To run a function backwards, fold the page.

An inverse function undoes what the function did. On a graph that turns out to be a reflection, and the reflection shows exactly when an inverse can exist.

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This is $y = x^2 - 4x + 7$. It takes $3$ to $4$, so the point $(3, 4)$ is on it. Running it backwards takes $4$ back to $3$: the point $(4, 3)$. Swapping the coordinates is a reflection in the line $y = x$.

Fold the whole curve and the result fails. The line $x = 4$ meets it twice, because $f(1)$ and $f(3)$ are both $4$: running backwards, the input $4$ would need two outputs.

Keep only $x \geq 2$, the half that starts at the vertex $(2, 3)$. Now no two inputs share an output, and its mirror image is a function: the inverse $f^{-1}$.

Every point swaps its coordinates, so the domain and range swap too. $f$ has domain $x \geq 2$ and range $f(x) \geq 3$; $f^{-1}$ has domain $x \geq 3$ and range $f^{-1}(x) \geq 2$. The endpoints $(2, 3)$ and $(3, 2)$ are mirror images.

Swap x and y. Check it's still a function.

Domain and range, composite functions, the inverse of a fraction, and twelve graded exam questions with full worked solutions are waiting.

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