Graphs of Inverse Functions

Intuition

In the Functions chapter you found inverses with algebra: write $y = f(x)$ , swap $x$ and $y$ , and solve for $y$ . That instruction to swap $x$ and $y$ was pure symbol pushing, but it hides a picture. Every inverse you build has a graph, and that graph is not something new to plot point by point. It is the graph of $f$ you already have, flipped across one special line, the diagonal $y = x$ . Fold the plane along that diagonal like a sheet of paper and the graph of $f$ lands exactly on the graph of $f^{-1}$ . This lesson derives why the fold works and shows how to sketch an inverse straight off a graph with no formula at all. It explains in pictures why the domain and range trade places, and why a function must pass the horizontal line test before it can have an inverse.

Learning goals

Reflect a graph across $y = x$ to draw its inverse
Swap the coordinates of any point you can read off
Trade domain for range when the axes swap
Recognize a fixed point where the graph meets the diagonal
Turn the horizontal line test into the vertical one by reflecting
Restrict to a one-to-one branch before inverting $y = x^2$

The swap of coordinates is a reflection

Recall the one idea the whole inverse rests on: $f^{-1}$ reverses $f$ by swapping each input with its output. On a graph that swap has a precise meaning. A point $(a, b)$ on the graph of $f$ records the single fact that the input $a$ produces the output $b$ , that is, $b = f(a)$ . The inverse reverses that fact: it takes the input $b$ back to the output $a$ , which is the point $(b, a)$ on the graph of $f^{-1}$ . So each point of $f$ has a partner on $f^{-1}$ with the two coordinates switched.

Written as a chain of equivalent statements, the correspondence is exact:

\begin{aligned} (a, b) \text{ on the graph of } f &\iff b = f(a) \iff a = f^{-1}(b) \\ &\iff (b, a) \text{ on the graph of } f^{-1}. \end{aligned}

The middle step is just the definition of the inverse: $b = f(a)$ says exactly the same thing as $a = f^{-1}(b)$ , because applying $f^{-1}$ to both sides of $b = f(a)$ gives $f^{-1}(b) = a$ . Read end to end, the chain says a point is on the graph of $f$ exactly when its coordinate-swap is on the graph of $f^{-1}$ .

That leaves one geometric question: what does switching the two coordinates of a point do to its position in the plane? It reflects the point across the line $y = x$ . The line $y = x$ is the set of points whose two coordinates are equal, the perfect diagonal through the origin at forty-five degrees. Swapping the coordinates of a point that is not on that line carries it to the mirror-image spot on the other side, the same distance away.

Swapping the coordinates of a point reflects it across the dashed line y = x. The point (a, b) and its swap (b, a) sit the same distance from the line on opposite sides, and the segment joining them is cut in half at a right angle by y = x, which is exactly what it means for the line to be their mirror.

Swapping coordinates reflects a point across $y = x$ #

Take a point $P = (a, b)$ with $a \ne b$ , so $P$ is not on the diagonal, and let $Q = (b, a)$ be its coordinate-swap. To show the line $y = x$ is the mirror that sends $P$ to $Q$ , it is enough to show that line is the perpendicular bisector of the segment $PQ$ . The perpendicular bisector is the line that meets $PQ$ at a right angle through its midpoint.

First, the midpoint of $PQ$ is

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

whose two coordinates are equal, so the midpoint lies on $y = x$ .

Second, the segment $PQ$ runs from $(a, b)$ to $(b, a)$ , so its slope is

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

while the line $y = x$ has slope $1$ . Two slopes whose product is $-1$ mark perpendicular lines, and $(1)(-1) = -1$ , so $PQ$ crosses $y = x$ at a right angle.

The diagonal passes through the midpoint of $PQ$ and is perpendicular to it, so it is the perpendicular bisector of $PQ$ . Reflecting $P$ across $y = x$ therefore lands exactly on $Q$ . A point already on the diagonal has $a = b$ , so its swap $(b, a) = (a, b)$ is itself and it does not move. Either way, switching the coordinates of a point is the same as reflecting it across $y = x$ .

