A1.4 Connect inverse functions with reverse processes

Ontario Grade 11 Mathematics

Characteristics of Functions

MCR3U – A1.4 | Connecting a Function to Its Inverse

Every time you zip up a jacket, you know exactly how to unzip it. Every time you multiply a number by 3, you know that dividing by 3 brings you back. Mathematics formalises this idea of "undoing" through inverse functions. In this lesson you will see that an inverse function is simply the original function run in reverse: the output becomes the new input, and the input becomes the new output. By the end of the lesson you will be able to find inverses, verify them, and decide when an inverse is itself a function — all using skills you already have from Grade 10.

What you will learn

Explain what an inverse function does in plain language as a reverse process.
Find the equation of an inverse by switching input and output and solving for the new output variable.
Verify that two functions are inverses by checking that each undoes the other.
Interpret inverse functions using mapping diagrams and tables of values.
Recognise when a function's inverse is also a function and explain why using the horizontal line test.

Prerequisite Bridge: Functions and Function Notation

A function is a rule that assigns exactly one output to every input. You write

f(x)

to mean 'the output of function

f

when the input is

x

.' For example, if

f(x) = 2x + 6

, then

f(1) = 8

,

f(2) = 10

, and so on. The set of all allowed inputs is the domain, and the set of all resulting outputs is the range.

A mapping diagram is a picture with two ovals — one for the domain and one for the range — with arrows showing which input maps to which output. You saw these in Grade 10. Keep that picture in mind because it will make the idea of reversing a function completely visual.

A function gives exactly one output for every input.
$f(x)$ is read 'f of x' and represents the output value.
Domain = set of inputs; Range = set of outputs.
A mapping diagram uses arrows from inputs to outputs.

What Does an Inverse Function Do?

Imagine a machine: you drop a number into the top, the machine processes it, and a result comes out the bottom. An inverse function is a second machine placed right after the first one. Its job is to take whatever came out of the first machine and recover the original number you dropped in.

Formally, if

f

takes input

x

and produces output

y

, then the inverse — written

f^{-1}

— takes

y

as its input and gives

x

back as its output. The notation

f^{-1}

is read 'f inverse.' Important warning:

f^{-1}(x)

does not mean

\frac{1}{f(x)}

. The superscript

-1

here labels the inverse function, not a reciprocal.

A clean way to see this: if

f(3) = 11

, then

f^{-1}(11) = 3

. The arrow has simply been flipped. Because of this flip, the domain of

f

becomes the range of

f^{-1}

, and the range of

f

becomes the domain of

f^{-1}

.

On a mapping diagram, you literally reverse every arrow. Every arrow that once pointed from left to right now points from right to left.

f^{-1}(f(x)) = x

$f^{-1}$ is the inverse of $f$ ; it reverses the mapping of $f$ .
If $f(a) = b$ , then $f^{-1}(b) = a$ .
The domain and range swap when you take the inverse.
$f^{-1}(x)$ does NOT mean $\frac{1}{f(x)}$ .

Finding an Inverse Equation: The Switch-and-Solve Method

To find the rule for

f^{-1}

, you follow three steps. First, replace

f(x)

with the letter

y

so the equation is easier to rearrange. Second, swap

x

and

y

everywhere in the equation — this is the algebraic way of saying 'reverse the roles of input and output.' Third, solve the new equation for

y

. That expression for

y

is

f^{-1}(x)

.

Why does swapping

x

and

y

work? Because 'input' and 'output' are just roles. The original function says 'given

x

, find

y

.' The inverse says 'given

y

, find

x

.' By relabelling

y

as the new

x

and the old

x

as the new

y

, you simply restate the same relationship from the opposite direction.

After finding

f^{-1}(x)

, always verify it by checking two things:

f^{-1}(f(x)) = x

and

f(f^{-1}(x)) = x

. Both must hold. If even one fails, you made an error somewhere.

f(f^{-1}(x)) = x

Replace $f(x)$ with $y$ , then swap $x$ and $y$ , then solve for $y$ .
Swapping $x$ and $y$ reverses which variable is the input and which is the output.
Verify: substituting $f$ into $f^{-1}$ (or vice versa) must return $x$ .
Write the final answer as $f^{-1}(x) = \ldots$

When Is the Inverse Also a Function? The Horizontal Line Test

Reversing a function's arrows does not always produce a new function. Remember: a function must give exactly one output for every input. If two different inputs of

f

produce the same output, then when you reverse the arrows, that one output would have to map to two different inputs — breaking the one-output rule.

The horizontal line test checks this visually on the graph of

f

. Draw horizontal lines across the graph. If every horizontal line touches the graph at most once, then

f

is called one-to-one, and its inverse is a function. If any horizontal line touches the graph more than once, the inverse is a relation but not a function.

For example, a straight line with a non-zero slope is always one-to-one, so its inverse is always a function. A parabola like

f(x) = x^2

is not one-to-one over all real numbers (the horizontal line

y = 4

hits it at both

x = 2

and

x = -2

), so its inverse over all real numbers is not a function.

