Inverse functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Inverse function existence criteria (one-to-one functions, horizontal line test), domain-range symmetry, algebraic inversion, graphical reflection, composition verification, and domain restriction for invertible functions aligned to AP Precalculus learning objectives.

You should already know: How to identify domain and range of any algebraic function. How to compute composition of two functions. How to test if a relation is a function via the vertical line test.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Inverse functions?

An inverse function reverses the input-output mapping of an original invertible function. If the original function $f$ takes input $x$ to output $y=f(x)$, then the inverse function $f^{-1}$ takes input $y$ to output $x=f^{-1}(y)$, satisfying the core composition identities $f(f^{-1}(y))=y$ and $f^{-1}(f(x))=x$.

The AP Precalculus CED lists inverse functions as a core foundational topic in Unit 2 (Exponential and Logarithmic Functions) with concepts appearing in approximately 6-8% of all exam questions across both MCQ and FRQ sections. A critical notation convention to note: $f^{-1}$ never means $\frac{1}{f}$, a common student misconception. Inverses only exist for functions that are one-to-one (injective), meaning each output corresponds to exactly one input. The skills learned in this chapter are the foundation for defining logarithms, the inverses of exponential functions, which are the core of Unit 2.

2. Existence of Inverses and the Horizontal Line Test

A function can only have a valid inverse over a domain if it is one-to-one on that domain. By definition, a function is one-to-one if whenever $f(a)=f(b)$, then $a=b$; that is, no two distinct inputs produce the same output.

The simplest graphical test for one-to-oneness is the horizontal line test: a function passes the test if no horizontal line intersects its graph more than once. If any horizontal line crosses the graph multiple times, the function is not one-to-one and cannot have an inverse over its entire domain. Algebraically, you can test for one-to-oneness by setting $f(a)=f(b)$ and simplifying to check if you must have $a=b$.

Many common functions (such as quadratics) are not one-to-one over their full domain, but we can restrict the domain to an interval where the function is strictly monotonic (always increasing or always decreasing), which guarantees it is one-to-one and thus invertible.

Worked Example

Problem: Determine if $f(x)=3x^3-6$ has an inverse over its entire domain $\mathbb{R}$. If not, explain why.

1. Apply the algebraic one-to-one test: set $f(a)=f(b)$ to get $3a^3-6=3b^3-6$.
2. Simplify the equation: add 6 to both sides and divide by 3 to get $a^3=b^3$, which simplifies to $a=b$ for all real $a,b$.
3. Confirm with the horizontal line test: $f(x)=3x^3-6$ is a strictly increasing cubic function, so every horizontal line intersects it exactly once.
4. Conclusion: $f$ is one-to-one over $\mathbb{R}$, so it has an inverse over its entire domain.

Exam tip: On MCQ questions asking to identify an invertible function from a set of graphs, draw 2-3 test horizontal lines across each graph to quickly check for multiple intersections — don't rely solely on memory of function shapes.

3. Finding Inverses Algebraically and Verifying with Composition

Once we confirm a function is invertible (or restrict it to an invertible domain), we can find its inverse algebraically using a standard 4-step process. The core properties of inverse functions also let us verify that our calculated inverse is correct. The step-by-step process is:

1. Replace $f(x)$ with $y$
2. Swap the positions of $x$ and $y$ (this reflects the domain-range swap between $f$ and $f^{-1}$)
3. Solve the new equation for $y$
4. Replace $y$ with $f^{-1}(x)$, and set the domain of $f^{-1}$ equal to the range of the original $f$.

By the definition of inverse functions, any valid inverse must satisfy both composition identities: $$f(f^{-1}(x))=x\text{and}{f}^{-1}(f(x))=x$$ Checking these identities is the most reliable way to confirm you did the algebra correctly, especially on FRQ questions where partial credit is given for verification.

Worked Example

Problem: Find the inverse of $f(x)=\frac{3x+2}{x-4}$ for $x>4$, and verify your result.

