What are composite functions?

The composite f g is the function (f g)(x) = f(g(x)): apply g first, then f to the result.

What is finding an inverse algebraically?

The inverse f^-1 "undoes" f: f^-1(f(x)) = x for x dom(f) and f(f^-1(y)) = y for y range(f).

What is wrong domain for the inverse?

The domain of f^-1 equals the range of f. For f(x) = e^x with range (0, ∞), the inverse ln x has domain (0, ∞).

NESA 2022 HSC Q18-style practice: Let f(x) = 2 x - 3 and g(x) = x^2. Find f(g(x)), g(f(x)), and the inverse function f^-1(x).

Composition is "outside applied to inside". f(g(x)) = f(x^2) = 2 x^2 - 3. g(f(x)) = g(2 x - 3) = (2 x - 3)^2 = 4 x^2 - 12 x + 9. To invert f, swap x and y and solve: x = 2 y - 3 y = (x + 3)/(2). f^-1(x) = (x + 3)/(2). Markers reward correct composition in both orders (they differ), the swap-and-solve method for the inverse, and a tidy final expression.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

When does a function have an inverse, and how do we form, evaluate and graph composite and inverse functions?

Form composite functions, determine when a function has an inverse, find and graph the inverse, and use restriction of domain to invert non-one-to-one functions

A focused answer to the HSC Maths Advanced dot point on composite and inverse functions. Composition order and domain, the horizontal line test, finding the inverse by swapping and solving, the reflection in $y = x$ , and restricting domains to invert non-one-to-one functions, with worked examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

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

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What this dot point is asking

NESA wants you to form composite functions $f \circ g$ and $g \circ f$ , determine when a function has an inverse using the horizontal line test or one-to-one criterion, find the inverse algebraically, sketch it as the reflection in $y = x$ , and restrict the domain of a non-one-to-one function so an inverse exists.

The answer

Composite functions

The composite $f \circ g$ is the function $(f \circ g)(x) = f(g(x))$ : apply $g$ first, then $f$ to the result.

Domain: $x$ must be in the domain of $g$ , and $g(x)$ must be in the domain of $f$ . In other words, the natural domain of $f \circ g$ is

\{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}.

Composition is not commutative: usually $f(g(x)) \neq g(f(x))$ .

Composition is associative: $(f \circ g) \circ h = f \circ (g \circ h) = f(g(h(x)))$ .

When does an inverse exist

A function $f$ has an inverse (on its given domain) if and only if it is one-to-one: every value of $y$ in the range comes from exactly one $x$ .

Equivalent test (the horizontal line test): every horizontal line meets the graph at most once. Strictly increasing or strictly decreasing functions are automatically one-to-one. Many "natural" functions like $x^2$ , $\sin x$ , $\cos x$ , and $|x|$ are not one-to-one on their natural domain and need their domain restricted.

Finding an inverse algebraically

The inverse $f^{-1}$ "undoes" $f$ : $f^{-1}(f(x)) = x$ for $x \in \text{dom}(f)$ and $f(f^{-1}(y)) = y$ for $y \in \text{range}(f)$ .

To find $f^{-1}$ from $y = f(x)$ :

Swap $x$ and $y$ .
Solve for $y$ in terms of $x$ .
State the domain of $f^{-1}$ , which is the range of $f$ .

The inverse graph

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ . Point $(a, b)$ on $f$ corresponds to $(b, a)$ on $f^{-1}$ . Horizontal asymptotes of $f$ become vertical asymptotes of $f^{-1}$ and vice versa.

If $f$ is increasing, so is $f^{-1}$ . If $f$ is decreasing, so is $f^{-1}$ .

See the inverse appear as a reflection, stage by stage

Take $f(x) = 2x - 3$ (whose inverse is $f^{-1}(x) = \frac{x + 3}{2}$ , found below). The four panels build the inverse purely by reflecting in $y = x$ , with no algebra, so you can see why swapping coordinates is the same as mirroring.