In a table of values, the same check applies: if any two rows share the same

y

-value but have different

x

-values, the inverse will not be a function for those input values.

The horizontal line test: every horizontal line must hit the graph at most once for the inverse to be a function.
A one-to-one function always produces an inverse that is also a function.
Non-one-to-one functions produce inverses that are relations, not functions.
You can check tables of values: repeated outputs in $f$ mean the inverse is not a function.

Graphs of a Function and Its Inverse

Because

x

and

y

swap roles, the graph of

f^{-1}

is the reflection of the graph of

f

across the line

y = x

. Every point

(a, b)

on the graph of

f

corresponds to the point

(b, a)

on the graph of

f^{-1}

.

This reflection property is a powerful check: if you sketch

f

and then reflect it over the line

y = x

, the result should match the equation you found for

f^{-1}

. The line

y = x

acts like a mirror placed diagonally.

Notice that the line

y = x

itself is its own inverse — reflecting it over itself gives the same line. Any point that lies on

y = x

(where the

x

-coordinate equals the

y

-coordinate) is unchanged by the reflection, so it sits on both graphs at once.

The graph of $f^{-1}$ is the reflection of the graph of $f$ over the line $y = x$ .
Every point $(a, b)$ on $f$ becomes the point $(b, a)$ on $f^{-1}$ .
This reflection is a visual check that your inverse equation is correct.

Comparing g(x) and Its Inverse g⁻¹(x) — Roles of Input and Output Swap

Input for g(x)	Output of g(x) = x/3 + 7	Input for g⁻¹(x)	Output of g⁻¹(x) = 3(x − 7)
-3	6	6	-3
0	7	7	0
6	9	9	6
12	11	11	12

Worked example

Example 1 — Finding and Verifying the Inverse of a Linear Function

Let

f(x) = 4x - 3

. Find

f^{-1}(x)

and verify that

f^{-1}(f(x)) = x

.

Write the function as an equation in x and y
Replace $f(x)$ with $y$ so the equation is easier to rearrange. This gives $y = 4x - 3$ .
$y = 4x - 3$
Swap x and y to reverse the input–output roles
Swapping $x$ and $y$ makes the old output the new input and the old input the new output — exactly what a reverse process does.
$x = 4y - 3$
Solve for y — isolate the new output
Add 3 to both sides to get $x + 3 = 4y$ , then divide both sides by 4.
$y = \frac{x + 3}{4}$
State the inverse function
Label the result with the proper inverse notation.
$f^{-1}(x) = \frac{x + 3}{4}$
Verify by composing f inverse with f
Substitute $f(x) = 4x - 3$ into $f^{-1}$ . Replace the $x$ in $f^{-1}(x)$ with the entire expression $4x - 3$ , then simplify.
$f^{-1}(f(x)) = \frac{(4x - 3) + 3}{4} = \frac{4x}{4} = x \ ✓$

Answer:

f^{-1}(x) = \frac{x + 3}{4}

Check: Substitute a value to double-check. Try

x = 2

:

f(2) = 4(2) - 3 = 5

. Then

f^{-1}(5) = \frac{5 + 3}{4} = \frac{8}{4} = 2

. The original input 2 is recovered, confirming the inverse is correct.

Worked example

Example 2 — Inverse of a Two-Step Function and a Table Check

Let

g(x) = \frac{x}{3} + 7

. (a) Find

g^{-1}(x)

. (b) Use a table of values to confirm the inverse relationship for three input–output pairs.

Write g(x) as y
Replace $g(x)$ with $y$ to set up an equation you can rearrange.
$y = \frac{x}{3} + 7$
Swap x and y
Swap $x$ and $y$ to reflect the reverse-process idea: the output of $g$ becomes the input of $g^{-1}$ .
$x = \frac{y}{3} + 7$
Subtract 7 from both sides
Isolate the term containing $y$ by subtracting 7 from both sides.
$x - 7 = \frac{y}{3}$
Multiply both sides by 3
Multiply both sides by 3 to fully isolate $y$ .
$y = 3(x - 7)$
State the inverse function
Write the result using inverse notation. You can also expand: $3x - 21$ .
$g^{-1}(x) = 3(x - 7)$
Verify with composition
Substitute $g(x) = \frac{x}{3} + 7$ into $g^{-1}$ and simplify to confirm the result is $x$ .
$g^{-1}(g(x)) = 3 (\frac{x}{3} + 7 - 7) = 3 · \frac{x}{3} = x \ ✓$

Answer:

g^{-1}(x) = 3(x - 7)

Check: Table check using three pairs: when

x = 0

,

g(0) = 0/3 + 7 = 7

, and

g^{-1}(7) = 3(7-7) = 0

✓. When

x = 6

,

g(6) = 6/3 + 7 = 9

, and

g^{-1}(9) = 3(9-7) = 6

✓. When

x = -3

,

g(-3) = -3/3 + 7 = 6

, and

g^{-1}(6) = 3(6-7) = -3

✓. Every original input is recovered.

