A function takes an input to an output. Its inverse runs the machine backward: it takes the output back to the input. For linear models, inverses answer questions like “how long did the job take if it cost \$200?”. Finding one uses the skill of solving for a variable.
Why it works
If $f(x) = 2x + 3$, then $f$ multiplies by $2$ and then adds $3$. To undo it, do the opposite operations in the opposite order: subtract $3$, then divide by $2$. So $f^{-1}(x) = \frac{x - 3}{2}$. Algebraically, swapping the roles of $x$ and $y$ in $y = 2x + 3$ and solving for $y$ does exactly this. On a graph, swapping $x$ and $y$ reflects every point across the line $y = x$: if $(1, 5)$ is on $f$, then $(5, 1)$ is on $f^{-1}$.
Key idea
To find $f^{-1}$: write $y = f(x)$, swap $x$ and $y$, and solve for $y$. Check that $f(f^{-1}(x)) = x$.
Worked examples
Worked example: Find an inverse
Find the inverse of $f(x) = 4x - 7$.
- $y = 4x - 7$. Swap: $x = 4y - 7$.
- Add $7$: $x + 7 = 4y$. Divide by $4$: $y = \frac{x + 7}{4}$.
- Check: $f\left(\frac{x + 7}{4}\right) = 4 \cdot \frac{x + 7}{4} - 7 = x$.
Answer: $f^{-1}(x) = \frac{x + 7}{4}$
Worked example: Use an inverse in context
A plumber's charge is $C(h) = 40h + 60$. Find $C^{-1}(200)$ and say what it means.
- $C^{-1}(200)$ is the number of hours that cost \$200: solve $40h + 60 = 200$.
- $40h = 140$, so $h = 3.5$.
Answer: $3.5$ hours: a \$200 bill means three and a half hours of work.
Common mistake
Taking the inverse of $f(x) = 2x + 3$ to be $\frac{1}{2x + 3}$. That is the reciprocal, $1$ divided by the output, which does not undo $f$. Check: $f(1) = 5$, so the inverse must send $5$ back to $1$. $\frac{5 - 3}{2} = 1$ works, but $\frac{1}{5}$ does not.