NESA HSC-style-style practice: Let f(x) = x^2 - 3. (i) Show that f does not have an inverse function by referring to the horizontal line test. (ii) The domain of f is now restricted to x 0. Find the rule for the inverse function of this restricted function, and state its domain. (iii) Verify that f(f^-1(x)) = x for the restricted function.

The contrast between the unrestricted (fails) and restricted (works) cases is the heart of this dot point. **Part (i): the horizontal line test.** A function has an inverse function exactly when no horizontal line meets its graph more than once. The parabola f(x) = x^2 - 3 is symmetric in the y-axis, so the horizontal line y = 1 meets it where x^2 - 3 = 1, that is x^2 = 4, so x = 2 and x = -2: two points. Because a horizontal line cuts the graph twice, f is many-to-one and its inverse is not a function. **Part (ii): restrict and invert.** With x 0 the graph is only the right-hand half of the parabola, which is one-to-one (each height occurs once), so an inverse function exists. Write y = x^2 - 3 with x 0, then swap: x = y^2 - 3 with y 0. Solve for y: y^2 = x + 3, and since y 0 we take the positive root, y = sqrt(x + 3). So f^-1(x) = sqrt(x + 3). Its domain is the range of the original function: the restricted f has range y -3, so the domain of f^-1 is x -3. **Part (iii): verify.** Substitute: f(f^-1(x)) = (sqrt(x + 3))^2 - 3 = (x + 3) - 3 = x, valid for x -3. The composition returns the input, confirming the inverse. Markers reward naming the line that cuts twice in part (i), keeping the restriction y 0 that selects the positive root in part (ii), and the clean simplification to x in part (iii). A common error is to drop the discussion and forget that the restriction is what makes the single root legitimate.

NSWMaths Extension 1Syllabus dot point

How do we form the inverse of a relation by swapping $x$ and $y$ , when is that inverse itself a function, and how do we find and verify a rule for $f^{-1}(x)$ ?

Form the inverse relation by reflecting $y = f(x)$ in the line $y = x$ , use the horizontal line test to decide whether the inverse is a function, find the rule for $f^{-1}(x)$ and verify it by showing $f(f^{-1}(x)) = x$ , swap the domain and range, and restrict a domain so that a many-to-one function gains an inverse

The Year 11 Extension 1 dot point on inverse relations and functions. Swap x and y to get the inverse (a reflection in y = x), use the horizontal line test to decide whether it is a function, find and verify f-inverse by composition, swap the domain and range, and restrict a domain so a many-to-one function gains an inverse.

Generated by Claude Opus 4.822 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The inverse relation of $y = f(x)$ is found by swapping $x$ and $y$ (and any restriction); its graph is the reflection of $y = f(x)$ in the line $y = x$ , so a point $(a, b)$ maps to $(b, a)$ . That inverse is a function if and only if $f$ is one-to-one, which you check with the horizontal line test: no horizontal line may cross $y = f(x)$ more than once. To find the rule, swap and solve for $y$ , then verify by showing $f(f^{-1}(x)) = x$ . The domain and range swap: the domain of $f^{-1}$ is the range of $f$ . A many-to-one function (like a parabola) has no inverse function until you restrict the domain to a piece where it is one-to-one, and the restriction picks the correct branch (and sign of a root). A rational $\tfrac{ax+b}{cx+d}$ is its own inverse exactly when $a + d = 0$ . Beware: $f^{-1}(x)$ is the inverse, not the reciprocal $\tfrac{1}{f(x)}$ , and the mirror is $y = x$ , not an axis.

What this dot point is asking

NESA gives you a relation, usually a function $y = f(x)$ , and asks you to build its inverse: the relation you get by reversing every input-output pair. Geometrically that is a single clean idea, reflect the graph in the line $y = x$ , and the rest of the topic is the consequences of that one reflection.

There are four skills bundled together here:

Form the inverse relation by swapping $x$ and $y$ (and solving for $y$ if you can).
Decide whether the inverse is a function using the horizontal line test on the original.
Find and verify the rule for $f^{-1}(x)$ , checking it by composition: $f(f^{-1}(x)) = x$ .
Swap the domain and range, and restrict a domain when the inverse would otherwise fail to be a function.

The reflection in $y = x$ is the thread through all of it: it is why the coordinates swap, why the domain and range swap, and why "no horizontal line twice" is the test that the inverse passes the vertical line test. The same "act on the picture, feature by feature" habit that drives sketching $y = |f(x)|$ and $y = f(|x|)$ and sketching $y = \tfrac{1}{f(x)}$ applies here, with the line $y = x$ as the mirror. This Year 11 page is the first-contact treatment; the Year 12 page inverse functions and composite functions extends it with composition and calculus.

