Graphing Inverse Functions

6 min read

Reference: Stewart §6.1

The Mirror Trick

If you place a mirror along the line $y = x$, the reflection of any point $(a, b)$ is the point $(b, a)$, with the coordinates swapped. This geometric fact is the key to graphing inverses: since $f(a) = b$ means $f^{-1}(b) = a$, the graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$.

No need to find a formula, just reflect!

Before You Start: Quick Self-Check

Can you answer these questions?

What is an inverse function? The function $f^{-1}$ that undoes $f$: if $f(a) = b$, then $f^{-1}(b) = a$.
How do you reflect a point $(3, 7)$ across the line $y = x$? Swap coordinates: $(7, 3)$.
What is the equation of the diagonal line through the origin at 45°? $y = x$.

If you struggled with these, review first

Before learning to graph inverses, make sure you understand the Inverse Function Definition.

Prerequisite Map

Quick Reference

Property	Value
Chapter	Chapter 6: Inverse Functions
Section	§6.1 Inverse Functions and Their Derivatives
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Reflection Principle

$$\boxed{\text{The graph of } f^{-1} \text{ is the reflection of the graph of } f \text{ about the line } y = x}$$

Why Reflection Works

If $(a, b)$ is on the graph of $f$, then $f(a) = b$.

By definition of inverse: $f^{-1}(b) = a$.

So $(b, a)$ is on the graph of $f^{-1}$.

The points $(a, b)$ and $(b, a)$ are reflections of each other across $y = x$.

Visualizing the Reflection

        y
        |          . (b,a) ← on graph of f⁻¹
        |        /
        |      / y = x
        |    /
        |  .
        | (a,b) ← on graph of f
        |
        +------------------ x

Step-by-Step: Graphing an Inverse

Draw the graph of $f$ (given or sketch from formula)
Draw the line $y = x$ (45° line through origin)
Reflect key points: For each point $(a, b)$ on $f$, plot $(b, a)$
Connect the reflected points following the same shape

Key Points to Reflect

Focus on:

Intercepts: $(0, b) \to (b, 0)$ and $(a, 0) \to (0, a)$
Turning points: maxima and minima
Asymptotes: vertical asymptotes become horizontal, and vice versa
Endpoints: if domain is restricted

Example: Graphing $f^{-1}$ for $f(x) = x^3$

Point on $f$	Reflected Point on $f^{-1}$
$(-1, -1)$	$(-1, -1)$
$(0, 0)$	$(0, 0)$
$(1, 1)$	$(1, 1)$
$(2, 8)$	$(8, 2)$

Points on the line $y = x$ (like $(0,0)$ and $(1,1)$) are unchanged by reflection!

Asymptote Behavior

When $f$ has a vertical asymptote at $x = c$, then $f^{-1}$ has a horizontal asymptote at $y = c$.

When $f$ has a horizontal asymptote at $y = c$, then $f^{-1}$ has a vertical asymptote at $x = c$.

Domain and Range Swap Visually

   Graph of f                    Graph of f⁻¹

   Domain: [0, 4]                Domain: [1, 3]
   Range: [1, 3]                 Range: [0, 4]

        y                             y
      3 |....______                 4 |
        |   /                         |____
        |  /                          |    \
      1 |./                         1 |.....\
        +----------x                  +----------- x
        0    4                        1     3

Practice Problems

Level 1 Reflecting Points

The graph of $f$ passes through the points $(1, 4)$, $(2, 7)$, and $(3, 5)$.

Find three points on the graph of $f^{-1}$.

Thought Process

Goal: Find three points on the graph of $f^{-1}$.

Key principle: Reflection across $y = x$ swaps coordinates.

Step-by-step:

Point $(1, 4)$ on $f$ → Swap → $(4, 1)$ on $f^{-1}$
Point $(2, 7)$ on $f$ → Swap → $(7, 2)$ on $f^{-1}$
Point $(3, 5)$ on $f$ → Swap → $(5, 3)$ on $f^{-1}$

Check: The $y$-coordinate of the original becomes the $x$-coordinate of the reflection.

Show Answer

Swap coordinates for each point:

$(1, 4) \to \boxed{(4, 1)}$
$(2, 7) \to \boxed{(7, 2)}$
$(3, 5) \to \boxed{(5, 3)}$

These three points lie on the graph of $f^{-1}$.

