In Algebra 1 you undid a linear function (Inverse of a Linear Function). Now you will invert many kinds of functions, using composition to see what "undo" means.
Key idea
If $f(a) = b$, then $f^{-1}(b) = a$: the inverse sends each output back to its input. To find a formula, write $y = f(x)$, swap $x$ and $y$, and solve for $y$. The graph of $f^{-1}$ is the mirror image of the graph of $f$ across the line $y = x$.
Why it works
An inverse reverses every input-output pair, so each point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$. Swapping the coordinates of every point is exactly a reflection across $y = x$. Solving for $y$ undoes the steps of $f$ in reverse order: $f(x) = 2x^3 - 5$ cubes, doubles, then subtracts 5, so $f^{-1}$ adds 5, halves, then takes the cube root.
Worked examples
Worked example: Undoing in reverse order
Find the inverse of $f(x) = 2x^3 - 5$.
- Write $y = 2x^3 - 5$ and swap: $x = 2y^3 - 5$.
- Add 5: $x + 5 = 2y^3$. Divide by 2: $y^3 = \frac{x + 5}{2}$.
- Take the cube root: $y = \sqrt[3]{\frac{x + 5}{2}}$.
- Check: $f(2) = 11$ and $f^{-1}(11) = \sqrt[3]{8} = 2$.
Answer: $f^{-1}(x) = \sqrt[3]{\frac{x + 5}{2}}$
Worked example: A value without a formula
$f(x) = 3x - 7$. Find $f^{-1}(8)$.
- $f^{-1}(8)$ is the input that $f$ sends to 8. Solve $3x - 7 = 8$.
- $3x = 15$, so $x = 5$. Check: $f(5) = 8$.
Answer: $f^{-1}(8) = 5$
Worked example: A fraction
Find the inverse of $f(x) = \frac{x + 1}{x - 2}$.
- Swap: $x = \frac{y + 1}{y - 2}$. Multiply by $y - 2$: $xy - 2x = y + 1$.
- Collect the $y$ terms: $xy - y = 2x + 1$.
- Factor out $y$: $y(x - 1) = 2x + 1$, so $y = \frac{2x + 1}{x - 1}$.
Answer: $f^{-1}(x) = \frac{2x + 1}{x - 1}$
Common mistake
Reading $f^{-1}(x)$ as $\frac{1}{f(x)}$. Here the $-1$ means "undo," not "one over." For $f(x) = 2x$, the inverse is $\frac{x}{2}$, while $\frac{1}{f(x)} = \frac{1}{2x}$. Check with a pair: $f(2) = 4$, so $f^{-1}(4)$ must be 2. $\frac{4}{2} = 2$ works; $\frac{1}{8}$ does not.