The answer

Forming the inverse relation: swap $x$ and $y$

A relation is just a set of ordered pairs $(x, y)$ . Its inverse relation is the set of all the reversed pairs $(y, x)$ : whatever the relation sent $x$ to, the inverse sends back. So the way to get the equation of the inverse is to swap $x$ and $y$ everywhere in the equation (and in any restriction), then, if possible, solve for $y$ to get a rule.

For example, the inverse of $y = 2x - 6$ is found by writing $x = 2y - 6$ and solving: $y = \dfrac{x + 6}{2}$ . The inverse of $y = x^3 + 1$ is $x = y^3 + 1$ , which solves to $y = \sqrt[3]{x - 1}$ .

Two facts fall straight out of the definition:

The inverse of the inverse is the original relation (swapping twice gets you back).
The graph of the inverse is the reflection of the original graph in the line $y = x$ , because reflecting in $y = x$ is exactly what exchanges the first and second coordinates of every point.

Notice the two named points: $(0, 1)$ on the cubic maps to $(1, 0)$ on the inverse, the coordinates simply swapping. Every point obeys the same rule, which is why the whole curve is a mirror image in $y = x$ .

The horizontal line test: is the inverse a function?

Swapping the pairs always gives an inverse relation, but it may not be an inverse function. A function is allowed only one $y$ for each $x$ (the vertical line test). When we reflect in $y = x$ , vertical lines become horizontal lines, so the inverse passes the vertical line test exactly when the original passes a horizontal line test.

The inverse of $y = f(x)$ is a function if and only if no horizontal line crosses the graph of $y = f(x)$ more than once, that is, the original function is one-to-one.

A one-to-one function never repeats a $y$ -value: different inputs give different outputs. A many-to-one function repeats at least one $y$ -value (some horizontal line cuts twice), and when reflected those two points share an $x$ -value, breaking the vertical line test for the inverse.

The line on the left is one-to-one, so its inverse is a function. The parabola on the right is many-to-one (the line $y = 1$ meets it at $x = 2$ and $x = -2$ ), so its inverse is the two-armed relation $y = \pm\sqrt{x + 3}$ , which is not a function.

Finding the rule for $f^{-1}(x)$ and verifying it

Once you know the inverse is a function, the recipe for its rule is the same swap, finished off by making $y$ the subject:

Start from $y = f(x)$ .
Swap $x$ and $y$ (and swap any restriction).
Solve for $y$ . The result is $y = f^{-1}(x)$ .
Verify by composition: show $f(f^{-1}(x)) = x$ (substituting the inverse into $f$ should collapse to $x$ ).

The verification is not optional decoration. It is the definition of "inverse" in action, and in an exam it both earns marks and catches algebra slips. If $f(f^{-1}(x))$ does not simplify to $x$ , you have made an error.

For the line $f(x) = 2x - 6$ : swap to $x = 2y - 6$ , solve to $f^{-1}(x) = \dfrac{x + 6}{2}$ , then check $f(f^{-1}(x)) = 2 \cdot \dfrac{x + 6}{2} - 6 = (x + 6) - 6 = x$ . It works, so the rule is right.

A rational function: $\dfrac{ax + b}{cx + d}$

The swap-and-solve method shines on a rational function, where the algebra is the whole task. Take $f(x) = \dfrac{2x + 3}{x - 1}$ . Swap: $x = \dfrac{2y + 3}{y - 1}$ . Multiply out: $x(y - 1) = 2y + 3$ , so $xy - x = 2y + 3$ . Collect the $y$ terms: $xy - 2y = x + 3$ , that is $y(x - 2) = x + 3$ . Divide: $f^{-1}(x) = \dfrac{x + 3}{x - 2}$ .

The asymptotes tell the same story as the reflection. The original has a vertical asymptote at $x = 1$ (zero denominator) and a horizontal asymptote at $y = 2$ (ratio of leading coefficients). Reflecting in $y = x$ swaps the roles of $x$ and $y$ , so the inverse has a vertical asymptote at $x = 2$ and a horizontal asymptote at $y = 1$ , which is exactly what the rule $\dfrac{x + 3}{x - 2}$ shows.

Domain and range swap

Because the inverse simply reverses every pair, the set of inputs and the set of outputs trade places:

the domain of $f^{-1}$ is the range of $f$ ;
the range of $f^{-1}$ is the domain of $f$ .

