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hscscience Maths Ext 1 · Y11
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Module 1 · L5 of 15 ~35 min ⚡ +95 XP available

Introduction to Inverse Functions

Every time you use a GPS to navigate, the device "undoes" a signal transformation to find your location. Every time a doctor reads a pH value, they're reversing a logarithm. Inverse functions are the mathematical machinery that lets us reverse any process, and in Extension 1, mastering them unlocks the door to inverse trig, implicit differentiation, and beyond. In this lesson you'll learn the three-step algebraic method that works every time.

Today's hook, If $f(x) = 2x$, you can clearly "undo" it by halving. But what if $f(x) = \dfrac{2x+1}{x-3}$? How do you undo that? By the end of this lesson you'll have a mechanical three-step method that cracks any such function, and you'll know exactly when it works and when it doesn't.
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Recall, your gut answer first

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Recall, your gut answer first
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If $f(x) = 2x$, what operation "undoes" $f$? Write down what you think $f^{-1}(x)$ would be, don't look it up, just reason it through. What would $f^{-1}(10)$ equal?

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Orientation and key terms

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The two moves
+5 XP to read

Every inverse function question in this module uses exactly two ideas. Lock swap $x$ and $y$, then solve for $y$ and check one-to-one using the horizontal line test into memory and you'll never be stuck.

Every inverse function task lives on one of two roads: verify the function is one-to-one (horizontal line test) before attempting, then swap and solve to find the explicit inverse.

Sequence for finding an inverse function: check one-to-one then swap and solve
$f^{-1}(f(x)) = x$
Always swap first
Write $y = f(x)$, then swap to $x = f(y)$. Solving for $y$ then gives you $f^{-1}(x)$ directly.
Domain of $f^{-1}$
The domain of $f^{-1}$ equals the range of $f$. Read this off the original function before swapping.
Always verify
Check by computing $f(f^{-1}(x)) = x$. One substitution confirms your whole answer.
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What you'll master
Know

Key facts

  • The definition of an inverse function: $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$
  • A function must be one-to-one to have an inverse that is also a function
  • Domain of $f^{-1}$ = range of $f$; range of $f^{-1}$ = domain of $f$
Understand

Concepts

  • Why the horizontal line test determines one-to-one status
  • The algebraic procedure: write $y=f(x)$, swap $x$ and $y$, solve for $y$
  • Why $f^{-1}(x)$ is NOT the same as $\dfrac{1}{f(x)}$
Can do

Skills

  • Find the inverse of linear functions algebraically
  • Find the inverse of rational functions by collecting $y$-terms
  • State the domain and range of $f^{-1}$
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Key terms
Inverse function $f^{-1}$A function that reverses the action of $f$, satisfying $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ for all $x$ in the appropriate domain.
One-to-one functionA function where each output corresponds to exactly one input. It passes the horizontal line test: any horizontal line cuts the graph at most once.
Horizontal line testA visual test: if any horizontal line meets the graph more than once, the function is not one-to-one and has no inverse function (over that domain).
Domain of $f^{-1}$Equals the range of the original function $f$. The outputs of $f$ become the inputs of $f^{-1}$.
Range of $f^{-1}$Equals the domain of the original function $f$. The inputs of $f$ become the outputs of $f^{-1}$.
Reflection in $y = x$The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. Every point $(a, b)$ on $f$ maps to $(b, a)$ on $f^{-1}$.
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What is an inverse function?

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What is an inverse function?
core concept

If $f$ is a one-to-one function, its inverse $f^{-1}$ is the function that reverses the action of $f$:

$$f^{-1}(f(x)) = x \qquad \text{and} \qquad f(f^{-1}(x)) = x$$

Think of $f$ as a machine and $f^{-1}$ as a machine that runs the original machine backwards. If $f$ turns $3$ into $10$, then $f^{-1}$ must turn $10$ back into $3$.

  • The domain of $f^{-1}$ equals the range of $f$
  • The range of $f^{-1}$ equals the domain of $f$

Critical condition: A function must be one-to-one to have an inverse that is also a function. This means it must pass the horizontal line test. If two different inputs give the same output, as with $f(x) = x^2$ where $f(2) = f(-2) = 4$, then no inverse function can exist over that domain (because $f^{-1}(4)$ would need to equal both $2$ and $-2$ simultaneously).

