Mathematical-Reasoning · Unit 23 · Video 3 · Interactive Practice

Decomposing a Function: Cut at the Last Step

IKey Formulas

Formula	Name	What it says
$(g \circ h)(x) = g\big(h(x)\big)$	Composite	$h$ (inner) acts first, $g$ (outer) acts second
$(3x + 2)^3 = g\big(h(x)\big)$ with $h(x) = 3x + 2$ , $g(x) = x^3$	Cut at the last step	Last step of the verbal description is the outer function, earlier steps form the inner one
$h: A \to B$ and $g: B \to C$	The sets fit	Codomain of the inner function equals the domain of the outer one, e.g. $f: \mathbb{R} \to \mathbb{R}^+$ , $h: \mathbb{R}^+ \to \mathbb{R}$ for $\ln(x^2 + 3)$
$(g \circ h)'(x) = g'\big(h(x)\big) \cdot h'(x)$	Chain rule	Uses exactly this split: $f'(x) = 3(3x + 2)^2 \cdot 3 = 9(3x + 2)^2$

Key Insight: Decomposing takes one step and two checks: cut at the last step, make the inner codomain equal the outer domain, and give the composite the original domain and codomain.

IIWhere to Cut

Every cut of the step chain gives a correct composite, but only the last-step cut leaves a one-operation outer function.

Function

IIIMake the Sets Fit

A decomposition is complete only when the inner codomain is the outer domain and the composite keeps the original sets.

Function Inner domain Inner codomain Outer domain Outer codomain

💡 Here $\mathbb{R}^+$ is the set of positive real numbers and $\mathbb{N} = \{1, 2, 3, \ldots\}$ . Each example has exactly one choice of the four sets that works.

IVQuiz Questions

Problem 1 · Cut at the Last Step

Given: $F: \mathbb{R} \to \mathbb{R}$ , $F(x) = \sqrt{x^2 + 7}$ . Which decomposition $F = g \circ h$ cuts at the last step of the verbal description?

✅ Correct! The last step is the square root, so the outer function is

g(x) = \sqrt{x}

and everything before it,

x^2 + 7

, is the inner function.

❌ Not the last step.

g(h(x)) = \sqrt{x^2 + 7}

is right, but the outer function

\sqrt{x + 7}

still takes two steps (add

7

, then take the square root).

❌ No progress. This composite is correct, but the inner function is the identity and the outer function is the whole formula.

❌ Order reversed.

g(h(x)) = (\sqrt{x})^2 + 7 = x + 7

for

x \ge 0

, which is not

F(x)

. Also

\sqrt{x}

is undefined for negative

x

Show solution

Verbal description of $F$ : choose $x$ , square it, add $7$ , take the square root. The last step is the square root, so

g(x) = \sqrt{x}, \qquad h(x) = x^2 + 7.

Check: $g(h(x)) = g(x^2 + 7) = \sqrt{x^2 + 7} = F(x)$ .

Sets: $x^2 + 7 \ge 7 > 0$ for every real $x$ , so take $h: \mathbb{R} \to \mathbb{R}^+$ and $g: \mathbb{R}^+ \to \mathbb{R}$ . Then $g \circ h: \mathbb{R} \to \mathbb{R}$ , as $F$ requires.

Options (a) and (b) are also correct composites, but their outer functions are not a single basic operation. Option (d) is not even a decomposition of $F$ .

Problem 2 · Which Function Goes on the Left?

Given: $F: \mathbb{R} \to \mathbb{R}$ , $F(x) = \sin(x^3)$ . Which pair gives $F = g \circ h$ ?

✅ Correct!

g(h(x)) = g(x^3) = \sin(x^3) = F(x)

. The sine is applied last, so it is the outer function, written on the left.

❌ The order is swapped.

g(h(x)) = (\sin x)^3

, which cubes last. In

F

the cube is taken first and the sine last, so the sine is the outer function

g

❌ Not quite.

g(h(x)) = \sin(3x)

. The inner step of

F

cubes

x

rather than multiplying it by

3

❌ Cubed twice.

g(h(x)) = \sin\big((x^3)^3\big) = \sin(x^9)

. The outer function must be only the last step.

Show solution

Verbal description of $F$ : choose $x$ , cube it, take the sine. The last step is the sine, so the outer function is $g(x) = \sin x$ and the inner function is $h(x) = x^3$ .

(g \circ h)(x) = g\big(h(x)\big) = g(x^3) = \sin(x^3) = F(x).