This is the reflection again: the $x$ -extent and the $y$ -extent of the graph are exchanged when you flip in $y = x$ . It is also how you write down the domain of an inverse that involves a square root, where the rule alone would not tell you the full story.

Exam technique

Markers reward a clear, complete method, not just a final rule. Lay it out like this.

Always swap $x$ and $y$ first, then solve. Write the line "swap $x$ and $y$ " or " $x = \ldots$ " so the marker sees the step that defines the inverse. Do not just declare an answer.
Verify with composition and say so. Show $f(f^{-1}(x)) = x$ . The simplification to $x$ is worth a mark and proves the rule.
Carry restrictions through the swap. A restriction on $x$ becomes a restriction on $y$ , which decides the sign of a square root and the inverse's domain. State the inverse's domain (it is the range of $f$ ).
Name the geometry. "Reflection in the line $y = x$ " earns marks on "describe the relationship" parts; a point $(a, b)$ maps to $(b, a)$ .
For a graph, plot $y = x$ as a dashed line and reflect a few key points (intercepts, asymptote crossings, endpoints), then draw the smooth reflected curve through them.

Restricting a domain so an inverse exists

When a function is many-to-one, its inverse is not a function, but you can often restrict the domain to a piece on which the function is one-to-one, and that piece has an inverse function. The classic case is a parabola: the whole of $f(x) = x^2 - 3$ fails the horizontal line test, but the right-hand branch $x \ge 0$ is increasing and one-to-one.

On that branch, swap $y = x^2 - 3$ ( $x \ge 0$ ) to $x = y^2 - 3$ ( $y \ge 0$ ); solving gives $y^2 = x + 3$ , and the restriction $y \ge 0$ forces the positive root, so $f^{-1}(x) = \sqrt{x + 3}$ . The restriction is what makes the single square root legitimate: without it, $y = \pm\sqrt{x + 3}$ is two-armed and not a function. (Choosing the left branch $x \le 0$ instead would give the other inverse, $f^{-1}(x) = -\sqrt{x + 3}$ .)

A quick way to spot functions that will always need restricting: any even function whose domain includes a non-zero number fails the horizontal line test, because $f(-a) = f(a)$ gives a repeated height. Parabolas and $y = x^4$ are typical. So is any polynomial with two or more distinct real linear factors, since it must turn back on itself between the roots.

A self-inverse function

A few functions are their own inverse: $f^{-1} = f$ , equivalently $f(f(x)) = x$ . Their graphs are unchanged by reflection in $y = x$ (they are symmetric in that line). The simplest is $f(x) = \dfrac{6}{x}$ : $f(f(x)) = \dfrac{6}{6/x} = x$ . More generally, a rational function $\dfrac{ax + b}{cx + d}$ is self-inverse exactly when $a + d = 0$ . For instance $f(x) = \dfrac{x + 3}{x - 1}$ has $a = 1$ and $d = -1$ , so $a + d = 0$ , and indeed substituting confirms $f(f(x)) = x$ for every $x$ in the domain. Recognising this lets you state the inverse instantly, with no swap-and-solve needed.

Combining the ideas, stage by stage

The full procedure, find the rule and present the graph, is best seen as a short sequence. Take the one-to-one line $f(x) = 2x - 6$ and build its inverse graph step by step.

Stage 1, plot the original $y = f(x)$: Draw $f(x) = 2x - 6$ , a line through $(3, 0)$ and $(0, -6)$ with gradient $2$ . It is one-to-one, so an inverse function exists.
Stage 2, draw the mirror line $y = x$: Add the dashed line $y = x$ . This is the mirror: the inverse graph will be the reflection of the line in it.
Stage 3, reflect the key points: Pick easy points on $f$ and swap their coordinates: $(1, -4) \to (-4, 1)$ , $(3, 0) \to (0, 3)$ and $(5, 4) \to (4, 5)$ . Each reflected point sits on the opposite side of $y = x$ , the same perpendicular distance away.
Stage 4, draw the inverse line: Join the reflected points to get $y = f^{-1}(x)$ , the line through $(0, 3)$ and $(-4, 1)$ . Solving algebraically confirms it: $f^{-1}(x) = \dfrac{x + 6}{2} = \dfrac{x}{2} + 3$ , gradient $\tfrac{1}{2}$ and $y$ -intercept $3$ , exactly the reflected line.

