The graphs of f and f⁻¹

Reflect in y = x, find where they meet

In 3 of 20 recent exams
Paper 1 or 2
2 to 4 marks

How it is set in the exam

Paper 1 (no calculator) (a sketch of $f^{-1}$ , or the meeting points by hand) and Paper 2 (calculator) (the meeting points on the GDC), 2–4 marks (Example 5.2(b), at 5, is at the long end). Typical (3 of 20 exam versions: twice on Paper 1 (no calculator), once on Paper 2 (calculator)). The meeting points are also the limits of the area between the graphs of $f$ and $f^{-1}$ in three more exam versions, all on Paper 2 (calculator), and a Paper 2 (calculator) question in a fourth asks for the point of a self-inverse graph nearest the origin, which in that question was where the graph meets $y = x$ .

Spot it in the exam

Exam phrases to spot (4)

“on the same axes, sketch the graph of $f^{-1}$ ”
“the graphs of $f$ and $f^{-1}$ intersect at the points $A$ and $B$ ; find their coordinates”
“solve $f(x) = f^{-1}(x)$ ”
“find the distance between the points where the graphs of $f$ and $f^{-1}$ meet”

In disguise (1)

an equation whose two sides are inverses of each other, not named as such, such as $x^2 - 4 = \sqrt{x + 4}$ for $x \ge 0$ : it asks where the graph of $x^2 - 4$ meets the graph of its inverse

Variants

a sketch of $f^{-1}$ from a given graph of $f$ (increasing or decreasing), with the intercepts, the endpoints and a point on $y = x$ , then the domain and range of $f^{-1}$ ; the meeting points found exactly (a rational function gives a quadratic); the meeting points on the GDC (an exponential and a log), then the distance between them.

Looks like it, but isn't (5)

the formula for $f^{-1}$ , or a constant from "given that $f^{-1}(x) = f(x)$ ": Find the inverse function, with its domain
$f(x) = g(x)$ where $g$ is not the inverse of $f$ : Solve graphically: intersections, how many solutions, above or below a level
a reflection in the $x$ -axis or the $y$ -axis: Transform a graph: new equation, image points, sketch, range
the area of the region enclosed by the graphs of $f$ and $f^{-1}$ : Area between two curves, with the meeting points found here as its limits
the point of a self-inverse graph nearest the origin: Optimisation: use the constraint, then optimise (minimise the distance on the GDC; the symmetry in $y = x$ pairs up the nearest points, but does not by itself put them on the line)

Worked example

Sketch f⁻¹ from the graph of f

Paper 1 (no calculator)

Question

The graph of $y = f(x)$ , for $-2 \le x \le 10$ , is an increasing curve that bends downwards (it is concave down). The table lists the points marked on it.

Point	On the graph
$(-2, 0)$	the left endpoint, on the $x$ -axis
$(0, 2)$	the $y$ -intercept
$(4, 4)$	on the line $y = x$
$(10, 6)$	the right endpoint

Point	$(-2, 0)$	$(0, 2)$	$(4, 4)$	$(10, 6)$
On the graph	the left endpoint, on the $x$ -axis	the $y$ -intercept	on the line $y = x$	the right endpoint

To see the picture: plot $y = \sqrt{4x + 9} - 1$ for $-2 \le x \le 10$ . The exam gives only the graph.

(a) On the same axes, sketch the graph of $y = f^{-1}(x)$ . [3]
(b) Write down the domain and the range of $f^{-1}$ . [2]

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Steps

Reflect:
each point $(a, b)$ on the graph of $f$ gives the point $(b, a)$ on the graph of $f^{-1}$ . Swap the coordinates of the key points: the intercepts, the endpoints and any other marked point. A point on $y = x$ stays where it is, and an asymptote $x = a$ becomes $y = a$ (and $y = b$ becomes $x = b$ ). An increasing graph stays increasing and its bend flips (concave down becomes concave up); a decreasing graph stays decreasing and keeps its bend (concave up stays concave up). The shape, with both swapped intercepts, earns the first sketch mark, and the other marks count only if it is earned.
2
Meet:
solve $f(x) = f^{-1}(x)$ .
- If $f$ is increasing on its whole domain, the graphs can only meet on $y = x$ , so $f(x) = x$ gives every meeting point. It is quicker, and markschemes accept it. Increasing on each branch separately is not enough.
- If $f$ is decreasing, or increasing only on each branch, the graphs can also meet off the line, so solve $f(x) = f^{-1}(x)$ itself.
- The exception: if $f(x) = \frac{ax + b}{cx + d}$ and $f$ is not its own inverse, $f(x) = x$ always gives the same equation as $f(x) = f^{-1}(x)$ , whether $f$ is increasing or decreasing (Example 5.2), and markschemes accept it.
- On , use zeros and intersections, and look for every meeting point.
Coordinates:
a meeting point on $y = x$ is $(a, a)$ . Keep only solutions in the domain of $f$ (and of $f^{-1}$ ): for $x^2 - 4$ with $x \ge 0$ , $x^2 - 4 = x$ gives $x = \frac{1 \pm \sqrt{17}}{2}$ , and the negative one is rejected.
Distance
between $(a, a)$ and $(b, b)$ : $\sqrt{(b - a)^2 + (b - a)^2} = \sqrt{2}\,\lvert b - a \rvert$ . Both $x$ -values must be seen first.

Common mistakes

Reflecting in an axis
instead of in $y = x$ .
"A reflection" with no line
say "a reflection in the line $y = x$ " (The idea of an inverse function).
Intercepts not swapped
the $y$ -intercept of $f$ becomes the $x$ -intercept of $f^{-1}$ , and the other way round.

More mistakes (6)

The wrong bend
reflecting in $y = x$ flips the bend of an increasing curve but keeps the bend of a decreasing one. With the wrong shape the first sketch mark is lost, and the others with it.
Using $f(x) = x$ for a decreasing $f$ , or one increasing only on each branch, that is not linear over linear
it can miss meeting points off the line. $f(x) = -\frac{1}{x^3}$ never meets $y = x$ , but it meets its inverse at $(1, -1)$ and $(-1, 1)$ .
Only one meeting point from the GDC
the mark for the points is lost, and so is the distance. Widen the window.
Only $x$ -values, or an extra root
coordinates need both values, and a root outside the domain of $f$ must go.
Rounding before the distance
use the stored values.
Sketching $f^{-1}$ beyond its domain
it starts and stops where the swapped endpoints of $f$ are.

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