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 confidencethat 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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