Finding Inverse Functions Algebraically

5 min read

Reference: Stewart §6.1

The Recipe for Reversing

You know that $f^{-1}$ “undoes” $f$. But how do you actually find a formula for $f^{-1}$? The key insight: if $y = f(x)$ describes the forward process, then solving for $x$ in terms of $y$ describes the reverse. After that, we just swap variable names to match standard notation.

Prerequisite Map

Quick Reference

Property	Value
Chapter	Chapter 6: Inverse Functions
Section	§6.1 Inverse Functions and Their Derivatives
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Three-Step Method

$$\boxed{\text{To find } f^{-1}:}$$

Step	Action	Why It Works
1	Write $y = f(x)$	Sets up the equation relating input and output
2	Solve for $x$ in terms of $y$	Reverses the process; now $x$ depends on $y$
3	Interchange $x$ and $y$	Standard convention: independent variable is $x$

The result is $y = f^{-1}(x)$.

Why Interchange $x$ and $y$?

After Step 2, you have $x$ written as a function of $y$. Mathematically, this is the inverse function: it tells you the input that produces a given output.

But conventionally, we like to write functions as “$y$ = something in $x$” with $x$ as the input. Swapping $x$ and $y$ is just a relabeling to match this convention. It doesn’t change the relationship, only the names.

Worked Example: Linear Function

Find the inverse of $f(x) = 3x + 7$.

Step 1: Write $y = 3x + 7$

Step 2: Solve for $x$: $$y - 7 = 3x$$ $$x = \frac{y - 7}{3}$$

Step 3: Interchange $x$ and $y$: $$y = \frac{x - 7}{3}$$

Result: $f^{-1}(x) = \frac{x - 7}{3}$

Verification: Check that $f(f^{-1}(x)) = x$: $$f\left(\frac{x-7}{3}\right) = 3 \cdot \frac{x-7}{3} + 7 = (x-7) + 7 = x \quad ✓$$

Worked Example: Cubic Function

Find the inverse of $f(x) = x^3 - 5$.

Step 1: Write $y = x^3 - 5$

Step 2: Solve for $x$: $$y + 5 = x^3$$ $$x = \sqrt[3]{y + 5}$$

Step 3: Interchange $x$ and $y$: $$y = \sqrt[3]{x + 5}$$

Result: $f^{-1}(x) = \sqrt[3]{x + 5}$

Worked Example: Rational Function

Find the inverse of $f(x) = \frac{x + 1}{x - 2}$ (for $x \neq 2$).

Step 1: Write $y = \frac{x + 1}{x - 2}$

Step 2: Solve for $x$ (this requires more algebra):

Multiply both sides by $(x - 2)$: $$y(x - 2) = x + 1$$

Expand: $$yx - 2y = x + 1$$

Collect $x$ terms on one side: $$yx - x = 2y + 1$$ $$x(y - 1) = 2y + 1$$ $$x = \frac{2y + 1}{y - 1}$$

Step 3: Interchange $x$ and $y$: $$y = \frac{2x + 1}{x - 1}$$

Result: $f^{-1}(x) = \frac{2x + 1}{x - 1}$ (for $x \neq 1$)

Domain Restrictions

When the original function has a restricted domain (to ensure it’s one-to-one), you must track what happens to that restriction.

Example: Find the inverse of $f(x) = x^2 + 4$ for $x \geq 0$.

Step 1: $y = x^2 + 4$

Step 2: $y - 4 = x^2$, so $x = \pm\sqrt{y - 4}$

Since the domain is $x \geq 0$, we take the positive root: $x = \sqrt{y - 4}$

Step 3: $y = \sqrt{x - 4}$

Result: $f^{-1}(x) = \sqrt{x - 4}$ with domain $x \geq 4$

Common Pitfalls

Mistake	Example	Correction
Forgetting to swap $x$ and $y$	Getting $x = \frac{y-7}{3}$ and stopping	Continue: swap to get $y = \frac{x-7}{3}$
Taking wrong root	$x^2 = 4 \Rightarrow x = 2$	Check domain: might need $x = -2$
Not checking domain	Finding inverse without noting restrictions	State domain of $f^{-1}$ (= range of $f$)

Practice Problems

Level 1 Linear Function

Find the inverse of $f(x) = 5x - 7$.

