Domain Restriction for Inverses
The inverse trig functions — $\arcsin$, $\arccos$, $\arctan$ — are some of the most powerful tools in Year 12 calculus. But they only exist because mathematicians chose a clever domain restriction. In this lesson you'll understand exactly what that means: how restricting a function's domain can transform a many-to-one curve into a one-to-one function with a proper inverse. This is the key that unlocks $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$.
Does $f(x) = x^2$ have an inverse? Without looking it up — think about the horizontal line test. If it doesn't have an inverse, can you change its domain so that it does? What would you restrict it to?
Every domain-restriction question in this lesson uses exactly two ideas. Lock find the vertex and restrict to one side and choose the correct square-root sign based on the restriction into memory and you'll solve every question.
Every domain-restriction task lives on one of two roads: identify the turning point (vertex) and restrict to one branch, then take the positive or negative square root based on which branch you chose.
Key facts
- Many common functions ($x^2$, $\sin x$, $\cos x$) are NOT one-to-one over their natural domains
- Standard restrictions: $x^2$ uses $x \ge 0$; $\sin x$ uses $[-\pi/2, \pi/2]$; $\cos x$ uses $[0, \pi]$
- Domain restriction creates a one-to-one function with a proper inverse
Concepts
- Why the horizontal line test fails for $x^2$ and $\sin x$ over their natural domains
- Why restricting to one branch of a parabola gives a unique inverse
- Why the sign of the square root (+ or −) depends on which branch was chosen
Skills
- Find the largest domain restriction that includes a specified point
- Find $f^{-1}(x)$ after restricting a quadratic, including choosing the correct sign
- State the domain and range of the resulting inverse function
Functions like $f(x) = x^2$ and $f(x) = \sin x$ are not one-to-one over their natural domains:
- $f(x) = x^2$: both $x = 2$ and $x = -2$ give $f(x) = 4$. A horizontal line at $y = 4$ cuts the parabola twice — the horizontal line test fails.
- $f(x) = \sin x$: infinitely many $x$-values give the same output ($\sin(0) = \sin(\pi) = \sin(2\pi) = 0$, etc.).
If we try to define an inverse, $f^{-1}(4)$ would need to equal both $2$ and $-2$ simultaneously — impossible for a function. The fix is to restrict the domain to a maximal interval where $f$ is one-to-one.
By restricting to $x \ge 0$ for $x^2$, every output now comes from exactly one input — and the inverse $f^{-1}(x) = \sqrt{x}$ (the positive square root) is well-defined.
f(x) = x^2 fails the HLT over all reals: restrict to x 0 or x 0; Restriction to x 0 gives inverse f^{-1}(x) = x; to x 0 gives f^{-1}(x) = -x
Pause — copy the domain-restriction rule for $f(x) = x^2$ into your book: restrict to $x \ge 0$ to get $f^{-1}(x) = \sqrt{x}$ (range $y \ge 0$), or to $x \le 0$ to get $f^{-1}(x) = -\sqrt{x}$ (range $y \le 0$).
Quick check: Why does $f(x) = x^2$ not have an inverse over all real $x$?
We just saw that $f(x) = x^2$ fails the horizontal line test over all reals, so we restrict to $x ge 0$ (or $x le 0$) before finding the inverse. That raises a question: for a general quadratic $(x-h)^2+k$, what is the standard restricted domain and what inverse formula does it produce? This card answers it → restrict to $x ge h$ (right branch) to get $f^{-1}(x) = h + sqrt{x-k}$, or to $x le h$ (left branch) for the negative root form.
These standard restrictions are used throughout the course and appear in HSC exams. Know them by heart:
| Function | Natural Domain | Common Restriction | Inverse |
|---|---|---|---|
| $f(x) = x^2$ | All real $x$ | $x \ge 0$ | $f^{-1}(x) = \sqrt{x}$, $x \ge 0$ |
| $f(x) = x^2$ | All real $x$ | $x \le 0$ | $f^{-1}(x) = -\sqrt{x}$, $x \ge 0$ |
| $f(x) = \sin x$ | All real $x$ | $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$ | $f^{-1}(x) = \arcsin x$ |
| $f(x) = \cos x$ | All real $x$ | $0 \le x \le \pi$ | $f^{-1}(x) = \arccos x$ |
For a general quadratic $f(x) = (x - h)^2 + k$ (vertex at $x = h$), restrict to either $x \ge h$ or $x \le h$. The choice depends on which side contains the required point.
Sign rule for the inverse: After completing the square and swapping, you get $(y - h)^2 = x - k$. Taking the square root:
- If $y \ge h$ (right-branch restriction): $y - h = +\sqrt{x - k}$, so $f^{-1}(x) = h + \sqrt{x - k}$
- If $y \le h$ (left-branch restriction): $y - h = -\sqrt{x - k}$, so $f^{-1}(x) = h - \sqrt{x - k}$
General quadratic (x-h)^2 + k: vertex at x = h; restrict to x h or x h; Right branch (x h): inverse is f^{-1}(x) = h + x-k, take positive root
Pause — copy the standard restriction table into your book: for $(x-h)^2+k$, restrict to $x \ge h$ giving $f^{-1}(x) = h + \sqrt{x-k}$, or to $x \le h$ giving $f^{-1}(x) = h - \sqrt{x-k}$.
