Graphs of a function and its inverse

6 minAS Level1.2.4

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can sketch $y = f(x)$ and $y = f^{-1}(x)$ on the same axes, explain why they are reflections of each other in $y = x$ , and find where they meet.

Swap the coordinates, flip the graph

Start with one point. If $f(2) = 7$ , then the point $(2, 7)$ is on the graph $y = f(x)$ .

The inverse undoes it: $f^{-1}(7) = 2$ . So the point $(7, 2)$ is on the graph $y = f^{-1}(x)$ .

That happens to every point:

$(a, b)$ on one graph becomes $(b, a)$ on the other.
Swapping the coordinates is exactly what a reflection in the line $y = x$ does.

(The line $y = x$ is the diagonal through $(0, 0)$ , $(1, 1)$ , $(2, 2)$ , … where both coordinates are equal.)

Definition

Graph of an inverse function

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ .

Why (a, b) and (b, a) are mirror images

A mirror line must do two things to the segment joining a point and its image:

cross it at right angles, and
cut it exactly in half.

Check both for the segment from $(a, b)$ to $(b, a)$ .

Right angles. Its gradient is "change in $y$ over change in $x$ ":

$\frac{a - b}{b - a} = -1$

(The top is minus the bottom. This needs $a \neq b$ ; if $a = b$ , the point is already on the mirror.)

Two lines are perpendicular when their gradients multiply to $-1$ . The line $y = x$ has gradient $1$ , and $1 \times (-1) = -1$ , so they're perpendicular.

In half. The midpoint averages the coordinates:

$\left(\frac{a + b}{2}, \frac{a + b}{2}\right)$

Both coordinates are equal, so the midpoint sits on $y = x$ . It's a perfect mirror.

Quick check

CalculateQuestion 1 · Recall

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