# Inverse functions Canonical: https://duckyhelper.com/learn/algebra-2/inverse-functions/ Updated: 2026-10-01 An inverse function undoes a function: if \(f(2) = 7\), then \(f^{-1}(7) = 2\). To find it, write \(y = f(x)\), swap x and y, and solve for y. The graph of \(f^{-1}\) is the graph of f reflected over the line \(y = x\). Check your answer by composing: \(f(f^{-1}(x))\) should equal x. Only one-to-one functions have inverses, so sometimes you restrict the domain first. ## The key idea A function sends inputs to outputs. Its inverse sends each output back to the input it came from. So inputs and outputs trade places, and every point \((a, b)\) on f becomes \((b, a)\) on \(f^{-1}\). $$ f(a) = b \iff f^{-1}(b) = a, \qquad f\big(f^{-1}(x)\big) = x = f^{-1}\big(f(x)\big) $$ The \(-1\) is not an exponent here. \(f^{-1}(x)\) means the inverse function, not \(\frac{1}{f(x)}\). ### Steps to find an inverse 1. Replace \(f(x)\) with y. 2. Swap x and y. 3. Solve the new equation for y. 4. Replace y with \(f^{-1}(x)\) and state any domain limits. 5. Check that \(f(f^{-1}(x)) = x\). ## Worked examples **Example 1: a linear function** Problem: Find the inverse of \(f(x) = 3x - 5\) and check it. 1. Write y for \(f(x)\). $$ y = 3x - 5 $$ 2. Swap x and y. $$ x = 3y - 5 $$ 3. Add 5 to both sides. $$ x + 5 = 3y $$ 4. Divide by 3. $$ y = \frac{x + 5}{3} $$ 5. Check: put \(\frac{x + 5}{3}\) into f. $$ 3 \cdot \frac{x + 5}{3} - 5 = x $$ Answer: \(f^{-1}(x) = \frac{x + 5}{3}\) **Example 2: a rational function** Problem: Find the inverse of \(f(x) = \frac{2x + 1}{x - 3}\). 1. Write \(y = \frac{2x + 1}{x - 3}\), then swap x and y. $$ x = \frac{2y + 1}{y - 3} $$ 2. Multiply both sides by \(y - 3\). $$ x(y - 3) = 2y + 1 $$ 3. Distribute. $$ xy - 3x = 2y + 1 $$ 4. Get every y term on one side and everything else on the other. $$ xy - 2y = 3x + 1 $$ 5. Factor out y. $$ y(x - 2) = 3x + 1 $$ 6. Divide by \(x - 2\). $$ y = \frac{3x + 1}{x - 2} $$ Answer: \(f^{-1}(x) = \frac{3x + 1}{x - 2}\), for \(x \ne 2\) **Example 3: restrict the domain first** Problem: \(f(x) = (x - 2)^2 + 1\) is not one-to-one, so restrict it to \(x \ge 2\). Find the inverse. 1. Swap x and y in \(y = (x - 2)^2 + 1\). $$ x = (y - 2)^2 + 1 $$ 2. Subtract 1. $$ x - 1 = (y - 2)^2 $$ 3. Take the square root. The outputs of the inverse are the old inputs, which were at least 2, so \(y - 2 \ge 0\) and only the positive root fits. $$ y - 2 = \sqrt{x - 1} $$ 4. Add 2. $$ y = 2 + \sqrt{x - 1} $$ Answer: \(f^{-1}(x) = 2 + \sqrt{x - 1}\), with domain \(x \ge 1\) **Example 4 (test-hard): an inverse value without the formula** Problem: Let \(f(x) = x^3 + x + 1\). Find \(f^{-1}(11)\). 1. \(f^{-1}(11)\) is the input that gives 11. Set up that equation instead of hunting for a formula. $$ x^3 + x + 1 = 11 $$ 2. Move 11 over. $$ x^3 + x - 10 = 0 $$ 3. Try small integers. \(x = 2\) works. $$ 2^3 + 2 - 10 = 0 $$ 4. f is always increasing (both \(x^3\) and x grow), so no other input gives 11. Answer: \(f^{-1}(11) = 2\) ## Common mistakes - **Writing \(\frac{1}{f(x)}\).** The inverse of \(3x - 5\) is not \(\frac{1}{3x - 5}\). Fix: swap x and y and solve. - **Forgetting to swap.** Solving \(y = 3x - 5\) for x gives the same function written backwards. Fix: swap first, then solve for y, so the answer is in terms of x. - **Leaving y on both sides.** In \(xy - 3x = 2y + 1\), collect every y term on one side and factor y out. - **Keeping both square roots.** With a restricted domain, only one sign makes sense. Fix: the outputs of \(f^{-1}\) must match the inputs you allowed for f. - **Skipping the check.