The graph of an inverse is the mirror image across y = x

Put the two halves together. Every point $(a, b)$ on the graph of $f$ corresponds to the point $(b, a)$ on the graph of $f^{-1}$ . Moving from $(a, b)$ to $(b, a)$ is a reflection across the line $y = x$ . Apply that reflection to every point at once, and the entire graph of $f$ is carried onto the entire graph of $f^{-1}$ . This is the central fact of the lesson.

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

Nothing about the reflection depends on having a formula for $f$ . Whatever the shape of $f$ , its inverse is the same shape flipped across the diagonal. If you can draw $f$ , you can draw $f^{-1}$ : reflect it across $y = x$ .

The line y = f(x) with slope 2 (dark) and its inverse (blue) with slope one half are mirror images across the dashed line y = x. The point (2, 3) on f reflects to (3, 2) on the inverse, and the short dashed segment joining them meets y = x at a right angle. Reading the inverse is reading f with the two coordinates swapped.

The figure shows it for a line. The function $f(x) = 2x - 1$ has a steep graph of slope $2$ ; its inverse $f^{-1}(x) = \dfrac{x + 1}{2}$ has the gentle graph of slope $\tfrac{1}{2}$ . The two lines are mirror images across the dashed diagonal. The point $(2, 3)$ sits on $f$ because $f(2) = 2(2) - 1 = 3$ , and its reflection $(3, 2)$ sits on $f^{-1}$ because $f^{-1}(3) = \dfrac{3 + 1}{2} = 2$ .

Worked example 1 Sketch the inverse of a line from its graph

The graph of $f(x) = 2x + 2$ is a line through $(-1, 0)$ , $(0, 2)$ , and $(1, 4)$ . Sketch the graph of $f^{-1}$ without first finding a formula for it.

By the reflection rule, each point of $f$ reflects across $y = x$ to a point of $f^{-1}$ with its coordinates swapped. Swap the coordinates of the three known points:

(-1, 0) \to (0, -1), \qquad (0, 2) \to (2, 0), \qquad (1, 4) \to (4, 1).

Plot $(0, -1)$ , $(2, 0)$ , and $(4, 1)$ and draw the line through them. That line is the graph of $f^{-1}$ , the mirror image of $f$ across the diagonal.

As a check, find the formula the algebraic way: from $y = 2x + 2$ , swap to $x = 2y + 2$ , and solve for $y = \dfrac{x - 2}{2}$ . Then

f^{-1}(0) = \frac{-2}{2} = -1, \qquad f^{-1}(2) = 0, \qquad f^{-1}(4) = 1,

which are exactly the three reflected points. The graph and the formula agree.

Reading an inverse straight off a graph

The reflection rule turns graphing an inverse into a mechanical task: take points you can read off $f$ and swap each one’s coordinates. You never need the formula for $f^{-1}$ , and often you never even need the formula for $f$ . That is what makes the graphical view so useful. A function given only by a curve, with no equation at all, still has an inverse you can plot point by point.

Worked example 2 Read four points of an inverse off a graph

A function $g$ is given only by its graph, a curve passing through $(-2, -3)$ , $(0, 1)$ , $(1, 3)$ , and $(3, 4)$ . The curve rises steadily from left to right, so it passes the horizontal line test and has an inverse. Find four points on the graph of $g^{-1}$ .

Swap the coordinates of each known point, since $g^{-1}$ is the reflection of $g$ across $y = x$ :

(-2, -3) \to (-3, -2), \quad (0, 1) \to (1, 0), \quad (1, 3) \to (3, 1), \quad (3, 4) \to (4, 3).

So $g^{-1}$ passes through $(-3, -2)$ , $(1, 0)$ , $(3, 1)$ , and $(4, 3)$ . Each new point is the mirror of an old one. For instance, $g(1) = 3$ becomes $g^{-1}(3) = 1$ : the input and output trade roles, and no formula was needed anywhere.

