Every time you unlock your phone with a passcode, encryption turns your code into something unreadable — and only the inverse process can turn it back. That is the power of an inverse function: it undoes what the original function did, perfectly and predictably.
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If a machine doubles a number and then adds 5, what operation would you need to perform on the output to get back to the original number? Can you write this "undoing" machine as a mathematical rule?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: A function can have two different outputs for the same input.
Right: By definition, a function has exactly one output for each input in its domain.
📚 Core Content
An inverse function undoes the effect of the original function. If $f$ takes an input $x$ and produces an output $y$, then $f^{-1}$ takes $y$ and returns the original $x$.
We write the inverse of $f$ as $f^{-1}$. The $-1$ is not an exponent — it is simply notation meaning "inverse." Do not confuse $f^{-1}(x)$ with $\frac{1}{f(x)}$.
For any function and its inverse:
$$f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$$
This means applying a function and then its inverse (or vice versa) returns you to exactly where you started.
Because an inverse swaps the $x$ and $y$ values of every point, its graph is the reflection of the original graph in the line $y = x$. If $(a, b)$ is on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$.
Not every function has an inverse over its entire natural domain. For an inverse to be a function, the original function must be one-to-one — meaning each output corresponds to exactly one input.
The classic example is $f(x) = x^2$. Both $f(2) = 4$ and $f(-2) = 4$. If we tried to find $f^{-1}(4)$, we would get two answers: $2$ and $-2$. Since a function can only have one output per input, $f(x) = x^2$ does not have an inverse unless we restrict its domain.
In general:
🧮 Worked Examples
🧪 Activities
1 $f(x) = 4x - 3$
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2 $f(x) = \dfrac{x}{2} + 5$
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3 $f(x) = \sqrt{x + 2}$
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4 $f(x) = \dfrac{3}{x + 1}$
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1 Find the decryption function $D(x)$, which is the inverse of $E(x)$.
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2 The letter C ($x = 3$) is encrypted. What is the encrypted code?
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3 An encrypted code of $22$ is received. Use your decryption function to find the original letter.
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Earlier you were asked: If a machine doubles a number and then adds 5, what operation would you need to perform on the output to get back to the original number? Can you write this "undoing" machine as a mathematical rule?
To undo "double and add 5," you must first subtract 5, then divide by 2. If the original function is $f(x) = 2x + 5$, the inverse is $f^{-1}(x) = \frac{x - 5}{2}$. Notice how the operations are reversed and the order is opposite: subtraction undoes addition, and division undoes multiplication. This is the essence of every inverse function — no matter how complex the original rule, the inverse always traces the steps backward.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Find the inverse of $f(x) = 5x + 2$. Use the swap-and-solve method and write your answer using inverse notation. 2 MARKS
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Answer in your workbook.
9. (a) Find the inverse of $f(x) = \dfrac{2x - 1}{3}$. (b) Verify that $f(f^{-1}(x)) = x$. Show all steps. 4 MARKS
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10. A student is asked to find the inverse of $f(x) = x^2 + 2$. They write: $y = x^2 + 2$, swap to get $x = y^2 + 2$, and solve to obtain $f^{-1}(x) = \sqrt{x - 2}$. Evaluate whether this answer is complete and correct. If there is an error or missing information, explain what it is and how it should be fixed. 4 MARKS
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Answer in your workbook.
1. $y = 4x - 3 \Rightarrow x = 4y - 3 \Rightarrow y = \frac{x + 3}{4}$. So $f^{-1}(x) = \frac{x + 3}{4}$.
2. $y = \frac{x}{2} + 5 \Rightarrow x = \frac{y}{2} + 5 \Rightarrow x - 5 = \frac{y}{2} \Rightarrow y = 2(x - 5) = 2x - 10$. So $f^{-1}(x) = 2x - 10$.
3. $y = \sqrt{x + 2} \Rightarrow x = \sqrt{y + 2} \Rightarrow x^2 = y + 2 \Rightarrow y = x^2 - 2$. So $f^{-1}(x) = x^2 - 2$ (with $x \geq 0$ since original range is $\geq 0$).
4. $y = \frac{3}{x + 1} \Rightarrow x = \frac{3}{y + 1} \Rightarrow x(y + 1) = 3 \Rightarrow y + 1 = \frac{3}{x} \Rightarrow y = \frac{3}{x} - 1 = \frac{3 - x}{x}$. So $f^{-1}(x) = \frac{3 - x}{x}$.
1. $y = 3x + 7 \Rightarrow x = 3y + 7 \Rightarrow y = \frac{x - 7}{3}$. So $D(x) = \frac{x - 7}{3}$.
2. $E(3) = 3(3) + 7 = 16$.
3. $D(22) = \frac{22 - 7}{3} = \frac{15}{3} = 5$. The 5th letter is E.
1. A — $y = 2x + 3 \Rightarrow x = 2y + 3 \Rightarrow y = \frac{x - 3}{2}$.
2. B — The inverse reverses inputs and outputs, interchanging domain and range.
3. B — $f^{-1}(x) = \frac{x + 6}{3}$, so $f^{-1}(3) = \frac{9}{3} = 3$.
4. C — Reflection in the line $y = x$.
5. C — $f(x) = x^2$ is not one-to-one without a domain restriction.
Q8 (2 marks): $y = 5x + 2 \Rightarrow x = 5y + 2 \Rightarrow 5y = x - 2 \Rightarrow y = \frac{x - 2}{5}$ [1]. Therefore $f^{-1}(x) = \frac{x - 2}{5}$ [1].
Q9 (4 marks):
(a) $y = \frac{2x - 1}{3} \Rightarrow x = \frac{2y - 1}{3} \Rightarrow 3x = 2y - 1 \Rightarrow 2y = 3x + 1 \Rightarrow y = \frac{3x + 1}{2}$So $f^{-1}(x) = \frac{3x + 1}{2}$ [2 marks for method and answer].
(b) $f(f^{-1}(x)) = f\left(\frac{3x + 1}{2}\right) = \frac{2\left(\frac{3x + 1}{2}\right) - 1}{3} = \frac{(3x + 1) - 1}{3} = \frac{3x}{3} = x$[2 marks for correct substitution and simplification].
Q10 (4 marks): The student's answer is incomplete [1]. The function $f(x) = x^2 + 2$ is not one-to-one over its natural domain because $f(a) = f(-a)$ [1]. Before finding an inverse, the domain must be restricted (e.g. to $x \geq 0$) [1]. Additionally, solving $x = y^2 + 2$ actually gives $y = \pm\sqrt{x - 2}$, but the restricted domain means we take only the positive root: $f^{-1}(x) = \sqrt{x - 2}$ with domain $x \geq 2$ [1].
Scale the platforms using your knowledge of inverse functions and their graphs. Pool: lessons 1–6.
Tick when you've finished all activities and checked your answers.