1. Step 1: Replace $f(x)$ with $y$: $y=\frac{3x+2}{x-4}$.
2. Step 2: Swap $x$ and $y$: $x=\frac{3y+2}{y-4}$.
3. Step 3: Solve for $y$: Multiply both sides by $(y-4)$: $x(y-4)=3y+2⟹xy-4x=3y+2⟹xy-3y=4x+2⟹y(x-3)=4x+2⟹y=\frac{4x+2}{x-3}$.
4. Step 4: Set the domain: The range of $f(x)$ for $x>4$ is $y\ne 3$, so $f^{-1}(x)=\frac{4x+2}{x-3}$ for $x>3$.
5. Verify: $f(f^{-1}(x))=\frac{3\left(\frac{4x+2}{x-3}\right)+2}{\left(\frac{4x+2}{x-3}\right)-4}=\frac{12x+6+2x-6}{4x+2-4x+12}=\frac{14x}{14}=x$, confirming the inverse is correct.

Exam tip: Always verify your inverse with at least one composition identity on FRQ questions to earn full credit, and never forget to write the domain of the inverse to match the original function's range.

4. Graphical Properties of Inverse Functions

The graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ over the line $y=x$. This makes sense because swapping $x$ and $y$ (the step we do when finding inverses algebraically) is exactly what reflecting over $y=x$ does to a graph.

Three key useful properties that follow from this reflection are:

1. A point $(a,b)$ on the graph of $f$ corresponds to a point $(b,a)$ on the graph of $f^{-1}$, so we can find $f^{-1}(b)$ directly if we know $f(a)=b$, without calculating the entire inverse.
2. If $f$ is strictly increasing, $f^{-1}$ is also strictly increasing; if $f$ is strictly decreasing, $f^{-1}$ is strictly decreasing.
3. Almost all intersections of $f$ and $f^{-1}$ (for the AP Precalculus scope) lie on the line $y=x$.

Worked Example

Problem: The graph of the invertible function $f(x)$ passes through the points $(-3,0)$, $(0,5)$, $(2,-1)$, and $(4,2)$. Find the value of $f^{-1}(2)$.

1. By the reflection property, $f^{-1}(k)=a$ if and only if $f(a)=k$.
2. We need $f^{-1}(2)$, so we look for the input $a$ such that $f(a)=2$.
3. From the given points, $f(4)=2$, so $a=4$.
4. Conclusion: $f^{-1}(2)=4$.

Exam tip: If you are asked for only a single value of $f^{-1}(k)$, never waste time calculating the entire inverse function. Just solve $f(x)=k$ for $x$ — that solution is your answer.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $f^{-1}(x)=\frac{1}{f(x)}$ when asked for the inverse of $f(x)$. Why: Students confuse inverse function notation $f^{-1}$ with exponent notation for reciprocals, where $x^{-1}=1/x$. Correct move: Memorize that $f^{-1}$ exclusively denotes the inverse function that reverses $f$, not the reciprocal, and confirm with the composition identity to avoid this mistake.
• Wrong move: Forgetting to restrict the original function's domain before inverting a non-one-to-one function, and forgetting to match the inverse's domain to the original's range. Why: Students focus on the algebraic inversion step and skip domain/range checks, which are required for full credit on FRQs. Correct move: After inverting, always set $\text{dom}(f^{-1})=\text{range}(f)$, where $f$ is the domain-restricted original function.
• Wrong move: Reflecting the graph of $f(x)$ over the x-axis ($y=0$) instead of $y=x$ to get $f^{-1}(x)$. Why: Students mix up reflection rules for different function transformations, confusing inverse reflection with vertical reflection. Correct move: Remember that swapping $x$ and $y$ maps $(a,b)$ to $(b,a)$, which only occurs when reflecting over $y=x$, not any other line.
• Wrong move: When finding the inverse of $f(x)=x^2$ for $x\ge 0$, writing the inverse as $f^{-1}(x)=\pm \sqrt{x}$. Why: Students remember that squaring has two square roots, and forget the original function is restricted to non-negative inputs, so the inverse must have non-negative outputs. Correct move: The range of $f^{-1}$ equals the domain of $f$, so we only take the non-negative root, giving $f^{-1}(x)=\sqrt{x}$ for $x\ge 0$.
• Wrong move: Only checking one composition identity to verify an inverse, and stopping, even for non-one-to-one functions. Why: Students assume that if one composition works, the other must, but one composition can equal $x$ over a restricted domain even if the inverse is incorrectly defined. Correct move: Check both composition identities for the full domain to confirm your inverse is correct.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

