Graphs of Inverse Functions Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

Reflect a graph across $y = x$ by swapping the coordinates of any point you can read off
Trade domain for range when the axes swap
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. Suppose $f(1) = 4$ . Then $f^{-1}(4) = 1$ : the point $(1, 4)$ on the graph of $f$ becomes the point $(4, 1)$ on the graph of $f^{-1}$ , its coordinates swapped. Plot both points and something jumps out: they sit on opposite sides of the diagonal $y = x$ , the line through the origin at forty-five degrees, like mirror images.

Swapping the coordinates of a point reflects it across the dashed line y = x. The point (1, 4) and its swap (4, 1) 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.

The diagram shows why: $(1, 4)$ and $(4, 1)$ sit exactly the same distance from the line $y = x$ , the set of points whose two coordinates are equal, on opposite sides, and the segment joining them meets the diagonal at a right angle. That is exactly what it means for $y = x$ to be the mirror between a point and its coordinate-swap.

The same swap happens at every point, not just this one. A point $(a, b)$ on the graph of $f$ records the fact that the input $a$ produces the output $b$ , that is, $b = f(a)$ . Applying $f^{-1}$ to both sides gives $f^{-1}(b) = a$ , which says the point $(b, a)$ is on the graph of $f^{-1}$ . So a point is on the graph of $f$ exactly when its coordinate-swap is on the graph of $f^{-1}$ .

By the same reasoning as the picture above, a line through the midpoint of a segment that meets it at a right angle is that segment’s mirror, so $y = x$ is always the mirror between $(a, b)$ and $(b, a)$ . A point already on the diagonal has $a = b$ , so its swap is itself and it does not move.

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 $f$ : if you can draw $f$ , you can draw its reflection by hand, no algebra needed. This works whenever $f^{-1}$ already exists, that is, whenever $f$ is one-to-one. (A later section shows what goes wrong, and how to fix it, when $f$ is not one-to-one.)

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. As long as it is one-to-one, a function given only by a curve, with no equation at all, still has an inverse you can plot point by point.

The curve y = g(x) (dark), given only by its graph, rises through (-2, -3), (0, 1), (1, 3), and (3, 4). It passes the horizontal line test, so reflecting it across the dashed line y = x gives the curve y = g inverse of x (blue) through the coordinate-swapped points (-3, -2), (1, 0), (3, 1), and (4, 3).

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

The graph above shows a function $g$ given only by its curve, no formula, 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

$f$ is one-to-one, and the point $(-4, 7)$ lies on its graph. 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 one-to-one 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, a fixed point of the reflection. So if $f(c) = c$ for some input $c$ , meaning $(c, c)$ is on the graph of $f$ , then $(c, c)$ stays put under the reflection and lands on the graph of $f^{-1}$ too, giving $f^{-1}(c) = c$ : a value the function leaves unchanged is left unchanged by the inverse as well. This is a handy sketching shortcut: wherever the graph of $f$ meets the diagonal $y = x$ , mark that point, since it is shared with $f^{-1}$ and does not move when you reflect.

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 anyway. The reflected curve is the mirror image of $f$ ‘s graph, and the offending horizontal line becomes a vertical line that crosses that 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 is not the graph of any function, meaning $f$ has no inverse function at all. The reflection is a function, namely $f^{-1}$ , 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 $(4, 2)$ . These three points are anchors, but by themselves they do not pin down the curve: many different curves pass through the same three points. What determines the shape is reflecting the entire right branch, not just a few points on it. Mirror the whole branch across $y = x$ and it lands on the square-root curve, rising steeply at first and then flattening. 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.

Check your understanding

$j(x) = x^2$ restricted to $x \ge 0$ is the right branch of the parabola, passing through $(0, 0)$ , $(1, 1)$ , and $(3, 9)$ . Which curve is the graph of $j^{-1}$ ?

A Only the right branch reflected: the curve through

(0, 0)

(1, 1)

(9, 3)

B The whole parabola reflected: the curve through

(0, 0)

(1, 1)

(9, 3)

, and also

(9, -3)

C The left branch reflected instead: the curve through

(0, 0)

(1, -1)

(9, -3)

j

is its own inverse on

x \ge 0

, so no reflection is needed

Common mistakes

Reflecting across the wrong line. The mirror for an inverse is the diagonal $y = x$ , so both coordinates trade places: $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$ . Negating just one coordinate instead, to $(a, -b)$ or $(-a, b)$ , reflects across an axis, not across $y = x$ ; for instance, reflecting across the $x$ -axis gives $y = -f(x)$ , a different graph entirely.
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.

Key takeaways

The graph of $f^{-1}$ is the mirror image of the graph of $f$ across the diagonal $y = x$ , because swapping the two coordinates of a point is exactly what that reflection does. A point already on $y = x$ doesn’t move, which gives you free anchor points when sketching.
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 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 Advanced. This lesson goes beyond core Algebra I. You can skip it.

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)

Core practice