Check your understanding

The point $(-4, 7)$ lies on the graph of $f$ . Which point must lie on the graph of $f^{-1}$ ?

(7, -4)

(-7, 4)

(4, -7)

(-4, 7)

Domain and range trade places by swapping the axes

The reflection also explains, in a single picture, why the domain and range swap when you invert. Reflecting across $y = x$ does more than move the curve. It swaps the two axes themselves: the horizontal axis (every point $(t, 0)$ ) reflects to the vertical axis (every point $(0, t)$ ), and the reverse. So a horizontal extent of a graph becomes a vertical extent after the flip, and a vertical extent becomes horizontal.

Now recall how a graph shows domain and range. The domain is the horizontal spread of the curve, read along the $x$ -axis; the range is its vertical spread, read along the $y$ -axis. When the graph of $f$ reflects to the graph of $f^{-1}$ , its horizontal spread becomes the new curve’s vertical spread. Likewise the vertical spread of $f$ becomes the new curve’s horizontal spread. In other words,

the domain of $f$ (a horizontal spread) becomes the range of $f^{-1}$ (a vertical spread), and
the range of $f$ (a vertical spread) becomes the domain of $f^{-1}$ (a horizontal spread).

This is exactly the domain-range swap you proved algebraically for inverses, now visible as the geometric fact that reflecting across $y = x$ trades the two axes.

Worked example 3 Track the domain and range through the reflection

The function $f(x) = \sqrt{x} + 1$ has domain $x \ge 0$ (the square root needs a nonnegative input) and range $y \ge 1$ (the smallest output is $\sqrt{0} + 1 = 1$ ). Find the domain and range of $f^{-1}$ from the reflection.

Reflecting across $y = x$ swaps the horizontal and vertical spreads, so the domain and range trade:

\text{domain of } f^{-1} = \text{range of } f, \qquad \text{range of } f^{-1} = \text{domain of } f.

The range of $f$ was $y \ge 1$ , so the domain of $f^{-1}$ is $x \ge 1$ . The domain of $f$ was $x \ge 0$ , so the range of $f^{-1}$ is $y \ge 0$ .

A check against the formula agrees: from $y = \sqrt{x} + 1$ , swap to $x = \sqrt{y} + 1$ , so $\sqrt{y} = x - 1$ and $y = (x - 1)^2$ for $x \ge 1$ . That inverse accepts only $x \ge 1$ and returns values $y \ge 0$ , matching the swapped domain and range.

Check your understanding

A function $h$ has domain $2 \le x \le 8$ and range $-1 \le y \le 5$ . What are the domain and range of $h^{-1}$ ?

A domain

-1 \le x \le 5

, range

2 \le y \le 8

B domain

2 \le x \le 8

, range

-1 \le y \le 5

C domain

-1 \le x \le 5

, range

-1 \le y \le 5

D domain

2 \le x \le 8

, range

2 \le y \le 8

Fixed points stay on the mirror

A point that already lies on the line $y = x$ has equal coordinates, $(c, c)$ , so swapping them changes nothing: its reflection is itself. Such a point is called a fixed point of the reflection. This has a handy consequence for sketching. Wherever the graph of $f$ crosses the diagonal $y = x$ , that crossing point is shared with the graph of $f^{-1}$ . That is because a point on the mirror is its own image.

Concretely, suppose $f(c) = c$ for some input $c$ , so that $(c, c)$ is on the graph of $f$ . Reflecting fixes that point, so $(c, c)$ is on the graph of $f^{-1}$ as well, meaning $f^{-1}(c) = c$ . A value the function leaves unchanged is left unchanged by the inverse too. When you sketch an inverse, marking where $f$ meets $y = x$ gives you anchor points that do not move, which pins down the reflected curve.

The horizontal line test becomes the vertical line test

Back in the Inverse Functions lesson you saw that a function has an inverse only when it is one-to-one. On a graph that condition is the horizontal line test: no horizontal line may cross the graph more than once. The reflection across $y = x$ shows why that is exactly the right test.