Common traps

Forgetting to swap: The inverse comes from swapping $x$ and $y$ then solving for $y$ , not from rearranging the original for $x$ and stopping. The final rule must be $y$ (or $f^{-1}(x)$ ) in terms of $x$ .
Calling every inverse a function: An inverse relation is always defined, but it is a function only if the original is one-to-one. Apply the horizontal line test before writing $f^{-1}$ .
Dropping the $\pm$ when inverting a square: Inverting $y = x^2 + c$ gives $y = \pm\sqrt{x - c}$ , which is two-valued. Only after the domain is restricted (e.g. $x \ge 0$ ) can you legitimately keep a single sign.
Losing the restriction in the swap: A restriction on $x$ becomes a restriction on $y$ after swapping, and that restriction is what picks the correct branch and sets the inverse's domain. Carry it through.
Confusing $f^{-1}(x)$ with $\dfrac{1}{f(x)}$: The notation $f^{-1}$ means the inverse function, not the reciprocal. They are completely different: the inverse of $f(x) = 2x - 6$ is $\dfrac{x + 6}{2}$ , whereas $\dfrac{1}{f(x)} = \dfrac{1}{2x - 6}$ .
Reflecting in the wrong line: The inverse is the reflection in $y = x$ , not in the $x$ -axis or $y$ -axis. Reflecting in the $x$ -axis gives $y = -f(x)$ ; only $y = x$ swaps the coordinates.

How exam questions ask about inverses

The wording tells you which skill to use.

"Find the inverse function $f^{-1}(x)$ ." Swap $x$ and $y$ , solve for $y$ , and (if asked) verify by composition. State the domain if a root or restriction is involved.
"Show that the inverse is / is not a function." Apply the horizontal line test to $y = f(x)$ : name a horizontal line that cuts once (it is) or twice (it is not).
"Verify that $f(f^{-1}(x)) = x$ ." Substitute your $f^{-1}(x)$ into $f$ and simplify; it must collapse to $x$ .
"State the domain (or range) of $f^{-1}$ ." The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .
"Restrict the domain of $f$ so that an inverse function exists." Choose a largest interval on which $f$ is one-to-one (e.g. $x \ge$ the turning-point value for a parabola), then invert on that piece.
"Sketch $y = f(x)$ and $y = f^{-1}(x)$ on the same axes." Draw $y = x$ dashed, reflect a few key points, and draw the smooth reflection.
"Show that $f$ is its own inverse." Compute $f(f(x))$ and show it equals $x$ (or, for a rational $\tfrac{ax+b}{cx+d}$ , check $a + d = 0$ ).
"Describe the geometric relationship between the graphs." They are reflections of each other in the line $y = x$ .

Worked example

Find and verify the inverse of a linear function

Find the inverse function of $f(x) = 2x - 6$ , verify it by composition, and state the relationship between the two graphs.

Name the method: The inverse comes from swapping $x$ and $y$ and solving for $y$ .
Swap and solve: Write $y = 2x - 6$ , then swap to $x = 2y - 6$ . Add $6$ : $x + 6 = 2y$ . Divide by $2$ : $y = \dfrac{x + 6}{2}$ . So $f^{-1}(x) = \dfrac{x + 6}{2}$ .
Verify by composition: $f(f^{-1}(x)) = 2 \cdot \dfrac{x + 6}{2} - 6 = (x + 6) - 6 = x$ . The composition returns the input, so the rule is correct.
State the geometry: The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ .
Final answer: $f^{-1}(x) = \dfrac{x + 6}{2}$ (equivalently $\dfrac{x}{2} + 3$ ); $f(f^{-1}(x)) = x$ ; the graphs are reflections in $y = x$ .

Invert a cubic of the form $x^3 + c$

Find the inverse function of $f(x) = x^3 + 1$ , and find the point on $y = f^{-1}(x)$ corresponding to the point $(0, 1)$ on $y = f(x)$ .

Check it is one-to-one: The cube function is strictly increasing, so $f(x) = x^3 + 1$ passes the horizontal line test and has an inverse function.
Swap and solve: Write $y = x^3 + 1$ , then swap: $x = y^3 + 1$ . Subtract $1$ : $x - 1 = y^3$ . Take the cube root, which is defined for every real number (so no restriction is needed): $y = \sqrt[3]{x - 1}$ . So $f^{-1}(x) = \sqrt[3]{x - 1}$ .
Verify by composition: $f(f^{-1}(x)) = \left(\sqrt[3]{x - 1}\right)^3 + 1 = (x - 1) + 1 = x$ .
Find the corresponding point: A point $(a, b)$ on $f$ maps to $(b, a)$ on $f^{-1}$ . The point $(0, 1)$ on $y = f(x)$ (since $f(0) = 1$ ) corresponds to $(1, 0)$ on $y = f^{-1}(x)$ . Check: $f^{-1}(1) = \sqrt[3]{1 - 1} = 0$ .
Final answer: $f^{-1}(x) = \sqrt[3]{x - 1}$ , and $(0, 1)$ on $f$ corresponds to $(1, 0)$ on $f^{-1}$ .

