Inverse Functions

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Quick Reference

Before You Start: Prerequisite Check

Try This First

A vending machine takes a code and gives back a snack. Suppose pressing $A1$ gives back a granola bar, $A2$ gives back pretzels, and $A3$ gives back trail mix.

Now imagine the snacks come back to the machine and you want to recover which code produced each one. The granola bar points back to $A1$. The pretzels point back to $A2$. The trail mix points back to $A3$.

Before reading any formula, answer this in your own words: what does the reversed machine do, and why does each snack have to point back to exactly one code?

The Idea: A Function Run Backward

A function is a process that takes an input and gives back an output. An inverse function runs that process backward: it takes the output and gives back the input that produced it.

Two pictures help hold the idea together.

Picture 1 (the machine). The original function $f$ turns an input $x$ into an output $y$. The inverse $f^{-1}$ takes that same $y$ and returns the original $x$.

$$x \;\xrightarrow{\;f\;}\; y \qquad\qquad y \;\xrightarrow{\;f^{-1}\;}\; x$$

Picture 2 (the table). Reversing a function swaps the role of input and output. Read the same pairs from right to left.

The left table sends $1 \to 4$. The right table sends $4 \to 1$. The inverse is the same set of pairs with the arrows turned around. Translate between the two pictures yourself: which column of the machine picture matches which column of the table?

A reversal only produces a function when every output came from a single input. A function with that property is called one-to-one: different inputs always give different outputs. The vending machine in the opener works as a reversible process for exactly this reason.

Prerequisite Map

Builds on

Unlocks

The scope lists no direct successor skill for this node yet. The idea returns later in the course when inverse trigonometric functions and logarithms are introduced as the inverses of trigonometric and exponential functions.

Quick Reference

Inverse Functions at a Glance

Key Concepts

The Official Definition

Inverse function. For any one-to-one function $f(x) = y$, a function $f^{-1}(x)$ is the inverse function of $f$ if $f^{-1}(y) = x$. The notation $f^{-1}$ is read “f inverse” and does not mean a reciprocal: $f^{-1}(x)$ is not equal to $\dfrac{1}{f(x)}$.

Source: OpenStax Precalculus 2e, Section 1.7 Inverse Functions.

The definition carries one warning worth stating on its own. The symbol $f^{-1}$ borrows the same superscript that means “reciprocal” in arithmetic, where $5^{-1} = \frac{1}{5}$. With a function name in front of it, the $-1$ is a label for reversal instead. The reciprocal $\frac{1}{f(x)}$ is a different object entirely.

Key Property: The Round Trip

Inverse composition property. If $f^{-1}$ is the inverse of $f$, then $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$, and $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$.

Source: OpenStax Precalculus 2e, Section 1.7 Inverse Functions.

This property is the round trip. Send an input through $f$ and then back through $f^{-1}$ and you land where you started. Send an input through $f^{-1}$ and then back through $f$ and the same thing happens.

The round trip is also the test. To check whether two functions are inverses, compose them both ways. If both compositions collapse to $x$, the functions are inverses. If either one does not, they are not.

$$f^{-1}(f(x)) = x \qquad\text{and}\qquad f(f^{-1}(x)) = x$$

Reading the Inverse Off a Graph

The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. Reversing a function swaps each pair $(a, b)$ for the pair $(b, a)$, and swapping coordinates is exactly what reflecting across $y = x$ does.

        y = x
          ╱
   b ──●  ╱
      │ ╲╱
      │ ╱╲
      │╱  ●── point (b, a) on the inverse
      ╱   │
     ╱    │
    ╱─────┴───
          a

A point $(a, b)$ on $f$ becomes the point $(b, a)$ on $f^{-1}$. The line $y = x$ acts as the mirror.

Finding an Inverse Algebraically

When a function is given by a formula, the inverse can often be found by solving for the input. One way to organize the work:

Step 1. Write the function as $y = f(x)$.

Step 2. Swap $x$ and $y$. (This records the role reversal: the old output becomes the new input.)

Step 3. Solve the new equation for $y$.

Step 4. Write the result as $f^{-1}(x) = y$.

Step 5. Check by composing both ways and confirming each composition gives back $x$.

Another way to see the same procedure: solve $y = f(x)$ for $x$ first, then rename the variables at the end. Both routes reach the same inverse. Which order feels clearer to you, and why?

Worked Examples

Worked Example 1: Verify an inverse with the round trip

Show that $f(x) = 3x - 6$ and $g(x) = \dfrac{x + 6}{3}$ are inverses.