Reflecting across the diagonal turns a horizontal line into a vertical line. A horizontal line $y = c$ is the set of points at height $c$ ; swapping coordinates sends each point $(t, c)$ to $(c, t)$ , and those make up the vertical line $x = c$ . So horizontal lines and vertical lines are mirror images of each other across $y = x$ .

Now watch what that does to the two line tests. Suppose some horizontal line crosses the graph of $f$ twice, so $f$ fails the horizontal line test and is not one-to-one. Reflect the whole picture. The graph of $f$ becomes the graph of $f^{-1}$ , and the offending horizontal line becomes a vertical line that crosses the reflected curve twice. A curve that some vertical line meets twice is not the graph of a function, by the vertical line test. So when $f$ fails the horizontal line test, its reflection fails the vertical line test and cannot be a function. The reflection is a function exactly when $f$ passed the horizontal line test to begin with. The two tests are mirror images, which is why one-to-one is precisely the condition for an inverse function to exist.

Why a function that fails the horizontal line test has no inverse graph. The solid curve y = x squared (dark) and its reflection across the dashed line y = x (blue) are mirror images. The horizontal line y = 2 crosses the parabola at two points, so the parabola fails the horizontal line test. Its mirror image, the vertical line x = 2, crosses the reflected curve at the two mirror points, so the reflection fails the vertical line test and is not the graph of a function.

Worked example 4 Which reflections are graphs of functions?

Compare $f(x) = x^3$ and $g(x) = x^2$ , both reflected across $y = x$ .

The cube $f(x) = x^3$ always rises: as $x$ increases, $x^3$ increases, so every horizontal line meets its graph exactly once and $f$ is one-to-one. Reflecting its graph across $y = x$ gives a curve that every vertical line meets exactly once. So that reflection is the graph of a genuine function, the inverse (the cube-root curve $y = \sqrt[3]{x}$ ). No restriction is needed.

The square $g(x) = x^2$ is different. The horizontal line $y = 4$ meets its graph twice, at $(-2, 4)$ and $(2, 4)$ , because

(-2)^2 = 2^2 = 4.

Reflecting, those two points become $(4, -2)$ and $(4, 2)$ , both on the reflected curve, so the vertical line $x = 4$ meets the reflection twice. The reflected sideways parabola is not the graph of a function, which is the picture of the fact that $x^2$ on all inputs has no inverse. The next section repairs it.

Check your understanding

The graph of a function $p$ is reflected across the line $y = x$ . The reflection is itself the graph of a function exactly when $p$ ...

A passes the horizontal line test (is one-to-one) B passes the vertical line test C has a

y

-intercept D is increasing and then decreasing

Graphing the inverse of a restricted function

When a function fails the horizontal line test, you cannot reflect its whole graph and get a function. The fix from the Inverse Functions lesson was to shrink the domain until the rule is one-to-one, then invert that piece. Graphically, that means reflecting only the surviving branch.

Take $f(x) = x^2$ again, and restrict it to $x \ge 0$ . The graph is now just the right half of the parabola, the branch in the first quadrant. That branch passes the horizontal line test: each height is reached by a single nonnegative input. Reflect that single branch across $y = x$ . Because the branch lives in the first quadrant, its reflection also lives in the first quadrant, and it is the graph of $y = \sqrt{x}$ . Restricting first, then reflecting, produces the square-root curve, exactly the inverse you found by algebra.

Reflecting one branch of a parabola gives a square root. Restricting y = x squared to inputs x greater than or equal to 0 keeps only the right branch (dark), which passes the horizontal line test. Reflecting that branch across the dashed line y = x gives the solid blue curve y = square root of x, the inverse. The point (2, 4) on the parabola reflects to (4, 2) on the square root.

