Mathematical-Reasoning · Unit 24 · Video 5 · Interactive Practice

Undo the Last Step First: The Inverse of a Composite

IKey Formulas

Statement	Meaning
$f(x) = y \iff f^{-1}(y) = x$	Inverse function notation, for a bijection $f : A \to B$ , $x \in A$ , $y \in B$
$\left(f^{-1} \circ f\right)(x) = x$ , $\ \left(f \circ f^{-1}\right)(y) = y$	An inverse undoes its function: $f^{-1} \circ f = I_A$ and $f \circ f^{-1} = I_B$
$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$	For bijections $f : A \to B$ and $g : B \to C$ : the step done last is undone first
$h(x) = e^{x^3}$ , $\ h^{-1}(y) = \sqrt[3]{\ln y}$	$h = g \circ f$ with $f(x) = x^3$ and $g(x) = e^x$ onto $\mathbb{R}^+$

Key Insight: To get back from $C$ to $A$ you must pass through $B$ : first $g^{-1}$ takes $C$ to $B$ , then $f^{-1}$ takes $B$ to $A$ . The other order, $g^{-1} \circ f^{-1}$ , is not even defined, since $f^{-1}$ accepts elements of $B$ , not of $C$ .

IICube, Exponentiate, and Back

Undoing $x \mapsto e^{x^3}$ works only when the logarithm, which undoes the last step, is applied first.

x 1 Undo order

IIITracing an Element Back Through Two Bijections

For each $z \in C$ , the route $g^{-1}$ then $f^{-1}$ ends at the same $x$ that $g \circ f$ sends to $z$ .

z Step 0

IVQuiz Questions

Problem 1 · Reading Inverse Notation

Given: the bijection $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = 2x + 3$ . What is $f^{-1}(11)$ ?

✅ Correct!

f^{-1}(11) = x

exactly when

f(x) = 11

, and

2 \cdot 4 + 3 = 11

❌ That is $f(11)$ . The inverse runs the other way: find the input whose output is

11

❌ Check the subtraction. Solving

2x + 3 = 11

means subtracting

3

, not adding it.

❌ $f^{-1}$ is not $\tfrac{1}{f}$ . The exponent names the inverse function, not a reciprocal.

Show solution

By inverse function notation, $f^{-1}(11) = x \iff f(x) = 11$ :

2x + 3 = 11 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4.

So $f^{-1}(11) = 4$ . In general $f^{-1}(y) = \tfrac{y - 3}{2}$ .

Problem 2 · The Inverse of e to the x Cubed

Given: $h : \mathbb{R} \to \mathbb{R}^+$ , $h(x) = e^{x^3}$ , a bijection. Which formula is $h^{-1}(y)$ for $y > 0$ ?

✅ Correct!

h = g \circ f

with

f(x) = x^3

applied first, so

h^{-1} = f^{-1} \circ g^{-1}

: take

\ln

first, then the cube root.

❌ Wrong order.

\ln\left(\sqrt[3]{y}\right) = \tfrac{1}{3}\ln y

; at

y = h(2) = e^8

it gives

\tfrac{8}{3}

, not

2

❌ Not quite. Undo each step with its inverse, last step first: the exponential is undone by

\ln

, the cube by the cube root.

Show solution

Write $h = g \circ f$ with $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^3$ and $g : \mathbb{R} \to \mathbb{R}^+$ , $g(x) = e^x$ . Both are bijections, with $f^{-1}(y) = \sqrt[3]{y}$ and $g^{-1}(y) = \ln y$ . By the theorem on the inverse of a composition,

h^{-1}(y) = \left(f^{-1} \circ g^{-1}\right)(y) = f^{-1}(\ln y) = \sqrt[3]{\ln y}.

Check at $x = 2$ : $h(2) = e^8$ , $\ln e^8 = 8$ , $\sqrt[3]{8} = 2$ .

Problem 3 · Inverting a Composite

Given: the bijections $f, g : \mathbb{R} \to \mathbb{R}$ , $f(x) = x + 2$ and $g(x) = 3x$ . What is $(g \circ f)^{-1}(12)$ ?

✅ Correct!

(f^{-1} \circ g^{-1})(12) = f^{-1}(4) = 2

, and indeed

(g \circ f)(2) = 3 \cdot 4 = 12

❌ That undoes the first step first.

\tfrac{10}{3} = g^{-1}\left(f^{-1}(12)\right)

; the last step, multiplying by

3

, must be undone first.

❌ That composes forward. The inverse asks which input

g \circ f

sends to

12

Show solution

$f^{-1}(y) = y - 2$ and $g^{-1}(y) = \tfrac{y}{3}$ . By the theorem, $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ , so

(g \circ f)^{-1}(12) = f^{-1}\left(g^{-1}(12)\right) = f^{-1}(4) = 2.

Check directly: $(g \circ f)(x) = 3(x + 2) = 3x + 6 = 12$ gives $x = 2$ .

Problem 4 · Does the Square Root Undo Squaring?

Given: the claim " $\sqrt{x^2} = x$ for every real number $x$ ". Which assessment is correct?

✅ Correct! The corollary needs a bijection. Squaring on

\mathbb{R}

is not one, and in fact

\sqrt{x^2} = |x|

❌ The corollary does not apply. It holds for a bijection and its inverse;

x \mapsto x^2

\mathbb{R}

sends

3

and

-3

to the same output.

❌ Not quite. The square root is a function on

[0, \infty)

; the trouble is with squaring on all of

\mathbb{R}

Show solution

Squaring $\mathbb{R} \to \mathbb{R}$ is not an injection ( $(-3)^2 = 3^2$ ), so it has no inverse function and the corollary says nothing. The claim fails at $x = -3$ : $\sqrt{(-3)^2} = \sqrt{9} = 3 \ne -3$ . In general $\sqrt{x^2} = |x|$ . Restricted to $[0, \infty) \to [0, \infty)$ , squaring is a bijection with inverse $\sqrt{\ }$ , and then $\sqrt{x^2} = x$ for $x \ge 0$ .

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