Decide whether the inverse is a function, with and without a restriction

For $f(x) = x^2 - 3$ , (a) show the inverse is not a function over the natural domain, and (b) restrict the domain to $x \ge 0$ and find the inverse function with its domain.

Part (a): apply the horizontal line test: The parabola $f(x) = x^2 - 3$ is symmetric in the $y$ -axis. The horizontal line $y = 1$ meets it where $x^2 - 3 = 1$ , that is $x^2 = 4$ , so $x = 2$ and $x = -2$ : two points. A horizontal line cuts twice, so $f$ is many-to-one and its inverse, $y = \pm\sqrt{x + 3}$ , is not a function.
Part (b): restrict and invert: With $x \ge 0$ the graph is the increasing right branch, which is one-to-one. Write $y = x^2 - 3$ with $x \ge 0$ , then swap: $x = y^2 - 3$ with $y \ge 0$ . Solve: $y^2 = x + 3$ , and the restriction $y \ge 0$ keeps the positive root, $y = \sqrt{x + 3}$ . So $f^{-1}(x) = \sqrt{x + 3}$ .
State the domain: The domain of $f^{-1}$ is the range of the restricted $f$ , which is $y \ge -3$ , so $f^{-1}$ has domain $x \ge -3$ .
Verify: $f(f^{-1}(x)) = \left(\sqrt{x + 3}\right)^2 - 3 = (x + 3) - 3 = x$ for $x \ge -3$ .
Final answer: (a) it fails the horizontal line test (e.g. $y = 1$ cuts at $x = \pm 2$ ); (b) $f^{-1}(x) = \sqrt{x + 3}$ with domain $x \ge -3$ .

Invert a rational function

Find the inverse function of $f(x) = \dfrac{2x + 3}{x - 1}$ , and state the asymptotes of $y = f^{-1}(x)$ .

Swap the variables: Write $y = \dfrac{2x + 3}{x - 1}$ , then swap: $x = \dfrac{2y + 3}{y - 1}$ .
Clear the fraction: Multiply both sides by $(y - 1)$ : $x(y - 1) = 2y + 3$ , so $xy - x = 2y + 3$ .
Collect the $y$ terms: Bring them together: $xy - 2y = x + 3$ , that is $y(x - 2) = x + 3$ . Divide by $(x - 2)$ : $y = \dfrac{x + 3}{x - 2}$ . So $f^{-1}(x) = \dfrac{x + 3}{x - 2}$ .
Read the asymptotes: The original $f$ has vertical asymptote $x = 1$ and horizontal asymptote $y = 2$ . The inverse swaps these, so $y = f^{-1}(x)$ has vertical asymptote $x = 2$ (where $x - 2 = 0$ ) and horizontal asymptote $y = 1$ (the ratio of leading coefficients).
Final answer: $f^{-1}(x) = \dfrac{x + 3}{x - 2}$ , with vertical asymptote $x = 2$ and horizontal asymptote $y = 1$ .

Show a function is its own inverse

Show that $f(x) = \dfrac{x + 3}{x - 1}$ , defined for $x \ne 1$ , satisfies $f(f(x)) = x$ , so it is its own inverse.

Set up the composition: Substitute $f(x) = \dfrac{x + 3}{x - 1}$ into the rule:
$f(f(x)) = \frac{\dfrac{x + 3}{x - 1} + 3}{\dfrac{x + 3}{x - 1} - 1}.$
Clear the inner fractions: Multiply the top and bottom of the big fraction by $(x - 1)$ . The numerator becomes $(x + 3) + 3(x - 1) = x + 3 + 3x - 3 = 4x$ . The denominator becomes $(x + 3) - (x - 1) = x + 3 - x + 1 = 4$ .
Simplify: So $f(f(x)) = \dfrac{4x}{4} = x$ .
Interpret: Because $f(f(x)) = x$ , the function is its own inverse; its graph is symmetric in the line $y = x$ . (This fits the $\tfrac{ax+b}{cx+d}$ rule with $a = 1$ , $d = -1$ , so $a + d = 0$ .)
Final answer: $f(f(x)) = x$ , so $f$ is its own inverse.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

HSC-style4 marksThe function

f(x) = 2x - 6

is defined for all real

x

. (i) Find the rule for the inverse function

f^{-1}(x)

. (ii) Verify your answer by showing that

f(f^{-1}(x)) = x

. (iii) State the geometric relationship between the graphs of

y = f(x)

and

y = f^{-1}(x)

Show worked answer →

This is the standard three-part inverse question: find the rule, verify by composition, name the geometry.