Predict first. The function $f$ multiplies by $3$ and then subtracts $6$. To reverse those steps, add $6$ and then divide by $3$. The formula for $g$ adds $6$ and divides by $3$, so predict that the two compositions will collapse to $x$.

Check both directions.

$$g(f(x)) = \frac{f(x) + 6}{3} = \frac{(3x - 6) + 6}{3} = \frac{3x}{3} = x$$

$$f(g(x)) = 3\,g(x) - 6 = 3\left(\frac{x + 6}{3}\right) - 6 = (x + 6) - 6 = x$$

Both compositions return $x$, so $f$ and $g$ are inverses. The prediction matched the computation: undoing “times $3$, minus $6$” really is “plus $6$, divide by $3$.”

Worked Example 2: Find an inverse from a formula

Find the inverse of $f(x) = 2x + 1$.

Step 1. Write $y = 2x + 1$.

Step 2. Swap $x$ and $y$: $$x = 2y + 1$$

Step 3. Solve for $y$: $$x - 1 = 2y \qquad\Longrightarrow\qquad y = \frac{x - 1}{2}$$

Step 4. Write the inverse: $$f^{-1}(x) = \frac{x - 1}{2}$$

Step 5. Check with the round trip: $$f^{-1}(f(x)) = \frac{(2x + 1) - 1}{2} = \frac{2x}{2} = x$$ $$f\left(f^{-1}(x)\right) = 2\left(\frac{x - 1}{2}\right) + 1 = (x - 1) + 1 = x$$

Both checks give $x$, so $f^{-1}(x) = \dfrac{x - 1}{2}$ is correct.

Worked Example 3: A function that is not one-to-one

Find the inverse of $f(x) = x^2$, or explain why no inverse function exists.

Predict first. Squaring sends both $3$ and $-3$ to $9$. If two different inputs share an output, the reversed process cannot decide which input to return. Predict that no inverse function exists without a restriction on the domain.

Check. Suppose an inverse existed. Then $f^{-1}(9)$ would have to equal both $3$ and $-3$, because $f(3) = 9$ and $f(-3) = 9$. A function returns one output for each input, so $f^{-1}(9)$ cannot be two numbers at once. The function $f(x) = x^2$ is not one-to-one on all real numbers, so it has no inverse function there.

A repair. Restrict the domain to $x \ge 0$. On that smaller domain, $f(x) = x^2$ is one-to-one, and its inverse is $f^{-1}(x) = \sqrt{x}$. The restriction throws away the negative half so that each output traces back to a single input. Why does keeping only $x \ge 0$ fix the problem, and would keeping only $x \le 0$ work just as well?

Common Misconceptions

Practice Problems

Mastery Checklist

Novice (Level 1-2):

• I can read $f^{-1}$ as “f inverse” and explain why it is not a reciprocal
• I can read inverse values off a table by reversing the input and output
• I can state what one-to-one means and why an inverse needs it

Competent (Level 3-4):

• I can find the inverse of a linear function and check it with a composition
• I can explain why a function like $x^2$ has no inverse without a domain restriction
• I can restrict a domain so that a non-invertible function becomes invertible

Proficient (Level 5):

• I can find the inverse of a rational function and state its domain
• I can prove an inverse relationship by composing in both directions
• I can explain why the domain and range of a function and its inverse swap

Mental Model

The “undo” model. A function is a set of instructions: take $x$, do something to it, report the result. The inverse is the set of instructions that puts everything back. If $f$ says “multiply by $3$, then subtract $6$,” then $f^{-1}$ says “add $6$, then divide by $3$,” carried out in the reverse order. The round trip $f^{-1}(f(x)) = x$ is the promise that following the instructions and then undoing them leaves the input untouched.

The model also explains the one-to-one requirement. Instructions can only be undone if they never erase information. Squaring erases the sign of the input, so there is no single instruction that recovers it, which is why $x^2$ has no inverse until the negative half of the domain is removed.

Connections

Looking back:

• Functions and Function Notation sets up the process view that an inverse reverses
• Function Composition supplies the round-trip test for inverses
• Transformation of Functions explains the reflection across $y = x$

Looking ahead:

• Logarithmic functions are introduced as the inverses of exponential functions
• Inverse trigonometric functions reverse sine, cosine, and tangent on restricted domains

Real-world connections:

• Converting Celsius to Fahrenheit and back uses a function and its inverse
• Encoding and decoding a message is a function paired with its inverse
• Converting a currency and then converting it back returns the original amount

Resources

• Textbook section: OpenStax Precalculus 2e, Section 1.7 Inverse Functions (online)
• Chapter review: OpenStax Precalculus 2e, Chapter 1 Key Concepts

Last updated: 2026-06-16