output changes are compositions on the outside, input changes on the inside. $I(x) = x$ changes nothing.
06
Next, in 2.8: inverse functions, functions that undo each other, so that $f(g(x)) = x$ .

Practice 33 problems Next: Chapter 2.8 Inverse Functions

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Composition of Functions

Building a Formula for f(g(x))

The Domain of a Composition

Decomposing a Function

Transformations Are Compositions

Practice

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