Common mistakes and how to avoid them

Writing

f^{-1}(x) = \frac{1}{f(x)}

because the

-1

looks like an exponent.

Correction: The

-1

in

f^{-1}

is a label for the inverse function, not an exponent.

f^{-1}(x)

means 'the inverse of

f

evaluated at

x

', which is completely different from the reciprocal

\frac{1}{f(x)}

.

Forgetting to swap

x

and

y

and instead just solving the original equation for

x

.

Correction: Solving the original equation for

x

is a useful step, but you must also relabel: the new

x

is what was

y

, and the new

y

is what was

x

. Both the swap and the solve are necessary.

Assuming every function automatically has an inverse that is also a function.

Correction: Only one-to-one functions have inverses that are functions. Always apply the horizontal line test to the graph of

f

first.

Reflecting the graph over the x-axis or y-axis instead of over the line

y = x

.

Correction: The graph of

f^{-1}

is obtained by reflecting

f

over the diagonal line

y = x

, not over either axis.

Only checking

f^{-1}(f(x)) = x

and not checking

f(f^{-1}(x)) = x

.

Correction: Both compositions must equal

x

. Check both directions to be fully confident the inverse is correct.

Lesson summary

An inverse function reverses the process of the original function: if $f$ maps $a$ to $b$ , then $f^{-1}$ maps $b$ back to $a$ .
To find $f^{-1}(x)$ : replace $f(x)$ with $y$ , swap $x$ and $y$ , then solve for $y$ .
Verify an inverse by confirming $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ .
The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$ , and every point $(a, b)$ becomes $(b, a)$ .
The horizontal line test determines whether the inverse of a function is itself a function; only one-to-one functions pass.
The domain and range of $f$ swap to become the range and domain of $f^{-1}$ , respectively.

Check your understanding

Question 1

If

f(5) = 12

, what is

f^{-1}(12)

?

5
12
60
Cannot be determined

Show answer and explanation

5

The inverse function reverses the input–output pair. Since

f(5) = 12

, the inverse maps 12 back to 5. So

f^{-1}(12) = 5

.

Question 2

What is

f^{-1}(x)

if

f(x) = 5x - 10

?

$\frac{x - 10}{5}$
$\frac{1}{5x - 10}$
$\frac{x + 10}{5}$
$5x + 10$

Show answer and explanation

\frac{x + 10}{5}

Replace

f(x)

with

y

:

y = 5x - 10

. Swap:

x = 5y - 10

. Solve:

x + 10 = 5y

, so

y = \frac{x + 10}{5}

. Therefore

f^{-1}(x) = \frac{x + 10}{5}

.

Question 3

A function

h

has the table of values:

(1, 4), (2, 4), (3, 9)

. Is the inverse of

h

a function?

Yes, because every input has an output.
No, because the outputs 4 and 4 are repeated, so reversing gives one input (4) mapping to two outputs (1 and 2).
Yes, because $h$ passes the vertical line test.
No, because $h$ only has three points.

Show answer and explanation

No, because the outputs 4 and 4 are repeated, so reversing gives one input (4) mapping to two outputs (1 and 2).

The output value 4 appears for two different inputs (1 and 2). When the arrows are reversed, the single input 4 would have to produce two outputs (1 and 2), which violates the definition of a function.

Question 4

Which transformation maps the graph of

f

to the graph of

f^{-1}

?

Reflection over the x-axis
Reflection over the y-axis
Reflection over the line $y = x$
Rotation of 90° about the origin

Show answer and explanation

Reflection over the line

y = x

Swapping

x

and

y

coordinates is exactly what a reflection over the line

y = x

does. Every point

(a, b)

on

f

moves to

(b, a)

on

f^{-1}

.

Key terms

Function: A rule that assigns exactly one output to every input in its domain.
Inverse function: A function that reverses the mapping of the original function; written $f^{-1}$ .
Domain: The complete set of allowed input values for a function.
Range: The complete set of output values produced by a function.
One-to-one function: A function in which every output value corresponds to exactly one input value; its inverse is also a function.
Horizontal line test: A visual test on a graph: if every horizontal line meets the graph at most once, the function is one-to-one and its inverse is a function.
Composition of functions: Applying one function to the result of another; for inverses, $f^{-1}(f(x)) = x$ .
Mapping diagram: A diagram using two ovals and arrows to show which inputs are paired with which outputs.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation A1.4. It is a study resource, not an official curriculum publication.

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A1.4 · Connect inverse functions with reverse processes

Ontario Grade 11 Mathematics

MCR3U – A1.4 | Connecting a Function to Its Inverse

What you will learn

Prerequisite Bridge: Functions and Function Notation

What Does an Inverse Function Do?

Finding an Inverse Equation: The Switch-and-Solve Method

When Is the Inverse Also a Function? The Horizontal Line Test

Graphs of a Function and Its Inverse

Comparing g(x) and Its Inverse g⁻¹(x) — Roles of Input and Output Swap

Example 1 — Finding and Verifying the Inverse of a Linear Function

Example 2 — Inverse of a Two-Step Function and a Table Check

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson