Mathematics 9709 · Pure Mathematics 1

The graph of a function and its inverse

You never need the formula for $\mathrm{f}^{-1}$ to draw it.
Its graph is just the graph of $\mathrm{f}$ flipped over one line. Get that line right and you get the sketch mark.

The graph of $\mathrm{f}^{-1}$ is the graph of $\mathrm{f}$ reflected in the line $y=x$ . No algebra needed. Just reflect it.

Reflecting in $y=x$ turns the graph on its side. So the graph's width (the domain) and height (the range) swap. That is the whole domain $\leftrightarrow$ range rule, all in one picture.

Reflect in y = x: here's why

Look at the curve below. $\mathrm{f}$ is in blue. $\mathrm{f}^{-1}$ is in green. The dashed diagonal is the mirror line $y=x$ . The two curves are exact reflections of each other in that line.

f in blue and its inverse in green: each point (a, b) on f goes to (b, a) on the inverse, a mirror image in the dashed line y = x.

The short reason: finding an inverse means swapping $x$ and $y$ . So every point $(a,b)$ on $\mathrm{f}$ becomes $(b,a)$ on $\mathrm{f}^{-1}$ . That is all the reflection is.

Why is swapping the coordinates the same as reflecting in y = x?

Take any point $(a,b)$ and its swapped partner $(b,a)$ . The line joining them has gradient $\dfrac{a-b}{b-a}=-1$ . So it is at right angles to $y=x$ (which has gradient $1$ ). Its midpoint is $\left(\tfrac{a+b}{2},\tfrac{a+b}{2}\right)$ , which sits on $y=x$ . So a point and its partner are on opposite sides of the line, the same distance away, at right angles. That is what “reflect in $y=x$ ” means. Do it to every point and the whole curve flips.

Two useful facts

The domain and range swap. The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ , and the other way round too. You can read both off the picture.
Say $\mathrm{f}$ equals its own inverse (a self-inverse function). Then its graph is the same on both sides of $y=x$ , because reflecting it does not change it. The $\mathrm{f}(x)=\frac{x+3}{x-1}$ from the last topic is one of these.

Worked example

Sketch $\mathrm{f}(x)=(x-1)^2-2$ for $1\le x\le 4$ , and its inverse

On $x\ge 1$ this is the right half of a parabola (the vertex comes from the completed-square form, from the Quadratics chapter): it rises from $(1,-2)$ up to $(4,7)$ , and is one-one, so an inverse exists.
To draw $\mathrm{f}^{-1}$ , reflect that curve in $y=x$ : the endpoints swap coordinates, $(1,-2)\to(-2,1)$ and $(4,7)\to(7,4)$ .
Plot the mirror image through those two swapped points, and draw the line $y=x$ so the symmetry is visible.

Answer

\mathrm{f}^{-1}\ \text{is the reflection of}\ \mathrm{f}\ \text{in}\ y=x

Where the marks go

Common mistake

The mirror line is

y=x

, not the

x

-axis or the

y

-axis. A neat diagonal at

45^\circ

through the origin gets the mark. Reflecting in an axis loses it. This is the most common mistake in the topic.

Common mistake

When asked to sketch

\mathrm{f}

and

\mathrm{f}^{-1}

on one diagram, draw the line $y=x$ and mark a swapped pair of end points. That makes the link “clear”, which is what gets the marks. (9709-style)

Now you try

A function $\mathrm{f}$ is one-one, and the point $(-7,-4)$ lies on its graph. Write down a point that must lie on the graph of $\mathrm{f}^{-1}$ , and state the mirror line that maps one graph onto the other. (9709-style)

Your turn— tap to reveal the worked answer (reflect in y = x)

Reflecting in $y=x$ swaps the coordinates, so $(-7,-4)$ on $\mathrm{f}$ becomes $(-4,-7)$ on $\mathrm{f}^{-1}$ .

Answer

(-4,\,-7)\ \text{on}\ \mathrm{f}^{-1};\ \ \text{mirror line}\ y=x

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The graph of a function and its inverse

You never need the formula for $\mathrm{f}^{-1}$ to draw it.
Its graph is just the graph of $\mathrm{f}$ flipped over one line. Get that line right and you get the sketch mark.

The graph of $\mathrm{f}^{-1}$ is the graph of $\mathrm{f}$ reflected in the line $y=x$ . No algebra needed. Just reflect it.

Reflecting in $y=x$ turns the graph on its side. So the graph's width (the domain) and height (the range) swap. That is the whole domain $\leftrightarrow$ range rule, all in one picture.

Reflect in y = x: here's why

Look at the curve below. $\mathrm{f}$ is in blue. $\mathrm{f}^{-1}$ is in green. The dashed diagonal is the mirror line $y=x$ . The two curves are exact reflections of each other in that line.

f in blue and its inverse in green: each point (a, b) on f goes to (b, a) on the inverse, a mirror image in the dashed line y = x.

The short reason: finding an inverse means swapping $x$ and $y$ . So every point $(a,b)$ on $\mathrm{f}$ becomes $(b,a)$ on $\mathrm{f}^{-1}$ . That is all the reflection is.

Why is swapping the coordinates the same as reflecting in y = x?

Two useful facts

The domain and range swap. The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ , and the other way round too. You can read both off the picture.
Say $\mathrm{f}$ equals its own inverse (a self-inverse function). Then its graph is the same on both sides of $y=x$ , because reflecting it does not change it. The $\mathrm{f}(x)=\frac{x+3}{x-1}$ from the last topic is one of these.

Worked example

Sketch $\mathrm{f}(x)=(x-1)^2-2$ for $1\le x\le 4$ , and its inverse

On $x\ge 1$ this is the right half of a parabola (the vertex comes from the completed-square form, from the Quadratics chapter): it rises from $(1,-2)$ up to $(4,7)$ , and is one-one, so an inverse exists.
To draw $\mathrm{f}^{-1}$ , reflect that curve in $y=x$ : the endpoints swap coordinates, $(1,-2)\to(-2,1)$ and $(4,7)\to(7,4)$ .
Plot the mirror image through those two swapped points, and draw the line $y=x$ so the symmetry is visible.

Answer

\mathrm{f}^{-1}\ \text{is the reflection of}\ \mathrm{f}\ \text{in}\ y=x

Where the marks go

Common mistake

The mirror line is

y=x

, not the

x

-axis or the

y

-axis. A neat diagonal at

45^\circ

through the origin gets the mark. Reflecting in an axis loses it. This is the most common mistake in the topic.

Common mistake

When asked to sketch

\mathrm{f}

and

\mathrm{f}^{-1}

on one diagram, draw the line $y=x$ and mark a swapped pair of end points. That makes the link “clear”, which is what gets the marks. (9709-style)

Now you try

Your turn— tap to reveal the worked answer (reflect in y = x)

Reflecting in $y=x$ swaps the coordinates, so $(-7,-4)$ on $\mathrm{f}$ becomes $(-4,-7)$ on $\mathrm{f}^{-1}$ .

Answer

(-4,\,-7)\ \text{on}\ \mathrm{f}^{-1};\ \ \text{mirror line}\ y=x