Level 2 Intercepts of the Inverse

The graph of $f$ has $x$-intercept $(3, 0)$ and $y$-intercept $(0, -2)$.

Find the intercepts of $f^{-1}$.

Thought Process

The intercepts are just points, so I reflect them:

$x$-intercept of $f$: $(3, 0)$ becomes $(0, 3)$ on $f^{-1}$, which is a $y$-intercept
$y$-intercept of $f$: $(0, -2)$ becomes $(-2, 0)$ on $f^{-1}$, which is an $x$-intercept

Intercept types swap under reflection!

Show Answer

Reflect the intercept points:

$(3, 0) \to (0, 3)$: The $y$-intercept of $f^{-1}$ is $\boxed{(0, 3)}$
$(0, -2) \to (-2, 0)$: The $x$-intercept of $f^{-1}$ is $\boxed{(-2, 0)}$

Notice: $x$-intercepts of $f$ become $y$-intercepts of $f^{-1}$, and vice versa.

Level 3 Sketching from a Graph

Sketch the graph of $f^{-1}$ given that the graph of $f$ passes through $(0, 1)$, $(1, 2)$, $(4, 3)$, and has a horizontal asymptote $y = 4$ as $x \to \infty$.

Thought Process

I’ll reflect the key points and the asymptote:

Points: $(0,1) \to (1,0)$, $(1,2) \to (2,1)$, $(4,3) \to (3,4)$
Horizontal asymptote $y = 4$ becomes vertical asymptote $x = 4$ for $f^{-1}$

The curve approaches $x = 4$ from the left as $y \to \infty$.

Show Answer

Key points on $f^{-1}$:

$(1, 0)$, from reflecting $(0, 1)$
$(2, 1)$, from reflecting $(1, 2)$
$(3, 4)$, from reflecting $(4, 3)$

Asymptote: Vertical asymptote at $x = 4$

Sketch:

        y
        |              . (3,4)
        |            .
        |          .    |
        |        .      | x = 4
        |      . (2,1)  |
        | . (1,0)       |
        +---------------+---- x
        1   2   3   4

The graph of $f^{-1}$ rises from $(1,0)$ and approaches the vertical asymptote $x = 4$.

Level 4 Self-Inverse Functions

The graph of a function $g$ is symmetric about the line $y = x$.

(a) What can you conclude about $g$ and $g^{-1}$? (b) Give an example of such a function. (c) If $g(3) = 7$, what is $g(7)$?

Thought Process

If a graph is symmetric about $y = x$, then reflecting it across $y = x$ gives the same graph. Since $f^{-1}$ is the reflection of $f$...

(a) The function equals its own inverse: $g = g^{-1}$

(b) I need a function that equals its inverse. Try $g(x) = \frac{1}{x}$ or $g(x) = -x$.

Show Answer

(a) If $g$ is symmetric about $y = x$, then reflecting $g$ gives the same graph. Since $g^{-1}$ is the reflection of $g$, we have:

$$\boxed{g^{-1} = g}$$

Such functions are called self-inverse or involutions.

(b) Examples of self-inverse functions:

$g(x) = x$ (identity)
$g(x) = -x$ (negation)
$g(x) = \frac{1}{x}$ (reciprocal)
$g(x) = \frac{1-x}{1+x}$ (Möbius transformation)

(c) If $g(3) = 7$, then by the inverse relationship: $$g^{-1}(7) = 3$$

Since $g = g^{-1}$: $$\boxed{g(7) = 3}$$

Level 5 Intersection of $f$ and $f^{-1}$

The graphs of $f$ and $f^{-1}$ intersect. Prove that their intersection points must lie on the line $y = x$.

Thought Process

Suppose $(a, b)$ is on both graphs. Then:

On $f$: $f(a) = b$
On $f^{-1}$: $f^{-1}(a) = b$

From the first: $f(a) = b$. From the second: $f^{-1}(a) = b$, which means $f(b) = a$.

So $f(a) = b$ and $f(b) = a$.

Hmm, this doesn’t immediately give $a = b$. Time to reconsider.