Did you get this? True or false: there is only one correct way to restrict the domain of $f(x) = x^2 - 4x + 7$ to make it one-to-one.
Worked examples · 3 in a row, reveal as you go
Restrict the domain of $f(x) = (x - 1)^2$ so that $f^{-1}$ contains the point $(4, 3)$. Find $f^{-1}(x)$ and state its domain.
Find the largest domain containing $x = 0$ for which $f(x) = x^2 - 4x + 5$ has an inverse. Find $f^{-1}(x)$ and state its domain.
Find the largest domain restriction for $f(x) = x^2 - 6x + 10$ that includes $x = 4$. Find $f^{-1}(x)$.
Fill the gap: For $f(x) = (x-2)^2 + 3$ restricted to $x \le 2$, the inverse is $f^{-1}(x) = 2$ $\sqrt{x-3}$ and the domain of $f^{-1}$ is $x \ge$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $f(x) = (x-4)^2 + 1$ restricted to $x \le 4$, the correct inverse is $f^{-1}(x) = 4 - \sqrt{x-1}$.
Activities · practice with the ideas
Restrict the domain of $f(x) = x^2 - 6x + 10$ to include $x = 4$, and find $f^{-1}(x)$. State the domain of $f^{-1}$.
Explain why $f(x) = \sin x$ does not have an inverse over its natural domain, and state the standard restriction used to define $\arcsin x$.
If $f(x) = (x + 2)^2 - 3$ is restricted to $x \le -2$, find $f^{-1}(x)$ and state its domain.
For $f(x) = (x-3)^2 + 2$ restricted to $x \ge 3$, find $f^{-1}(6)$.
In your own words, explain why we must choose the negative square root when restricting a parabola to its left branch. Give an example to illustrate.
Earlier you were asked whether $f(x) = x^2$ has an inverse and whether changing the domain could fix it. Now you have the full framework. Why do we need $y \ge 1$ (not just $y \ge 0$) when taking the square root in Worked Example 1? What would go wrong if we chose the wrong side?
Odd one out: Which of the following requires the LARGEST domain restriction to make it one-to-one?
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. Restrict the domain of $f(x) = x^2 - 6x + 10$ to make it one-to-one, with the restriction including $x = 4$. Find $f^{-1}(x)$ and state its domain. (3 marks)
Q2. Explain why $f(x) = \sin x$ does not have an inverse over its natural domain, and state the standard restriction used to define $\arcsin x$. (2 marks)
Q3. If $f(x) = (x + 2)^2 - 3$ is restricted to $x \le -2$, find $f^{-1}(x)$ and state its domain. (3 marks)
Comprehensive answers (click to reveal)
Activity 1: 1. $f(x)=(x-3)^2+1$, vertex at $x=3$; include $x=4 \Rightarrow$ restrict to $x \ge 3$; $f^{-1}(x) = 3 + \sqrt{x-1}$, domain $x \ge 1$ · 2. $\sin x$ repeats every $2\pi$: e.g. $\sin(0)=\sin(\pi)=0$, HLT fails. Standard restriction: $[-\pi/2, \pi/2]$ · 3. Left branch ($x \le -2$): swap gives $(y+2)^2 = x+3$; take $-\sqrt{}$ since $y \le -2$; $f^{-1}(x) = -2 - \sqrt{x+3}$, domain $x \ge -3$ · 4. $f^{-1}(x) = 3 + \sqrt{x-2}$; $f^{-1}(6) = 3 + \sqrt{4} = 5$ · 5. Left branch means $y \le h$, so $y - h \le 0$: squaring both sides of $y - h = \pm\sqrt{\ldots}$ always gives $+$, but to keep $y \le h$ we need the minus sign. Example: $(x-2)^2, x \le 2 \Rightarrow f^{-1}(x) = 2 - \sqrt{x}$.
Q1 (3 marks): $f(x) = (x-3)^2 + 1$ [0.5], vertex $x=3$, restrict to $x \ge 3$ since $4 > 3$ [0.5]. Swap: $x = (y-3)^2 + 1$, $y \ge 3$; $(y-3)^2 = x-1$; $y-3 = +\sqrt{x-1}$ [1]; $f^{-1}(x) = 3 + \sqrt{x-1}$, domain $x \ge 1$ [1].
Q2 (2 marks): $\sin x$ is not one-to-one: $\sin(0) = \sin(\pi) = 0$ (or similar), HLT fails [1]. Standard restriction: $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$ [1].
Q3 (3 marks): Vertex at $x = -2$; restrict to $x \le -2$ [0.5]. Swap: $x = (y+2)^2 - 3$, $y \le -2$; $(y+2)^2 = x+3$ [1]; $y+2 = -\sqrt{x+3}$ (negative root since $y \le -2$) [1]; $f^{-1}(x) = -2 - \sqrt{x+3}$, domain $x \ge -3$ [0.5].
Five timed questions on domain restriction. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering domain restriction questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.