If $f(x)=4-3x^3$ is invertible, what is $f^{-1}(-20)$? A) $-2$ B) $2$ C) $\frac{1}{2}$ D) $-\frac{16}{3}$

Worked Solution: To find $f^{-1}(-20)$, we use the inverse function property: $f^{-1}(k)$ is the solution to $f(x)=k$, so we do not need to find the entire inverse. Set $4-3x^3=-20$, subtract 4 from both sides: $-3x^3=-24$, so $x^3=8$, which gives $x=2$. This means $f(2)=-20$, so $f^{-1}(-20)=2$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=x^2+6x+10$, with domain restriction $x\ge -3$. (a) Explain why $f(x)$ is invertible over this domain. (b) Find $f^{-1}(x)$ and state the domain of $f^{-1}$. (c) Verify your inverse using the composition identity $f(f^{-1}(x))=x$.

Worked Solution: (a) Rewrite $f(x)$ by completing the square: $f(x)=(x+3{)}^2+1$. For $x\ge -3$, $f(x)$ is strictly increasing, which means it is one-to-one, passes the horizontal line test, and is therefore invertible. (b) To find the inverse: set $y=(x+3{)}^2+1$, swap $x$ and $y$: $x=(y+3{)}^2+1$. Rearrange: $(y+3{)}^2=x-1$. Since the range of $f^{-1}$ is $y\ge -3$, we take the positive square root: $y+3=\sqrt{x-1}⟹f^{-1}(x)=\sqrt{x-1}-3$. The range of the original $f(x)$ is $[1,\infty )$, so the domain of $f^{-1}(x)$ is $[1,\infty )$. (c) Verify the composition: $$f(f^{-1}(x))=f\left(\sqrt{x-1}-3\right)={\left(\sqrt{x-1}-3+3\right)}^2+1={\left(\sqrt{x-1}\right)}^2+1=(x-1)+1=x$$ This confirms the inverse is correct.

Question 3 (Application / Real-World Style)

The kinetic energy $E$ (in joules) of a 2 kg object moving in a straight line is given by $E(v)=v^2$, where $v$ is the speed of the object in meters per second, for $0\le v\le 20$. Explain why the inverse function $v(E)$ exists, find $v(E)$, and use it to find the speed of the object when its kinetic energy is 64 joules.

Worked Solution: $E(v)=v^2$ is strictly increasing for $v\ge 0$ on the given domain, so it is one-to-one and the inverse exists. To find the inverse: set $E=v^2$, swap $E$ and $v$, then solve for the new $v$: $v=E^2⟹E=\sqrt{v}$ (we take the positive root because speed is non-negative), so $v(E)=\sqrt{E}$ for $0\le E\le 400$. Substitute $E=64$: $v(64)=\sqrt{64}=8$ m/s. In context, a 2 kg object with 64 joules of kinetic energy is moving at 8 meters per second.

7. Quick Reference Cheatsheet

8. What's Next

Inverse functions are the foundational prerequisite for the next core topic in Unit 2: logarithms. By definition, a logarithm is the inverse function of an exponential function, so every key property of logarithms comes directly from the inverse function properties we covered in this chapter. Without mastering how to check for invertibility, find inverses, and use the domain-range swap property, you will not be able to correctly simplify logarithmic expressions, solve exponential and logarithmic equations, or interpret logarithmic scales in real-world contexts. Beyond Unit 2, inverse functions are also critical for inverse trigonometric functions (covered in Unit 3 of AP Precalculus) and for understanding derivatives of inverse functions in future AP Calculus courses.

• Logarithms and their properties
• Solving exponential and logarithmic equations
• Inverse trigonometric functions