Thought Process

This is a linear function, so the inverse will also be linear. I’ll use the three-step method:

Write $y = 5x - 7$
Solve for $x$: add 7, then divide by 5
Swap $x$ and $y$

Show Answer

Step 1: $y = 5x - 7$

Step 2: $$y + 7 = 5x$$ $$x = \frac{y + 7}{5}$$

Step 3: Swap $x$ and $y$: $$y = \frac{x + 7}{5}$$

Answer: $f^{-1}(x) = \frac{x + 7}{5}$

Level 2 Root Function

Find the inverse of $g(x) = \sqrt{x + 3}$ (for $x \geq -3$).

Thought Process

The function takes a square root, so the inverse should involve squaring. I need to track the domain carefully: the range of $g$ becomes the domain of $g^{-1}$.

Since $\sqrt{x+3} \geq 0$ for all valid inputs, the range of $g$ is $[0, \infty)$.

Show Answer

Step 1: $y = \sqrt{x + 3}$

Step 2: Square both sides: $$y^2 = x + 3$$ $$x = y^2 - 3$$

Step 3: Swap $x$ and $y$: $$y = x^2 - 3$$

Domain of inverse: Since range of $g$ is $[0, \infty)$, domain of $g^{-1}$ is $x \geq 0$.

Answer: $g^{-1}(x) = x^2 - 3$ for $x \geq 0$

Level 3 Quadratic with Restriction

Find the inverse of $h(x) = x^2 - 4x$ for $x \geq 2$.

Thought Process

This is a parabola opening upward with vertex at $x = 2$. On the domain $x \geq 2$, the function is increasing.

To solve $y = x^2 - 4x$ for $x$, I’ll complete the square or use the quadratic formula. Since $x \geq 2$, I’ll need to choose the appropriate root.

Show Answer

Step 1: $y = x^2 - 4x$

Step 2: Complete the square: $$y = (x^2 - 4x + 4) - 4 = (x-2)^2 - 4$$ $$y + 4 = (x-2)^2$$ $$x - 2 = \pm\sqrt{y + 4}$$

Since $x \geq 2$, we have $x - 2 \geq 0$, so take the positive root: $$x = 2 + \sqrt{y + 4}$$

Step 3: Swap $x$ and $y$: $$y = 2 + \sqrt{x + 4}$$

Domain: The vertex of the parabola is at $(2, -4)$, so range of $h$ is $[-4, \infty)$.

Answer: $h^{-1}(x) = 2 + \sqrt{x + 4}$ for $x \geq -4$

Level 4 Rational Function

Find the inverse of $f(x) = \frac{3x - 2}{x + 4}$ (for $x \neq -4$).

Thought Process

For rational functions of the form $\frac{ax + b}{cx + d}$, I solve for $x$ by:

Multiplying both sides by the denominator
Collecting all $x$ terms on one side
Factoring out $x$
Dividing to isolate $x$

Show Answer

Step 1: $y = \frac{3x - 2}{x + 4}$

Step 2: Multiply by $(x + 4)$: $$y(x + 4) = 3x - 2$$ $$yx + 4y = 3x - 2$$

Collect $x$ terms: $$yx - 3x = -2 - 4y$$ $$x(y - 3) = -4y - 2$$ $$x = \frac{-4y - 2}{y - 3}$$

This can be rewritten as: $x = \frac{-2(2y + 1)}{y - 3}$

Step 3: Swap $x$ and $y$: $$y = \frac{-4x - 2}{x - 3}$$

Answer: $f^{-1}(x) = \frac{-4x - 2}{x - 3}$ for $x \neq 3$

Verification: Note that $f(x) \to 3$ as $x \to \pm\infty$, so 3 is a horizontal asymptote and is not in the range. This matches the domain restriction $x \neq 3$ for $f^{-1}$.

Level 5 Self-Inverse Investigation

(a) Find the inverse of $f(x) = \frac{1 - x}{1 + x}$ (for $x \neq -1$).

(b) What do you notice? Explain why this happens geometrically.

Thought Process

(a) Standard three-step method for a rational function.

(b) If $f = f^{-1}$, then $f(f(x)) = x$. Geometrically, self-inverse functions have graphs that are symmetric about the line $y = x$.

(c) For $g$ to be self-inverse, $g(g(x)) = x$. I’ll compute $g(g(x))$ and set it equal to $x$, then determine what values of $a$ make this work.