Real-world link. Encryption algorithms are designed to be one-to-one so they have inverses (decryption). If an encryption function were many-to-one, multiple plaintext messages would produce the same ciphertext, and decryption would be ambiguous. Inverse functions literally secure the internet.

Inverse function: f^{-1} undoes f, f^{-1}(f(x)) = x and f(f^{-1}(x)) = x; Condition: f must be one-to-one (pass the horizontal line test)

Pause, copy the inverse function definition into your book: $f^{-1}$ undoes $f$ so $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$; a function has an inverse only if it is one-to-one (horizontal line test).

Quick check: Which statement about $f^{-1}(x)$ is correct?

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Finding an inverse algebraically, the four steps
core concept

We just saw that $f^{-1}$ exists only when $f$ passes the horizontal line test (one-to-one). That raises a question: given a function that passes the test, what is the algebraic procedure for finding $f^{-1}(x)$? This card answers it → write $y = f(x)$, swap $x$ and $y$, solve for $y$, then write $f^{-1}(x)$ with its domain.

There is one reliable method for finding $f^{-1}(x)$ algebraically:

  1. Write $y = f(x)$.
  2. Swap $x$ and $y$ to get $x = f(y)$.
  3. Solve for $y$ in terms of $x$.
  4. Write $f^{-1}(x) = \ldots$ and state its domain (= range of $f$).

This works because swapping $x$ and $y$ is precisely the algebraic version of reflecting the graph in the line $y = x$, which is exactly what finding the inverse does geometrically.

For rational functions, step 3 requires collecting all $y$-terms on one side, factoring out $y$, and dividing. This is the step most students rush, take your time with the algebra.

$$\text{Write } y = f(x) \;\longrightarrow\; \text{swap } x,y \;\longrightarrow\; \text{solve for } y = f^{-1}(x)$$

Step 1: Write y = f(x); Step 2: Swap, replace every x with y and every y with x, giving x = f(y)

Pause, copy the four-step inverse method into your book: (1) write $y = f(x)$; (2) swap $x$ and $y$; (3) solve for $y$; (4) write $f^{-1}(x) = $ result, stating the domain.

Did you get this? True or false: the domain of $f^{-1}$ always equals the domain of $f$.

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · LINEAR FUNCTION

Find the inverse of $f(x) = 3x - 4$. Verify your answer.

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Write $y = 3x - 4$. Swap $x$ and $y$: $\quad x = 3y - 4$
Swapping $x$ and $y$ is the algebraic equivalent of reflecting in $y = x$.
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PROBLEM 2 · RATIONAL FUNCTION

Find the inverse of $f(x) = \dfrac{2x + 1}{x - 3}$ for $x \ne 3$. State the domain of $f^{-1}$.

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Write $y = \dfrac{2x + 1}{x - 3}$. Swap: $x = \dfrac{2y + 1}{y - 3}$
Step 1–2 of the method. Now we need to solve this for $y$, it requires collecting $y$ terms.
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PROBLEM 3 · EVALUATING THE INVERSE

If $f(x) = 2x - 3$, find $f^{-1}(7)$ without first finding $f^{-1}(x)$ explicitly.

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We need to find $x$ such that $f(x) = 7$, i.e. $2x - 3 = 7$
$f^{-1}(7) = x$ means "what input gives $f(x) = 7$?" This is an efficient shortcut when you only need one value.
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Fill the gap: To find $f^{-1}(x)$, write $y = f(x)$, then swap $x$ and $y$, then to get $f^{-1}(x)$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Confusing $f^{-1}(x)$ with $\dfrac{1}{f(x)}$
The most common error: writing $f^{-1}(x) = \dfrac{1}{f(x)}$. The $-1$ in $f^{-1}$ is a notation for "inverse function", it is NOT an exponent. $f^{-1}(x)$ means "undo $f$"; $\dfrac{1}{f(x)} = [f(x)]^{-1}$ means "reciprocal of $f$". These are almost never equal.
Trap 02
Forgetting the domain restriction
Students find the formula for $f^{-1}(x)$ but forget to state its domain. For example, $f^{-1}(x) = \dfrac{3x+1}{x-2}$ requires $x \ne 2$, this is an essential part of the answer and will cost marks if omitted in an exam.
Trap 03
Rushing the algebra for rational functions
When swapping gives $x = \dfrac{2y+1}{y-3}$, students often try to divide too early. The correct method is: multiply out, expand, collect $y$ terms, factor, then divide. Skipping the collect-and-factor step leads to wrong answers.