Part (i): find the inverse: Write $y = 2x - 6$ , then swap $x$ and $y$ to get the inverse relation $x = 2y - 6$ . Solve for $y$ : add $6$ to both sides to get $x + 6 = 2y$ , then divide by $2$ , giving $y = \dfrac{x + 6}{2}$ . So $f^{-1}(x) = \dfrac{x + 6}{2}$ , which can also be written $f^{-1}(x) = \dfrac{x}{2} + 3$ .
Part (ii): verify by composition: Substitute $f^{-1}(x)$ into $f$ : $f(f^{-1}(x)) = 2 \cdot \dfrac{x + 6}{2} - 6 = (x + 6) - 6 = x$ . Because composing the two functions returns the input unchanged, $f^{-1}$ really is the inverse. (Checking the other order, $f^{-1}(f(x)) = \dfrac{(2x - 6) + 6}{2} = \dfrac{2x}{2} = x$ , confirms it both ways.)
Part (iii): the geometry: The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ . Each point $(a, b)$ on $y = f(x)$ corresponds to the point $(b, a)$ on $y = f^{-1}(x)$ .

Markers reward the swap-and-solve in part (i), the substitution that simplifies to $x$ in part (ii), and the words "reflection in the line $y = x$ " in part (iii).

HSC-style5 marksLet

f(x) = x^2 - 3

. (i) Show that

f

does not have an inverse function by referring to the horizontal line test. (ii) The domain of

f

is now restricted to

x \ge 0

. Find the rule for the inverse function of this restricted function, and state its domain. (iii) Verify that

f(f^{-1}(x)) = x

for the restricted function.

Show worked answer →

The contrast between the unrestricted (fails) and restricted (works) cases is the heart of this dot point.

Part (i): the horizontal line test: A function has an inverse function exactly when no horizontal line meets its graph more than once. The parabola $f(x) = x^2 - 3$ is symmetric in the $y$ -axis, so the horizontal line $y = 1$ meets it where $x^2 - 3 = 1$ , that is $x^2 = 4$ , so $x = 2$ and $x = -2$ : two points. Because a horizontal line cuts the graph twice, $f$ is many-to-one and its inverse is not a function.
Part (ii): restrict and invert: With $x \ge 0$ the graph is only the right-hand half of the parabola, which is one-to-one (each height occurs once), so an inverse function exists. Write $y = x^2 - 3$ with $x \ge 0$ , then swap: $x = y^2 - 3$ with $y \ge 0$ . Solve for $y$ : $y^2 = x + 3$ , and since $y \ge 0$ we take the positive root, $y = \sqrt{x + 3}$ . So $f^{-1}(x) = \sqrt{x + 3}$ . Its domain is the range of the original function: the restricted $f$ has range $y \ge -3$ , so the domain of $f^{-1}$ is $x \ge -3$ .
Part (iii): verify: Substitute: $f(f^{-1}(x)) = \left(\sqrt{x + 3}\right)^2 - 3 = (x + 3) - 3 = x$ , valid for $x \ge -3$ . The composition returns the input, confirming the inverse.

Markers reward naming the line that cuts twice in part (i), keeping the restriction $y \ge 0$ that selects the positive root in part (ii), and the clean simplification to $x$ in part (iii). A common error is to drop the $\pm$ discussion and forget that the restriction is what makes the single root legitimate.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFind the rule for the inverse function of

f(x) = 3x + 12

, and verify your answer by showing that

f^{-1}(f(x)) = x

Show worked solution →

Swap and solve: Write $y = 3x + 12$ , then swap $x$ and $y$ : $x = 3y + 12$ . Subtract $12$ : $x - 12 = 3y$ . Divide by $3$ : $y = \dfrac{x - 12}{3}$ . So $f^{-1}(x) = \dfrac{x - 12}{3}$ .
Verify by composition: Substitute $f(x) = 3x + 12$ into $f^{-1}$ : $f^{-1}(f(x)) = \dfrac{(3x + 12) - 12}{3} = \dfrac{3x}{3} = x$ . Because the composition returns the input, the rule is correct.
Answer: $f^{-1}(x) = \dfrac{x - 12}{3}$ , and $f^{-1}(f(x)) = x$ as required.