In $g \circ h$ the function on the right acts first, so the inner function sits next to the input. Both $h$ and $g$ can be taken from $\mathbb{R}$ to $\mathbb{R}$ , because cubing and sine are defined for every real number.

Problem 3 · Make the Sets Fit

Given: $F: \mathbb{R} \to \mathbb{R}$ , $F(x) = \ln(x^4 + 1)$ , decomposed with inner $f(x) = x^4 + 1$ and outer $h(x) = \ln x$ . Which choice of sets makes $F = h \circ f$ ? ( $\mathbb{R}^+$ is the set of positive reals.)

✅ Correct!

x^4 + 1 \ge 1 > 0

, so

f

lands in

\mathbb{R}^+

, which is exactly where

\ln

is defined, and

h \circ f: \mathbb{R} \to \mathbb{R}

matches

F

❌ The outer function fails.

\ln(-1)

and

\ln 0

do not exist, so

h: \mathbb{R} \to \mathbb{R}

is not a function.

❌ The sets do not fit. The codomain of

f

\mathbb{R}^+

but the domain of

h

\mathbb{R}

. Besides,

\ln

is not defined on all of

\mathbb{R}

❌ Wrong domain for the composite. These pieces fit, but

h \circ f: \mathbb{R}^+ \to \mathbb{R}

has a different domain from

F: \mathbb{R} \to \mathbb{R}

, so it is a different function.

Show solution

Outer domain. $\ln x$ is defined only for $x > 0$ , so $h: \mathbb{R}^+ \to \mathbb{R}$ .

Fit. The codomain of $f$ must be $\mathbb{R}^+$ . It can be: $x^4 \ge 0$ , so $x^4 + 1 \ge 1 > 0$ for every real $x$ , giving $f: \mathbb{R} \to \mathbb{R}^+$ .

Composite. $h \circ f: \mathbb{R} \to \mathbb{R}$ and $(h \circ f)(x) = \ln(x^4 + 1) = F(x)$ for every $x \in \mathbb{R}$ , so $F = h \circ f$ .

Problem 4 · Decomposing on the Integers

Given: $f: \mathbb{Z} \to \mathbb{Z}$ , $f(x) = |2x - 5|$ . Cutting at the last step gives inner $k(x) = 2x - 5$ and outer $g(x) = |x|$ , with $k: \mathbb{Z} \to B$ and $g: B \to \mathbb{Z}$ . Which set $B$ makes $f = g \circ k$ ? ( $\mathbb{N} = \{1, 2, 3, \ldots\}$ .)

✅ Correct!

2x - 5

is an integer whenever

x

is, and the absolute value of an integer is an integer, so

k: \mathbb{Z} \to \mathbb{Z}

and

g: \mathbb{Z} \to \mathbb{Z}

❌ The inner function leaves $\mathbb{N}$ .

k(2) = -1

and

k(0) = -5

are not natural numbers, so

k: \mathbb{Z} \to \mathbb{N}

is not a function.

❌ The outer function leaves $\mathbb{Z}$ .

k: \mathbb{Z} \to \mathbb{R}

is fine, but

g: \mathbb{R} \to \mathbb{Z}

is not a function, since

g\big(\tfrac{1}{2}\big) = \tfrac{1}{2} \notin \mathbb{Z}

❌ Negative outputs.

k(2) = -1

is not a positive real number, so

k: \mathbb{Z} \to \mathbb{R}^+

is not a function.

Show solution

Inner codomain. $k$ takes negative values, e.g. $k(0) = -5$ and $k(2) = -1$ , so $B$ must contain negative numbers. That rules out $\mathbb{N}$ and $\mathbb{R}^+$ .

Outer codomain. $g: B \to \mathbb{Z}$ must send every element of $B$ to an integer. With $B = \mathbb{R}$ , $g\big(\tfrac{1}{2}\big) = \tfrac{1}{2} \notin \mathbb{Z}$ , so $\mathbb{R}$ fails.

$B = \mathbb{Z}$ works. If $x \in \mathbb{Z}$ then $2x - 5 \in \mathbb{Z}$ , and $|n| \in \mathbb{Z}$ for $n \in \mathbb{Z}$ . So $g \circ k: \mathbb{Z} \to \mathbb{Z}$ and $(g \circ k)(x) = |2x - 5| = f(x)$ , hence $f = g \circ k$ .

Check one input: $k(1) = -3$ , $g(-3) = 3$ , and $f(1) = |2 - 5| = 3$ .

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