Inverse functions — finding f⁻¹(x), horizontal line test: Exercises

This page is a self-test ladder. Cover the solutions, try each problem, then reveal. Difficulty climbs from "can you spot it" (L1) to "can you build the whole thing" (L5). Every solution is fully worked — no skipped algebra.

Parent: Inverse functions (parent topic). If any word below feels new, that note builds it from zero.

Before we start, three ideas we will lean on constantly:

Recall What "inverse" and "one-to-one" mean (quick refresh)

An inverse function $f^{- 1}$ undoes $f$ : if $f$ sends $x \to y$ , then $f^{- 1}$ sends $y \to x$ . A function is one-to-one if different inputs never collide onto the same output. Only one-to-one functions can be reversed — see Bijective Functions. The horizontal line test: draw every horizontal line $y = c$ ; if none hits the graph more than once, the function is one-to-one.

Level 1 — Recognition

Goal: read a graph or equation and decide invertibility. No algebra yet.

Exercise 1.1

Look at the four graphs below. For each, state whether it passes the horizontal line test and is therefore invertible.

Figure — Inverse functions — finding f⁻¹(x), horizontal line test

Recall Solution 1.1

A horizontal line is a flat line at some fixed height $y = c$ . If any such line crosses the curve twice or more, the function fails and is not invertible.

(A) Line $f (x) = 2 x - 1$ : every horizontal line hits it exactly once. Passes → invertible.
(B) Parabola $f (x) = x^{2}$ : the line $y = 4$ hits at $x = - 2$ and $x = 2$ . Fails → not invertible.
(C) Cubic $f (x) = x^{3}$ : always rising, each height reached once. Passes → invertible.
(D) Sideways-S / sine wave: a single line like $y = 0$ crosses many times. Fails → not invertible.

Why the test is exactly this: a horizontal line collects all points sharing one $y$ -value. Two crossings = two inputs $x_{1} \neq = x_{2}$ giving the same output — the very definition of not one-to-one.

Exercise 1.2

True or false: $f (x) = 5$ (a constant function) is invertible.

Recall Solution 1.2

False. The graph is a flat horizontal line at height $5$ . The horizontal line $y = 5$ lies on top of it — infinitely many intersections. Every input maps to $5$ , so you can never recover which $x$ you started from. Not one-to-one, no inverse.

Level 2 — Application

Goal: run the swap method on standard functions.

Exercise 2.1

Find $f^{- 1} (x)$ for $f (x) = 4 x + 9$ .

Recall Solution 2.1

Step 1 (write): $y = 4 x + 9$ . Step 2 (swap $x$ and $y$ ): $x = 4 y + 9$ . Why swap? Because input and output trade roles under an inverse — the swap literally encodes that. Step 3 (solve for $y$ ): $x - 9 = 4 y \Rightarrow y = \frac{x - 9}{4} .$ Step 4: $f^{- 1} (x) = \frac{x - 9}{4}$ . Step 5 (verify via Function Composition): $f (\frac{x - 9}{4}) = 4 \cdot \frac{x - 9}{4} + 9 = (x - 9) + 9 = x . ✓$

Exercise 2.2

Find $f^{- 1} (x)$ for $f (x) = \frac{x}{5} - 2$ .

Recall Solution 2.2

$y = \frac{x}{5} - 2 \Rightarrow x = \frac{y}{5} - 2$ (swap). $x + 2 = \frac{y}{5} \Rightarrow y = 5 (x + 2) = 5 x + 10.$ $f^{- 1} (x) = 5 x + 10.$ Check: $f (5 x + 10) = \frac{5 x + 10}{5} - 2 = (x + 2) - 2 = x . ✓$

Exercise 2.3

Find $f^{- 1} (x)$ for $f (x) = 3 x - 1$ .

Recall Solution 2.3

$y = 3 x - 1 \Rightarrow x = 3 y - 1$ . $x + 1 = 3 y \Rightarrow y = (x + 1)^{3} .$ Why cube? Cubing is the exact undo of a cube root — the inverse of a step is applied to reverse it. $f^{- 1} (x) = (x + 1)^{3}, domain R .$ Cube root is defined and one-to-one over all reals, so no domain restriction is needed.

Level 3 — Analysis

Goal: handle rational functions and track domain/range swaps — see Domain and Range.

Exercise 3.1

Find $f^{- 1} (x)$ for $f (x) = \frac{x + 4}{x - 1}$ , $x \neq = 1$ . State the domain of $f^{- 1}$ .