Actually, if the point is on both curves, it should equal its reflection. A point equals its reflection across $y = x$ only if... it’s on the line $y = x$.

Show Answer

Proof:

Let $(a, b)$ be an intersection point of the graphs of $f$ and $f^{-1}$.

Step 1: Since $(a, b)$ is on the graph of $f$, we have $f(a) = b$.

Step 2: Since $(a, b)$ is on the graph of $f^{-1}$, we have $f^{-1}(a) = b$, which by definition means $f(b) = a$.

Step 3: From Steps 1 and 2: $f(a) = b$ and $f(b) = a$.

Step 4: Apply $f^{-1}$ to $f(a) = b$: we get $a = f^{-1}(b)$.

Apply $f^{-1}$ to $f(b) = a$: we get $b = f^{-1}(a)$.

Step 5: From Step 2, we also have $f^{-1}(a) = b$. So $b = f^{-1}(a)$.

From Step 4, $a = f^{-1}(b)$. Substituting $b = f^{-1}(a)$: $$a = f^{-1}(f^{-1}(a))$$

Alternative direct argument: The point $(a, b)$ on $f^{-1}$ means it’s the reflection of some point on $f$ across $y = x$. If $(a, b)$ is also on $f$, then its reflection $(b, a)$ is on $f^{-1}$. For $(a, b)$ and $(b, a)$ to both equal the same intersection point, we need $(a, b) = (b, a)$, which requires $a = b$.

Therefore, any intersection point has the form $(a, a)$, which lies on $y = x$. $\square$

CCI-Style Conceptual Questions

Question 1: If the graph of $f$ has a vertical asymptote at $x = 2$, what feature does the graph of $f^{-1}$ have?

(A) A vertical asymptote at $x = 2$ (B) A horizontal asymptote at $y = 2$ (C) A hole at $x = 2$ (D) An $x$-intercept at $(2, 0)$

Answer

(B) A horizontal asymptote at $y = 2$

Asymptotes reflect too: vertical becomes horizontal, and the position swaps from $x = c$ to $y = c$. The vertical asymptote $x = 2$ reflects to horizontal asymptote $y = 2$.

Question 2: A point $(p, p)$ lies on the graph of $f$ where $f$ is one-to-one. This point also lies on:

(A) The graph of $f^{-1}$ only (B) The line $y = x$ only (C) Both the graph of $f^{-1}$ and the line $y = x$ (D) Neither

Answer

(C) Both the graph of $f^{-1}$ and the line $y = x$

The point $(p, p)$ clearly lies on $y = x$ (since both coordinates are equal).

When reflected across $y = x$, the point $(p, p)$ maps to itself. So it’s also on $f^{-1}$.

Points where the graph of $f$ crosses $y = x$ are fixed points that appear on both $f$ and $f^{-1}$.

Common Misconceptions

Common misconception

the graph of $f^{-1}$ is the reflection of $f$ across the $y$-axis, not the line $y = x$.

This is the concept-image-conflicts-definition error. Reflection across the $y$-axis replaces each point $(a, b)$ with $(-a, b)$, which does not swap input and output roles. Reflection across $y = x$ replaces $(a, b)$ with $(b, a)$, which is exactly what the inverse relationship $f^{-1}(b) = a$ requires. For $f(x) = 2^x$, reflecting across the $y$-axis produces $2^{-x}$, which is still an exponential; reflecting across $y = x$ produces $\log_2 x$, which is the actual inverse.

Mastery Checklist

I can reflect a graph across the line $y = x$
I can find points on $f^{-1}$ by swapping coordinates from $f$
I understand how intercepts transform under reflection
I can convert asymptote information between $f$ and $f^{-1}$
I recognize that self-inverse functions are symmetric about $y = x$
I can sketch $f^{-1}$ given the graph of $f$

Mental Model

The Mirror Along the Diagonal:

Imagine placing a two-way mirror along the line $y = x$. Looking through from the “$f$ side,” you see the graph of $f$. Looking through from the “$f^{-1}$ side,” you see the reflection, which is exactly the graph of $f^{-1}$.

Every point $(a, b)$ has a “mirror twin” $(b, a)$ on the other side.

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Finding Inverse Functions	Chapter 6	Derivative of Inverse Functions

Last updated: 2026-01-22