foundation3 marksThe function

g(x) = x^3 - 8

is one-to-one. Find

g^{-1}(x)

, and state the coordinates of the point on

y = g^{-1}(x)

that corresponds to the point

(2, 0)

y = g(x)

Show worked solution →

Swap and solve: Write $y = x^3 - 8$ , then swap: $x = y^3 - 8$ . Add $8$ : $x + 8 = y^3$ . Take the cube root (defined for every real number, so no restriction is needed): $y = \sqrt[3]{x + 8}$ . So $g^{-1}(x) = \sqrt[3]{x + 8}$ .
Find the corresponding point: A point $(a, b)$ on $y = g(x)$ maps to $(b, a)$ on the inverse. Here $g(2) = 2^3 - 8 = 0$ , so $(2, 0)$ lies on $y = g(x)$ ; swapping the coordinates gives $(0, 2)$ on $y = g^{-1}(x)$ . Checking: $g^{-1}(0) = \sqrt[3]{0 + 8} = \sqrt[3]{8} = 2$ , confirming $(0, 2)$ .
Answer: $g^{-1}(x) = \sqrt[3]{x + 8}$ , and the point $(2, 0)$ on $g$ corresponds to $(0, 2)$ on $g^{-1}$ .

core4 marksFind the rule for the inverse function of

h(x) = \dfrac{3x - 1}{x + 2}

, and state the equations of the vertical and horizontal asymptotes of

y = h^{-1}(x)

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Swap the variables. Write $y = \dfrac{3x - 1}{x + 2}$ , then swap $x$ and $y$ : $x = \dfrac{3y - 1}{y + 2}$ .

Clear the fraction and collect $y$ . Multiply both sides by $(y + 2)$ : $x(y + 2) = 3y - 1$ , so $xy + 2x = 3y - 1$ . Gather the $y$ terms on one side: $xy - 3y = -1 - 2x$ , that is $y(x - 3) = -(2x + 1)$ . Divide by $(x - 3)$ : $y = \dfrac{-(2x + 1)}{x - 3} = \dfrac{2x + 1}{3 - x}$ .

So $h^{-1}(x) = \dfrac{2x + 1}{3 - x}$ .

Read the asymptotes. The original $h$ has a vertical asymptote at $x = -2$ (zero denominator) and a horizontal asymptote at $y = 3$ (ratio of leading coefficients). The inverse swaps these roles, so $y = h^{-1}(x)$ has a vertical asymptote at $x = 3$ and a horizontal asymptote at $y = -2$ . This matches the rule $\dfrac{2x + 1}{3 - x}$ directly: the denominator $3 - x$ is zero at $x = 3$ , and for large $x$ the value tends to $\dfrac{2x}{-x} = -2$ .

Answer: $h^{-1}(x) = \dfrac{2x + 1}{3 - x}$ , with vertical asymptote $x = 3$ and horizontal asymptote $y = -2$ .

core4 marksThe function

f(x) = (x - 2)^2

is restricted to the domain

x \ge 2

. Find

f^{-1}(x)

, state its domain and range, and verify that

f(f^{-1}(x)) = x

Show worked solution →

Check it is one-to-one: With $x \ge 2$ the graph is the right-hand half of the parabola with vertex $(2, 0)$ , which is increasing, so it is one-to-one and an inverse function exists.
Swap and solve: Write $y = (x - 2)^2$ with $x \ge 2$ , then swap: $x = (y - 2)^2$ with $y \ge 2$ . Take the square root: $y - 2 = \pm\sqrt{x}$ , and because the restriction forces $y \ge 2$ , we keep the positive root $y - 2 = \sqrt{x}$ , so $y = 2 + \sqrt{x}$ . Thus $f^{-1}(x) = 2 + \sqrt{x}$ .
State domain and range: The restricted $f$ has domain $x \ge 2$ and range $y \ge 0$ (the smallest value is $f(2) = 0$ ). Swapping gives the inverse domain $x \ge 0$ and inverse range $y \ge 2$ .
Verify by composition: $f(f^{-1}(x)) = \left((2 + \sqrt{x}) - 2\right)^2 = \left(\sqrt{x}\right)^2 = x$ , valid for $x \ge 0$ .
Answer: $f^{-1}(x) = 2 + \sqrt{x}$ , domain $x \ge 0$ , range $y \ge 2$ , and $f(f^{-1}(x)) = x$ as required.