Stage 1, draw the function and the mirror line. Plot $y = f(x) = 2x - 3$ and the dashed line $y = x$ . The line $y = x$ is the mirror; every point will be flipped across it. Mark the $y$ -intercept $(0, -3)$ to track later.

Stage 2, reflect a single point. Take a point on $f$ , say $(2, 1)$ . Reflecting in $y = x$ swaps its coordinates to give $(1, 2)$ . The connector between them crosses $y = x$ at right angles, which is exactly what "reflection in $y = x$ " means.

Stage 3, reflect the whole graph. Reflect every point of $f$ in the same way and the images trace out a new straight line: $y = f^{-1}(x) = \frac{x + 3}{2}$ . The original $f$ is now muted; its mirror image is the inverse, in accent. Because $f$ is increasing, so is its inverse.

Stage 4, read off the swapped intercepts. The reflection swaps the axes, so the $y$ -intercept of $f$ at $(0, -3)$ becomes the $x$ -intercept of $f^{-1}$ at $(-3, 0)$ . This is the graphical face of "domain and range swap" and is a quick way to sketch an inverse without finding its equation.

Restricting the domain

For a non-one-to-one function $f$ , choose a domain on which $f$ is one-to-one, then invert. Different restrictions give different inverses.

$f(x) = x^2$ : restrict to $x \ge 0$ to get $f^{-1}(x) = \sqrt{x}$ . Restricting to $x \le 0$ gives $f^{-1}(x) = -\sqrt{x}$ .
$f(x) = \sin x$ : standard restriction is $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ . Inverse is $\arcsin x$ on $[-1, 1]$ .
$f(x) = \cos x$ : standard restriction is $0 \le x \le \pi$ . Inverse is $\arccos x$ on $[-1, 1]$ .

Operations and inverses

The inverse of a composition reverses the order:

(f \circ g)^{-1} = g^{-1} \circ f^{-1},

provided both inverses exist on appropriate domains. Think of putting on socks then shoes: to undo, take off shoes then socks.

How exam questions ask about composite and inverse functions

These dot points are examined with a small set of recurring instructions. Translate each one to the method:

"Find $f(g(x))$ and $g(f(x))$ ." Substitute the inner function into the outer one, both ways, and simplify. They usually differ, so compute both; do not assume symmetry (2022 HSC Q18).
"State the domain of $f \circ g$ ." Two conditions: $x$ in the domain of $g$ , and $g(x)$ in the domain of $f$ . The binding one is whichever restricts $x$ more.
"Find $f^{-1}(x)$ " or "find the inverse function." Swap $x$ and $y$ , solve for $y$ , and state the domain of $f^{-1}$ (which is the range of $f$ ).
"Does $f$ have an inverse?" or "is $f$ one-to-one?" Apply the horizontal line test, or argue $f$ is strictly increasing/decreasing. If it fails, the answer is no on that domain.
"State a restriction so that $f$ has an inverse, then find it." Pick an interval on which $f$ is monotonic (for $x^2$ , take $x \ge 0$ or $x \le 0$ ), then invert; the restriction decides the sign of the square root (2021 HSC Q19).
"Sketch $y = f(x)$ and $y = f^{-1}(x)$ on the same axes." Draw $f$ , draw $y = x$ dashed, and reflect; mark that intercepts and asymptotes swap axes. You do not need the inverse's equation to draw it.
"Show that $g = f^{-1}$ " or "verify the inverse." Compute $f(g(x))$ (and/or $g(f(x))$ ) and show it simplifies to $x$ .
"Where do $f$ and $f^{-1}$ meet?" For an increasing $f$ , intersections lie on $y = x$ , so solve $f(x) = x$ .

Worked example

Six examples covering composition with a domain check, finding inverses of linear and exponential functions, restricting domains, verifying an inverse by composition, and inverting a real Australian conversion formula.

Form a composition and find its domain

Let $f(x) = \sqrt{x}$ (domain $x \ge 0$ ) and $g(x) = x - 4$ . Apply $g$ first, then $f$ : this means substituting $g(x)$ into $f$ .

f(g(x)) = \sqrt{g(x)} = \sqrt{x - 4}.

