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.
Functions like $f(x) = x^2$ and $f(x) = \sin x$ are not one-to-one over their natural domains:
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:
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)
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].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
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