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.
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?
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.
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 $f$
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 $\dfrac{1}{f(x)}$
Skills
- Find the inverse of linear functions algebraically
- Find the inverse of rational functions by collecting $y$-terms
- State the domain and range of $f^{-1}$
If $f$ is a one-to-one function, its inverse $f^{-1}$ is the function that reverses the action of $f$:
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).
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?
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:
- Write $y = f(x)$.
- Swap $x$ and $y$ to get $x = f(y)$.
- Solve for $y$ in terms of $x$.
- 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.
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 $f$.
Worked examples · 3 in a row, reveal as you go
Find the inverse of $f(x) = 3x - 4$. Verify your answer.
Find the inverse of $f(x) = \dfrac{2x + 1}{x - 3}$ for $x \ne 3$. State the domain of $f^{-1}$.
Expand: $xy - 3x = 2y + 1$
Factor: $y(x - 2) = 3x + 1$
If $f(x) = 2x - 3$, find $f^{-1}(7)$ without first finding $f^{-1}(x)$ explicitly.
Fill the gap: To find $f^{-1}(x)$, write $y = f(x)$, then swap $x$ and $y$, then to get $f^{-1}(x)$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $f(x) = 5x + 2$, then $f^{-1}(x) = \dfrac{1}{5x+2}$.
Activities · practice with the ideas
Find the inverse of $f(x) = 4x + 1$. State the domain of $f^{-1}$.
Find the inverse of $f(x) = \dfrac{x}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$.
Given $f(x) = 2x - 3$, find $f^{-1}(7)$ and $f^{-1}(-1)$.
Find the inverse of $f(x) = 5x + 2$. Verify by computing $f(f^{-1}(x))$.
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.
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?
Odd one out: Which of these functions does NOT have an inverse function over its natural domain?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the inverse of $f(x) = 5x + 2$. Show all working. (2 marks)
Q2. Find the inverse of $f(x) = \dfrac{x}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$. (3 marks)
Q3. If $f(x) = 2x - 3$, find $f^{-1}(7)$. Then explain why the answer satisfies $f(f^{-1}(7)) = 7$. (2 marks)
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].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inverse function questions. Lighter alternative to the boss.
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