Recall Solution 3.1

Swap: $x = \frac{y + 4}{y - 1}$ . Clear the fraction: $x (y - 1) = y + 4 \Rightarrow x y - x = y + 4.$ Collect $y$ -terms (so we can factor $y$ out): $x y - y = x + 4 \Rightarrow y (x - 1) = x + 4 \Rightarrow y = \frac{x + 4}{x - 1} .$ $f^{- 1} (x) = \frac{x + 4}{x - 1}, x \neq = 1.$ Surprise: $f$ is its own inverse (self-inverse / involution). Reflecting its graph across $y = x$ lands it back on itself. Domain reasoning: $f$ 's horizontal asymptote sits at $y = 1$ (ratio of leading coefficients $1/1$ ), so $1$ is never an output of $f$ → $1$ is excluded from $f^{- 1}$ 's domain. Matches $x \neq = 1$ above.

Exercise 3.2

Find $f^{- 1} (x)$ for $f (x) = \frac{3 x - 2}{x + 5}$ , $x \neq = - 5$ . State the excluded input of $f^{- 1}$ .

Recall Solution 3.2

Swap: $x = \frac{3 y - 2}{y + 5}$ . $x (y + 5) = 3 y - 2 \Rightarrow x y + 5 x = 3 y - 2.$ $x y - 3 y = - 2 - 5 x \Rightarrow y (x - 3) = - (5 x + 2) .$ $y = \frac{- ( 5 x + 2 )}{x - 3} = \frac{- 5 x - 2}{x - 3} = \frac{5 x + 2}{3 - x} .$ $f^{- 1} (x) = \frac{5 x + 2}{3 - x}, x \neq = 3.$ Why $x \neq = 3$ : $f$ 's horizontal asymptote is at $y = 3$ (leading coefficients $3/1$ ), so $3$ is never output by $f$ ; hence it is barred from $f^{- 1}$ 's domain.

Level 4 — Synthesis

Goal: combine restriction, transcendental inverses, and composition.

Exercise 4.1

$f (x) = (x - 3)^{2} + 1$ is not one-to-one over $R$ . Restrict its domain to make it invertible, then find $f^{- 1} (x)$ and state both domains.

Recall Solution 4.1

Why restrict: the parabola has vertex at $(3, 1)$ opening upward. Any horizontal line above the vertex hits twice (once left, once right of $x = 3$ ). It fails the test. Choose a branch: take the right half, ==domain $x \geq 3$ == (the branch where the curve only rises). On it, each height is reached once. Find inverse: $y = (x - 3)^{2} + 1, x \geq 3 \Rightarrow x = (y - 3)^{2} + 1.$ $x - 1 = (y - 3)^{2} \Rightarrow y - 3 = + x - 1$ — we take the positive root because our chosen branch has $y - 3 = x - 3 \geq 0$ . $f^{- 1} (x) = 3 + x - 1 .$ Domains (Domain and Range): original domain $[3, \infty)$ → inverse range $[3, \infty)$ ; original range $[1, \infty)$ → inverse domain $[1, \infty)$ . So $f^{- 1} : [1, \infty) \to [3, \infty)$ .

Exercise 4.2

Find the inverse of $f (x) = 2 e^{x} - 3$ . (See Exponential and Logarithmic Functions.)

Recall Solution 4.2

$y = 2 e^{x} - 3 \Rightarrow x = 2 e^{y} - 3$ . $x + 3 = 2 e^{y} \Rightarrow e^{y} = \frac{x + 3}{2} .$ Why the natural log now? $ln$ is the exact inverse of $e^{(\cdot)}$ — it answers "which exponent gives this value?". It is the only tool that peels $y$ out of the exponent. $y = ln (\frac{x + 3}{2}) .$ $f^{- 1} (x) = ln (\frac{x + 3}{2}), x > - 3.$ Domain $x > - 3$ : $e^{y} > 0$ always, so $\frac{x + 3}{2} > 0 \Rightarrow x > - 3$ . This is exactly the range of the original $f$ (outputs of $2 e^{x} - 3$ start just above $- 3$ ).

Exercise 4.3

Given $f (x) = x^{3} + 1$ and $g (x) = 2 x$ , find $(f \circ g)^{- 1} (x)$ . Use the rule $(f \circ g)^{- 1} = g^{- 1} \circ f^{- 1}$ .

Recall Solution 4.3

Direct route: $(f \circ g) (x) = f (2 x) = (2 x)^{3} + 1 = 8 x^{3} + 1$ . Invert: $y = 8 x^{3} + 1 \Rightarrow x = 8 y^{3} + 1 \Rightarrow y^{3} = \frac{x - 1}{8} \Rightarrow y = \frac{3 x - 1}{2} .$ $(f \circ g)^{- 1} (x) = \frac{3 x - 1}{2} .$ Check with the composition rule (order reverses — you undo the last action first): $f^{- 1} (x) = 3 x - 1$ , $g^{- 1} (x) = \frac{x}{2}$ , so $g^{- 1} (f^{- 1} (x)) = \frac{3 x - 1}{2} . ✓$ Same answer.

Level 5 — Mastery

Goal: reason about existence, symmetry, and fixed points — not just crank algebra.

Exercise 5.1

A function equals its own inverse ( $f = f^{- 1}$ ) exactly when its graph is symmetric about the line $y = x$ . Show that $f (x) = \frac{2 x + 1}{x - 2}$ is self-inverse, and find where its graph crosses $y = x$ .

Recall Solution 5.1

Self-inverse check: swap and solve. $x = \frac{2 y + 1}{y - 2} \Rightarrow x (y - 2) = 2 y + 1 \Rightarrow x y - 2 x = 2 y + 1.$ $x y - 2 y = 2 x + 1 \Rightarrow y (x - 2) = 2 x + 1 \Rightarrow y = \frac{2 x + 1}{x - 2} = f (x) .$ Since $f^{- 1} = f$ , the function is self-inverse. Its graph is its own reflection across $y = x$ . Crossings with $y = x$ : set $f (x) = x$ : $\frac{2 x + 1}{x - 2} = x \Rightarrow 2 x + 1 = x^{2} - 2 x \Rightarrow x^{2} - 4 x - 1 = 0.$ $x = \frac{4 \pm 16 + 4}{2} = 2 \pm 5 .$ These two fixed points $x = 2 + 5 \approx 4.236$ and $x = 2 - 5 \approx - 0.236$ are the only inputs $f$ leaves unchanged — they sit exactly on the mirror line $y = x$ .

Exercise 5.2

For which value(s) of $k$ is $f (x) = x^{2} + k x + 4$ not one-to-one on all of $R$ , and does restricting to $x \geq - \frac{k}{2}$ always fix it?

Recall Solution 5.2

Any real $k$ makes $f$ non-injective on $R$ : a parabola $x^{2} + k x + 4$ always opens upward with a single vertex at $x = - \frac{k}{2}$ , so a horizontal line above the vertex always meets it twice. There is no $k$ that rescues a full-line parabola. The restriction $x \geq - \frac{k}{2}$ keeps only the branch to the right of the vertex, where the curve strictly increases. On a strictly increasing branch, distinct inputs give distinct outputs → one-to-one → invertible. So yes, the restriction always works, for every $k$ . (Restricting to $x \leq - \frac{k}{2}$ works too — the left, strictly-decreasing branch.)

Exercise 5.3

$sin$ is not invertible over $R$ , yet $arcsin$ exists. Explain in one paragraph how restriction rescues it, and give the domain and range of $arcsin$ . (See Trigonometric Functions.)

Recall Solution 5.3

$sin$ repeats every $2 π$ , so a horizontal line like $y = 0$ hits its graph infinitely often — hopelessly non-injective. We restrict the domain to $[- \frac{π}{2}, \frac{π}{2}]$ , the single rising piece from trough to crest. On that one branch $sin$ climbs monotonically from $- 1$ to $1$ , passing the horizontal line test, so an inverse exists there. That inverse is == $arcsin$ ==, with domain $[- 1, 1]$ (the outputs $sin$ can produce) and range $[- \frac{π}{2}, \frac{π}{2}]$ (the restricted inputs). This is the exact same "restrict a symmetric graph to one branch" trick used for $x^{2}$ in Exercise 4.1 — see Bijective Functions.

Recall One-line self-quiz

Why does restricting a domain rescue an inverse? ::: It deletes the duplicate inputs so each output has a unique source — the graph then passes the horizontal line test. $(f \circ g)^{- 1} = ?$ ::: $g^{- 1} \circ f^{- 1}$ — undo the last-applied function first. A function meets $y = x$ at its ... ? ::: fixed points, where $f (x) = x$ .