The domain requires two conditions: $x$ must be in the domain of $g$ (all of $\mathbb{R}$ ) and $g(x)$ must be in the domain of $f$ (that is, $g(x) \ge 0$ ). The binding constraint is $x - 4 \ge 0$ , giving $x \ge 4$ .

Domain of $f \circ g$ : $[4, \infty)$ .

Find the inverse of a linear function

For $f(x) = 5 - 2x$ , use the swap-and-solve method. Write $y = 5 - 2x$ , swap $x$ and $y$ to get $x = 5 - 2y$ , then solve for $y$ :

2y = 5 - x \implies y = \frac{5 - x}{2}.

f^{-1}(x) = \frac{5 - x}{2}.

Both $f$ and $f^{-1}$ are decreasing linear functions, and their graphs are reflections of each other in the line $y = x$ .

Find the inverse of an exponential function

For $f(x) = e^x + 1$ , the domain is $\mathbb{R}$ and the range is $(1, \infty)$ (since $e^x > 0$ always, so $e^x + 1 > 1$ always). The inverse will therefore have domain $(1, \infty)$ .

Swap $x$ and $y$ : $x = e^y + 1$ . Solve for $y$ by isolating the exponential, then taking the natural log:

e^y = x - 1 \implies y = \ln(x - 1).

f^{-1}(x) = \ln(x - 1), \quad \text{domain } (1, \infty).

Restrict the domain to create an invertible function

The function $f(x) = (x-1)^2$ on all of $\mathbb{R}$ is not one-to-one: for example, $f(0) = f(2) = 1$ . A horizontal line at $y = 1$ hits the graph twice.

Restrict to $x \ge 1$ : on this half of the parabola, $f$ is strictly increasing with range $[0, \infty)$ , so it passes the horizontal line test.

Now swap and solve: $x = (y-1)^2$ with $y \ge 1$ , so $y - 1 = \sqrt{x}$ (the positive square root, because the restriction forces $y \ge 1$ ).

f^{-1}(x) = 1 + \sqrt{x}, \quad \text{domain } [0, \infty).

Verify by composition

To check that $f^{-1}(x) = \frac{5-x}{2}$ is the correct inverse of $f(x) = 5 - 2x$ , compute $f(f^{-1}(x))$ and confirm it equals $x$ :

f\!\left(\frac{5-x}{2}\right) = 5 - 2 \cdot \frac{5-x}{2} = 5 - (5 - x) = x. \checkmark

Invert an Australian conversion formula

A driver wants the reverse of the speed-conversion rule that turns kilometres per hour into metres per second. The forward rule is $f(v) = \frac{5}{18}v$ (a $100$ km/h freeway speed becomes $f(100) = \frac{500}{18} \approx 27.8$ m/s). Find $f^{-1}$ , which converts metres per second back to kilometres per hour.

Write $y = \frac{5}{18}v$ , swap the variables to get $v = \frac{5}{18}y$ , and solve:

y = \frac{18}{5}v.

So $f^{-1}(v) = \frac{18}{5}v = 3.6\,v$ . Check it against the school-zone limit: $40$ km/h forward gives $f(40) = \frac{200}{18} \approx 11.1$ m/s, and the inverse returns $f^{-1}(11.1) = 3.6 \times 11.1 \approx 40$ km/h. The inverse of "multiply by $\frac{5}{18}$ " is "multiply by $\frac{18}{5}$ ", as expected for a linear scaling.

Final answers: domain of $f \circ g$ is $[4, \infty)$ ; $f^{-1}(x) = \frac{5-x}{2}$ for $f(x) = 5-2x$ ; $f^{-1}(x) = \ln(x-1)$ for $f(x) = e^x+1$ ; $f^{-1}(x) = 1+\sqrt{x}$ on $[0, \infty)$ for the restricted parabola; $f^{-1}(v) = \frac{18}{5}v$ for the km/h to m/s rule.

