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What is the inverse of a function, and how do you find and verify it?

Topic 2.8 Inverse Functions: determine whether a function has an inverse, find the inverse by swapping input and output, and verify an inverse using composition and the reflection over the line y = x.

A focused answer to AP Precalculus Topic 2.8, covering one-to-one functions and the horizontal line test, finding an inverse by swapping variables, verifying with composition, and the reflection over y = x.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. When a function has an inverse
  3. Finding an inverse
  4. Verifying with composition
  5. Reflection over y = x
  6. Try this

What this topic is asking

The College Board (Topic 2.8) wants you to work with inverse functions. An inverse f−1f^{-1} undoes ff: it swaps inputs and outputs. You must decide whether a function has an inverse (it must be one-to-one, passing the horizontal line test), find the inverse by swapping and solving, and verify it using composition or the reflection over the line y=xy = x.

When a function has an inverse

Functions that are not one-to-one (such as x2x^2 over all reals) can still be inverted on a restricted domain where they are one-to-one.

Finding an inverse

The swap-and-solve routine works for any one-to-one function, including the exponential functions whose inverses are logarithms (Topics 2.10 and 2.11).

Verifying with composition

Reflection over y = x

The graph of f−1f^{-1} is the mirror image of the graph of ff across the line y=xy = x. Each point (a,b)(a, b) on ff corresponds to (b,a)(b, a) on f−1f^{-1}. This reflection explains why the domain and range swap, and it gives a quick visual check: if you reflect a function's graph over y=xy = x and the result is also a function (passes the vertical line test), the original was one-to-one and the reflection is its inverse.

A distinction worth stating is that the inverse function f−1f^{-1} is not the reciprocal 1f\frac{1}{f}. The notation f−1f^{-1} means "the function that undoes ff", not "ff raised to the power −1-1". For f(x)=2x+6f(x) = 2x + 6, the inverse is x−62\frac{x - 6}{2}, while the reciprocal is 12x+6\frac{1}{2x + 6}, a completely different function. Keeping these apart is the single most common source of error on inverse questions, and it matters even more for exponentials, whose inverse is a logarithm rather than a reciprocal.

Try this

Q1. Does f(x)=x2f(x) = x^2 on all real numbers have an inverse? Why or why not? [1 point]

  • Cue. No: it fails the horizontal line test (for example f(2)=f(−2)=4f(2) = f(-2) = 4), so it is not one-to-one over all reals.

Q2. A point (3,7)(3, 7) lies on ff. What point must lie on f−1f^{-1}? [1 point]

  • Cue. (7,3)(7, 3): the inverse swaps the coordinates.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). What is the inverse of f(x)=2x+6f(x) = 2x + 6? (A) f−1(x)=x−62f^{-1}(x) = \frac{x - 6}{2} (B) f−1(x)=x2+6f^{-1}(x) = \frac{x}{2} + 6 (C) f−1(x)=2x−6f^{-1}(x) = 2x - 6 (D) f−1(x)=12x+6f^{-1}(x) = \frac{1}{2x + 6}
Show worked answer →

The correct answer is (A), f−1(x)=x−62f^{-1}(x) = \frac{x - 6}{2}.

Swap xx and yy in y=2x+6y = 2x + 6 to get x=2y+6x = 2y + 6, then solve for yy: x−6=2yx - 6 = 2y, so y=x−62y = \frac{x - 6}{2}. Choice (D) is the reciprocal of ff, which is not the inverse; the inverse undoes the function, it does not invert its value.

AP 2024 (style)3 marksSection II (free response, calculator allowed). Let f(x)=x3−1f(x) = x^3 - 1. (a) Find f−1(x)f^{-1}(x). (b) Verify your inverse by computing f(f−1(x))f(f^{-1}(x)).
Show worked answer →

A 3-point question on finding and verifying an inverse.

(a) Inverse (2 points): set y=x3−1y = x^3 - 1, swap to x=y3−1x = y^3 - 1, solve y3=x+1y^3 = x + 1, so f−1(x)=x+13f^{-1}(x) = \sqrt[3]{x + 1}.
(b) Verify (1 point): f(f−1(x))=(x+13)3−1=(x+1)−1=xf(f^{-1}(x)) = \left(\sqrt[3]{x + 1}\right)^3 - 1 = (x + 1) - 1 = x. Since the composition returns xx, the inverse is confirmed.

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