Worked example 5 Graph the inverse of $f(x) = x^2$ on $x \ge 0$

With the domain restricted to $x \ge 0$ , plot a few points on the branch and swap coordinates. Three easy points on $y = x^2$ with $x \ge 0$ are

(0, 0), \qquad (1, 1), \qquad (2, 4).

Reflect each across $y = x$ by swapping coordinates:

(0, 0) \to (0, 0), \qquad (1, 1) \to (1, 1), \qquad (2, 4) \to (4, 2).

The points $(0, 0)$ and $(1, 1)$ sit on the diagonal, so they are fixed and do not move, while $(2, 4)$ reflects to $(4, 2)$ . Draw a smooth curve through $(0, 0)$ , $(1, 1)$ , and $(4, 2)$ : it rises steeply at first and then flattens, the shape of $y = \sqrt{x}$ . The reflection of the restricted parabola is the square-root graph, so $f^{-1}(x) = \sqrt{x}$ , matching the algebra where $x = y^2$ with $y \ge 0$ solves to $y = \sqrt{x}$ .

Had you tried to reflect the entire parabola instead, the left branch would have reflected to the lower half of a sideways parabola. The reflection of the whole parabola would fail the vertical line test. Restricting to one branch is what keeps the reflection a function.

Common mistakes

Reflecting across the wrong line. The mirror for an inverse is the diagonal $y = x$ , not the $x$ -axis or the $y$ -axis. Reflecting across the $x$ -axis gives $y = -f(x)$ , a different graph entirely; only $y = x$ swaps input and output.
Swapping only one coordinate. A point $(a, b)$ on $f$ reflects to $(b, a)$ on $f^{-1}$ , with both coordinates traded. Negating one coordinate, to $(a, -b)$ or $(-a, b)$ , is a reflection across an axis, not across $y = x$ .
Reading the diagonal off a squashed graph. The line $y = x$ only looks like a forty-five-degree mirror when the $x$ - and $y$ -axes use the same scale. With different scales the reflection is distorted and the picture misleads, so keep equal scales when you reflect.
Reflecting a graph that is not one-to-one. If $f$ fails the horizontal line test, reflecting its whole graph gives a curve that fails the vertical line test, so it is not a function. Restrict the domain to one branch first, then reflect.
Forgetting to swap domain and range. Because the axes trade places, the domain of $f^{-1}$ is the range of $f$ and its range is the domain of $f$ . Do not carry an interval across unchanged.
Expecting the two graphs to meet only on $y = x$ . The graphs of $f$ and $f^{-1}$ always share any point where $f$ crosses the diagonal, but for a decreasing function they can also cross off the diagonal. The guaranteed shared points are the ones on $y = x$ .

Key takeaways

A point $(a, b)$ is on the graph of $f$ exactly when $(b, a)$ is on the graph of $f^{-1}$ . Swapping the two coordinates of a point reflects it across the line $y = x$ .
The graph of $f^{-1}$ is the mirror image of the graph of $f$ across the diagonal $y = x$ . To sketch an inverse, reflect the graph, or swap the coordinates of points you can read off.
Reflecting across $y = x$ trades the two axes, so the domain and range swap. The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .
A point on the line $y = x$ is its own reflection, so wherever $f$ meets the diagonal, $f^{-1}$ meets it too (a fixed point), and $f^{-1}(c) = c$ there.
A horizontal line reflects to a vertical line, so $f$ passes the horizontal line test exactly when its reflection passes the vertical line test. That is why only one-to-one functions have inverse functions.
To graph the inverse of a function that fails the test, restrict it to a one-to-one branch first, then reflect. For example, the right branch of $y = x^2$ (where $x \ge 0$ ) reflects to $y = \sqrt{x}$ .

Graphs of Inverse Functions

Learning goals

The swap of coordinates is a reflection

The graph of an inverse is the mirror image across y = x

Reading an inverse straight off a graph

Domain and range trade places by swapping the axes

Fixed points stay on the mirror

The horizontal line test becomes the vertical line test

Graphing the inverse of a restricted function

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)