Common traps

Composing in the wrong order: $f(g(x))$ applies $g$ first; $g(f(x))$ applies $f$ first. These give different functions in general.
Confusing the inverse with the reciprocal: $f^{-1}(x)$ means the inverse function, not $\frac{1}{f(x)}$ . For example, if $f(x) = x + 2$ , then $f^{-1}(x) = x - 2$ , not $\frac{1}{x + 2}$ .
Forgetting to restrict: Writing $y = \sqrt{x}$ as "the inverse of $y = x^2$ " without specifying $x \ge 0$ misses half the picture: the original function must be made one-to-one first.
Wrong domain for the inverse: The domain of $f^{-1}$ equals the range of $f$ . For $f(x) = e^x$ with range $(0, \infty)$ , the inverse $\ln x$ has domain $(0, \infty)$ .
Mixing up which side gets the $\pm$: When solving $x = y^2$ for $y$ , the choice of $\sqrt{x}$ or $-\sqrt{x}$ is determined by the restricted domain of $f$ , not arbitrary.

Exam technique

The marks here come from order, domains, and the reflection. To collect them:

For composition, write the substitution explicitly. Show $f(g(x)) = f(\,\text{[the rule for } g]\,)$ before simplifying, so a marker sees you applied the inner function first.
Always state the domain (and range) when asked for an inverse. "Find the inverse" almost always carries a mark for the domain of $f^{-1}$ , which is the range of $f$ . Leaving it off is a common one-mark loss.
Justify one-to-one, do not just assert it. Say "strictly increasing, so one-to-one" or "the horizontal line $y = c$ meets the graph twice, so not one-to-one". A bare yes/no rarely earns the reasoning mark.
Use the reflection to sketch fast. With $y = x$ drawn, reflect a couple of key points (intercepts, an endpoint) rather than tabulating the inverse; mark that domain and range swap.
Keep $f^{-1}$ and $\frac{1}{f}$ separate. $f^{-1}(x)$ is the inverse function; $\frac{1}{f(x)}$ is the reciprocal. They are almost never equal.
Let the restriction fix the sign. When you invert a restricted parabola, the restriction ( $x \ge 1$ , say) tells you to take $+\sqrt{\,}$ rather than $-\sqrt{\,}$ ; state which and why.

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.

2022 HSC Q184 marksLet

f(x) = 2 x - 3

and

g(x) = x^2

. Find

f(g(x))

g(f(x))

, and the inverse function

f^{-1}(x)

Show worked answer →

Composition is "outside applied to inside".

$f(g(x)) = f(x^2) = 2 x^2 - 3$ .

$g(f(x)) = g(2 x - 3) = (2 x - 3)^2 = 4 x^2 - 12 x + 9$ .

To invert $f$ , swap $x$ and $y$ and solve: $x = 2 y - 3 \implies y = \frac{x + 3}{2}$ .

$f^{-1}(x) = \frac{x + 3}{2}$ .

Markers reward correct composition in both orders (they differ), the swap-and-solve method for the inverse, and a tidy final expression.

2021 HSC Q194 marksThe function

f(x) = x^2

is not one-to-one on

\mathbb{R}

. State a domain restriction that makes

f

one-to-one, find the inverse on that restricted domain, and state the domain and range of the inverse.

Show worked answer →

Restrict to $x \ge 0$ . On this domain, $f$ is increasing and one-to-one, with range $[0, \infty)$ .

Inverse: swap $x$ and $y$ in $y = x^2$ to get $x = y^2$ with $y \ge 0$ , so $y = \sqrt{x}$ .

$f^{-1}(x) = \sqrt{x}$ on domain $[0, \infty)$ with range $[0, \infty)$ .

Markers expect a domain choice that yields a one-to-one function, a correct inverse derived by swap-and-solve with the right sign, and a domain and range that match.

Practice questions

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

foundation2 marksLet

f(x) = 3x + 1

and

g(x) = x^2

. Find

f(g(x))

and

g(f(x))

Show worked solution →

Substitute the inner function into the outer one, both ways. For $f(g(x))$ , feed $g(x) = x^2$ into $f$ :

f(g(x)) = f(x^2) = 3x^2 + 1.