Did you get this? True or false: if $f(x) = 5x + 2$, then $f^{-1}(x) = \dfrac{1}{5x+2}$.

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Activities

Activities · practice with the ideas

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Find the inverse of $f(x) = 4x + 1$. State the domain of $f^{-1}$.

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Find the inverse of $f(x) = \dfrac{x}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$.

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Given $f(x) = 2x - 3$, find $f^{-1}(7)$ and $f^{-1}(-1)$.

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Find the inverse of $f(x) = 5x + 2$. Verify by computing $f(f^{-1}(x))$.

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Explain in your own words why $f^{-1}(x)$ is NOT the same as $\dfrac{1}{f(x)}$. Give a numerical example to support your explanation.

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Revisit your thinking

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Revisit your thinking

Earlier you wrote down what $f^{-1}(x)$ would be for $f(x) = 2x$. Now you have the full method, was your gut answer right? Why must a function be one-to-one to have an inverse that is also a function?

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Odd one out: Which of these functions does NOT have an inverse function over its natural domain?

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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 42 marks

Q1. Find the inverse of $f(x) = 5x + 2$. Show all working. (2 marks)

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ApplyBand 4–53 marks

Q2. Find the inverse of $f(x) = \dfrac{x}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$. (3 marks)

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AnalyseBand 52 marks

Q3. If $f(x) = 2x - 3$, find $f^{-1}(7)$. Then explain why the answer satisfies $f(f^{-1}(7)) = 7$. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $y = 4x+1 \Rightarrow x = 4y+1 \Rightarrow f^{-1}(x) = \frac{x-1}{4}$, domain all real $x$ · 2. $y = \frac{x}{x+2} \Rightarrow x(y+2) = y \Rightarrow xy+2x = y \Rightarrow y(x-1) = -2x \Rightarrow f^{-1}(x) = \frac{-2x}{x-1}$, domain $x \ne 1$ · 3. $f^{-1}(x) = \frac{x+3}{2}$; $f^{-1}(7) = 5$; $f^{-1}(-1) = 1$ · 4. $f^{-1}(x) = \frac{x-2}{5}$; $f(f^{-1}(x)) = 5 \cdot \frac{x-2}{5} + 2 = x$ ✓ · 5. $f^{-1}(x)$ undoes $f$; $\frac{1}{f(x)}$ is the reciprocal, e.g. for $f(x)=2x$: $f^{-1}(4)=2$ but $\frac{1}{f(4)} = \frac{1}{8}$.

Q1 (2 marks): $y = 5x+2 \Rightarrow x = 5y+2 \Rightarrow y = \frac{x-2}{5}$ [1]. $f^{-1}(x) = \frac{x-2}{5}$ [1].

Q2 (3 marks): $y = \frac{x}{x+2}$, swap: $x = \frac{y}{y+2}$ [0.5]. Multiply: $x(y+2) = y \Rightarrow xy + 2x = y \Rightarrow y(x-1) = -2x$ [1]. $f^{-1}(x) = \frac{-2x}{x-1}$ [1]; domain $x \ne 1$ [0.5].

Q3 (2 marks): $f^{-1}(x) = \frac{x+3}{2}$, so $f^{-1}(7) = 5$ [1]. $f(5) = 2(5)-3 = 7$ ✓, the inverse undoes $f$, so $f$ must map $f^{-1}(7)$ back to $7$ by definition [1].

Module overview · Extension 1 · Checkpoint 1 · Module quiz