Show Answer

(a) Find inverse of $f(x) = \frac{1-x}{1+x}$:

Step 1: $y = \frac{1-x}{1+x}$

Step 2: $$y(1+x) = 1-x$$ $$y + yx = 1 - x$$ $$yx + x = 1 - y$$ $$x(y + 1) = 1 - y$$ $$x = \frac{1-y}{1+y}$$

Step 3: Swap: $y = \frac{1-x}{1+x}$

Result: $f^{-1}(x) = \frac{1-x}{1+x} = f(x)$

(b) The function is self-inverse: $f^{-1} = f$.

Geometrically, the graph of $f(x) = \frac{1-x}{1+x}$ is symmetric about the line $y = x$. When you reflect it across $y = x$, you get the same graph back.

(c) Compute $g(g(x))$: $$g(g(x)) = g\left(\frac{a-x}{1+ax}\right) = \frac{a - \frac{a-x}{1+ax}}{1 + a \cdot \frac{a-x}{1+ax}}$$

Numerator: $\frac{a(1+ax) - (a-x)}{1+ax} = \frac{a + a^2x - a + x}{1+ax} = \frac{x(1+a^2)}{1+ax}$

Denominator: $\frac{(1+ax) + a(a-x)}{1+ax} = \frac{1+ax+a^2-ax}{1+ax} = \frac{1+a^2}{1+ax}$

$$g(g(x)) = \frac{x(1+a^2)/(1+ax)}{(1+a^2)/(1+ax)} = x$$

This equals $x$ for all values of $a$ (as long as $1 + a^2 \neq 0$, which is always true for real $a$).

Conclusion: $g(x) = \frac{a-x}{1+ax}$ is self-inverse for all real values of $a$.

CCI-Style Conceptual Questions

Question 1: If $f(x) = x^3 + 5$, what is the first step in finding $f^{-1}$?

(A) Take the cube root of $x$ (B) Write $y = x^3 + 5$ and solve for $x$ (C) Swap $f$ and $f^{-1}$ (D) Subtract 5 from $f(x)$

Answer

(B) The first step is to write $y = x^3 + 5$ and then solve for $x$ in terms of $y$. This gives $x = \sqrt[3]{y-5}$. Then swap $x$ and $y$ to get $f^{-1}(x) = \sqrt[3]{x-5}$.

Question 2: When finding the inverse of $f(x) = x^2$ for $x \leq 0$, you solve $y = x^2$ and get $x = \pm\sqrt{y}$. Why do you choose $x = -\sqrt{y}$?

(A) The negative is always correct for square roots (B) The domain restriction $x \leq 0$ requires the negative root (C) It doesn’t matter which root you choose (D) You should always use the positive root

Answer

(B) The domain restriction $x \leq 0$ means we’re only considering the left half of the parabola. When we solve for $x$, we must honor this restriction, so we take the negative root: $f^{-1}(x) = -\sqrt{x}$.

Common Misconceptions

Common misconception

$f^{-1}(x)$ means $\frac{1}{f(x)}$, the reciprocal of $f$.

This is the action-view-of-function error. The notation $f^{-1}$ denotes the inverse function, not a power: $f^{-1}(x)$ is the input that produces output $x$, not $[f(x)]^{-1}$. For $f(x) = x^3 + 1$, the inverse is $f^{-1}(x) = \sqrt[3]{x-1}$, while $\frac{1}{f(x)} = \frac{1}{x^3+1}$; these are entirely different functions with different domains and ranges.

Mastery Checklist

I can state the three-step method for finding inverses
I can find inverses of linear functions
I can find inverses of polynomial functions (with appropriate roots)
I can find inverses of rational functions
I correctly handle domain restrictions when finding inverses
I can verify my answer using composition ($f(f^{-1}(x)) = x$)
I understand why we interchange $x$ and $y$ at the end

Mental Model

The “Unwrapping” Picture:

Think of a function as wrapping a gift: $f(x) = 3x + 7$ means “triple it, then add 7.”

Finding the inverse is like writing instructions to unwrap: “subtract 7, then divide by 3.”

The order reverses! Whatever $f$ did last, $f^{-1}$ undoes first.

$f$: $x \xrightarrow{\times 3} 3x \xrightarrow{+7} 3x+7$
$f^{-1}$: $x \xrightarrow{-7} x-7 \xrightarrow{\div 3} \frac{x-7}{3}$

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Inverse Function Definition	Chapter 6	Graphing Inverse Functions

Last updated: 2026-01-23