exam4 marksShow that the function

f(x) = \dfrac{2x + 1}{x - 2}

, defined for

x \ne 2

, is its own inverse, that is

f(f(x)) = x

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Set up the composition: A function is its own inverse when $f(f(x)) = x$ . Compute $f(f(x))$ by substituting $f(x) = \dfrac{2x + 1}{x - 2}$ wherever $x$ appears in the rule:
$f(f(x)) = \frac{2 \cdot \dfrac{2x + 1}{x - 2} + 1}{\dfrac{2x + 1}{x - 2} - 2}.$
Clear the inner fractions: Multiply the top and bottom of the large fraction by $(x - 2)$ . The numerator becomes $2(2x + 1) + 1 \cdot (x - 2) = 4x + 2 + x - 2 = 5x$ . The denominator becomes $(2x + 1) - 2(x - 2) = 2x + 1 - 2x + 4 = 5$ .
Simplify: So $f(f(x)) = \dfrac{5x}{5} = x$ .
Interpret: Because $f(f(x)) = x$ , the function $f$ is its own inverse; equivalently, reflecting its graph in the line $y = x$ produces the same curve. (This is the general pattern for $\dfrac{ax + b}{cx + d}$ : it is self-inverse exactly when $a + d = 0$ , and here $a = 2$ , $d = -2$ , so $a + d = 0$ .)
Answer: $f(f(x)) = x$ , so $f$ is its own inverse.

exam5 marksConsider

f(x) = x^2 + 1

. (i) Explain why

f

does not have an inverse function over its natural domain. (ii) The domain is restricted to

x \le 0

. Find the rule for the inverse function and state its domain. (iii) Verify that

f^{-1}(f(x)) = x

for this restricted domain.

Show worked solution →

Part (i): why there is no inverse: A function has an inverse function only if it is one-to-one, that is, no horizontal line meets the graph more than once. The parabola $f(x) = x^2 + 1$ is symmetric in the $y$ -axis, so for instance the horizontal line $y = 2$ meets it where $x^2 + 1 = 2$ , giving $x^2 = 1$ and $x = 1$ or $x = -1$ : two points. The graph fails the horizontal line test, so $f$ is many-to-one and has no inverse function.
Part (ii): restrict and invert: With $x \le 0$ the graph is the left-hand half of the parabola, which is decreasing and therefore one-to-one. Write $y = x^2 + 1$ with $x \le 0$ , then swap: $x = y^2 + 1$ with $y \le 0$ . Solve: $y^2 = x - 1$ , so $y = \pm\sqrt{x - 1}$ , and because the restriction forces $y \le 0$ we take the negative root, $y = -\sqrt{x - 1}$ . So $f^{-1}(x) = -\sqrt{x - 1}$ . Its domain is the range of the restricted $f$ : the smallest value is $f(0) = 1$ , so the range is $y \ge 1$ and the inverse domain is $x \ge 1$ .
Part (iii): verify: For $x \le 0$ , $f(x) = x^2 + 1 \ge 1$ , so it lies in the domain of $f^{-1}$ . Then $f^{-1}(f(x)) = -\sqrt{(x^2 + 1) - 1} = -\sqrt{x^2} = -|x|$ . Since $x \le 0$ , $|x| = -x$ , so $-|x| = -(-x) = x$ . The composition returns the input.
Answer: (i) it fails the horizontal line test (e.g. $y = 2$ cuts twice); (ii) $f^{-1}(x) = -\sqrt{x - 1}$ , domain $x \ge 1$ ; (iii) $f^{-1}(f(x)) = -\sqrt{x^2} = -|x| = x$ because $x \le 0$ .

What this dot point is asking

The answer

Forming the inverse relation: swap xxx and yyy

The horizontal line test: is the inverse a function?

Finding the rule for f−1(x)f^{-1}(x)f−1(x) and verifying it

A rational function: ax+bcx+d\dfrac{ax + b}{cx + d}cx+dax+b​

Domain and range swap

Restricting a domain so an inverse exists

A self-inverse function

Combining the ideas, stage by stage

How exam questions ask about inverses

Exam-style practice questions

Practice questions

Related dot points

Forming the inverse relation: swap $x$ and $y$

Finding the rule for $f^{-1}(x)$ and verifying it

A rational function: $\dfrac{ax + b}{cx + d}$