** A quick composition catches most algebra slips. Fix: plug a number like \(x = 4\) into f, then into your inverse, and see if you get 4 back. ## Quick methods > **Tip: The horizontal line test** > > If any horizontal line hits a graph more than once, the function is not one-to-one and has no inverse on that domain. A parabola fails; a line that is not horizontal passes. This is a reliable visual check. > **Tip: Undo the steps in reverse** > > For simple functions, list what f does to x and undo it backwards. \(f(x) = 3x - 5\) multiplies by 3 then subtracts 5, so \(f^{-1}\) adds 5 then divides by 3: \(\frac{x + 5}{3}\). ## Practice **5 practice questions** 1. What is the inverse of \(f(x) = \frac{x + 4}{2}\)? A. \(f^{-1}(x) = 2x - 4\) B. \(f^{-1}(x) = 2x + 4\) C. \(f^{-1}(x) = \frac{x - 4}{2}\) D. \(f^{-1}(x) = \frac{2}{x + 4}\) Answer: \(f^{-1}(x) = 2x - 4\). Swap: \(x = \frac{y + 4}{2}\). Multiply by 2: \(2x = y + 4\), so \(y = 2x - 4\). \(\frac{2}{x + 4}\) is the reciprocal, which is not the inverse. 2. f is one-to-one and \(f(3) = 10\). Which point must be on the graph of \(f^{-1}\)? A. \((3, 10)\) B. \((10, 3)\) C. \((-3, -10)\) D. \((3, -10)\) Answer: \((10, 3)\). \((3, 10)\) is on f, and the inverse swaps inputs and outputs, so \((10, 3)\) is on \(f^{-1}\). That is the reflection of \((3, 10)\) over \(y = x\). 3. Which function does NOT have an inverse function on its whole domain? A. \(f(x) = 2x + 7\) B. \(g(x) = x^3\) C. \(h(x) = x^2 + 1\) D. \(k(x) = \sqrt{x}\) Answer: \(h(x) = x^2 + 1\). \(h(2) = h(-2) = 5\), so two inputs share an output and h fails the horizontal line test. The other three never repeat an output. 4. What is the inverse of \(f(x) = x^3 - 1\)? A. \(f^{-1}(x) = \sqrt[3]{x + 1}\) B. \(f^{-1}(x) = \sqrt[3]{x} + 1\) C. \(f^{-1}(x) = \frac{1}{x^3 - 1}\) D. \(f^{-1}(x) = \sqrt[3]{x - 1}\) Answer: \(f^{-1}(x) = \sqrt[3]{x + 1}\). Swap: \(x = y^3 - 1\), so \(y^3 = x + 1\) and \(y = \sqrt[3]{x + 1}\). Adding 1 after the cube root undoes the steps in the wrong order. 5. If \(f(x) = 5x - 7\), what is \(f^{-1}(18)\)? Answer: 5. Find the input that gives 18: \(5x - 7 = 18\), so \(5x = 25\) and \(x = 5\). Check: \(f(5) = 25 - 7 = 18\). ## Frequently asked questions ### Is f^-1(x) the same as 1/f(x)? No. \(f^{-1}\) is the function that undoes f, while \(\frac{1}{f(x)}\) is the reciprocal of the output. For \(f(x) = 2x\), the inverse is \(\frac{x}{2}\), but the reciprocal is \(\frac{1}{2x}\). The notation is confusing, so read it from context. ### How do I check that two functions are inverses? Compose them both ways. If \(f(g(x)) = x\) and \(g(f(x)) = x\) for every allowed x, they are inverses. A quick spot check: put a number into one, put the result into the other, and you should get your number back. ### What does one-to-one mean? Every output comes from exactly one input. If two different inputs give the same output, like \(x^2\) at 3 and \(-3\), the inverse would not know which input to return. Restricting the domain, for example to \(x \ge 0\), fixes that. ### How are the graphs of a function and its inverse related? They are mirror images over the line \(y = x\), because every point \((a, b)\) becomes \((b, a)\). If you fold the graph paper along \(y = x\), the two graphs land on top of each other. ## Related - [Transformations of functions](https://duckyhelper.com/learn/algebra-2/function-transformations/) - [Logarithms](https://duckyhelper.com/learn/algebra-2/logarithms/) - [Functions, function notation, domain and range](https://duckyhelper.com/learn/algebra-1/functions/) - [Functions and function notation on the SAT](https://duckyhelper.com/learn/sat-math/functions/) - [Algebra 2 study guides](https://duckyhelper.com/learn/algebra-2/) ## Try asking Ducky - "Why do we swap x and y? It feels like cheating." - "Can you check my inverse by composing it with the original?" - "Why does only the positive square root work in the restricted one?" ## Get DuckyHelper Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. [Try it free in your browser](https://app.duckyhelper.com/?utm_source=duckyhelper.com&utm_medium=learn) or [Get the Mac app](https://duckyhelper.com/download/)