For $g(f(x))$ , feed $f(x) = 3x + 1$ into $g$ :

g(f(x)) = g(3x + 1) = (3x + 1)^2 = 9x^2 + 6x + 1.

Marker's note: one mark for $f(g(x)) = 3x^2 + 1$ , one mark for $g(f(x)) = 9x^2 + 6x + 1$ correctly expanded. Writing the same answer for both (assuming composition is commutative) loses a mark.

foundation3 marksFind the inverse function of

f(x) = 4x - 7

, then verify your answer by showing that

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

Show worked solution →

Swap $x$ and $y$ , then solve. Write $y = 4x - 7$ and swap the variables:

x = 4y - 7 \quad\Rightarrow\quad 4y = x + 7 \quad\Rightarrow\quad y = \frac{x + 7}{4}.

So $f^{-1}(x) = \dfrac{x + 7}{4}$ .

Verify by composition. Substitute $f^{-1}(x)$ into $f$ :

f\!\left(\frac{x + 7}{4}\right) = 4 \cdot \frac{x + 7}{4} - 7 = (x + 7) - 7 = x. \checkmark

Marker's note: one mark for the swap-and-solve method, one for $f^{-1}(x) = \frac{x + 7}{4}$ , one for the verification simplifying to $x$ . A verification that stops before the final $= x$ does not earn the last mark.

core3 marksLet

f(x) = \sqrt{x - 2}

with domain

x \ge 2

and

g(x) = x^2 + 1

. Find

f(g(x))

and state its natural domain.

Show worked solution →

Substitute $g$ into $f$ .

f(g(x)) = \sqrt{g(x) - 2} = \sqrt{(x^2 + 1) - 2} = \sqrt{x^2 - 1}.

Apply both domain conditions. First $x$ must be in the domain of $g$ , which is all of $\mathbb{R}$ . Second, $g(x)$ must be in the domain of $f$ , so $g(x) \ge 2$ :

x^2 + 1 \ge 2 \quad\Rightarrow\quad x^2 \ge 1 \quad\Rightarrow\quad x \le -1 \text{ or } x \ge 1.

The natural domain of $f \circ g$ is $(-\infty, -1] \cup [1, \infty)$ .

Marker's note: one mark for $\sqrt{x^2 - 1}$ , one for setting up $g(x) \ge 2$ , one for the domain $x \le -1$ or $x \ge 1$ . Reading the domain straight off $\sqrt{x^2 - 1} \ge 0$ (giving the same answer here by luck) misses the required $g(x) \in \text{dom}(f)$ reasoning and is penalised when the two disagree.

core4 marksThe function

f(x) = e^{2x} - 3

has domain all real numbers. (a) State the range of

f

. (b) Find

f^{-1}(x)

and state its domain.

Show worked solution →

Part (a): find the range. Since $e^{2x} > 0$ for every real $x$ , we have $e^{2x} - 3 > -3$ . As $x \to \infty$ , $f(x) \to \infty$ , and as $x \to -\infty$ , $f(x) \to -3$ from above. So the range is $(-3, \infty)$ .

Part (b): invert by swap-and-solve. Write $y = e^{2x} - 3$ and swap:

x = e^{2y} - 3 \quad\Rightarrow\quad e^{2y} = x + 3.

Take natural logs and solve for $y$ :

2y = \ln(x + 3) \quad\Rightarrow\quad y = \frac{1}{2}\ln(x + 3).

So $f^{-1}(x) = \dfrac{1}{2}\ln(x + 3)$ . Its domain is the range of $f$ , namely $(-3, \infty)$ .

Marker's note: one mark for the range $(-3, \infty)$ in (a); one for isolating $e^{2y} = x + 3$ , one for $f^{-1}(x) = \frac{1}{2}\ln(x + 3)$ , one for the domain $(-3, \infty)$ in (b). Forgetting the factor of $\frac{1}{2}$ (writing $\ln(x + 3)$ ) is the common slip.

core3 marksA graph shows two curves drawn on the same axes together with the dashed line

y = x

. The first curve is an increasing exponential-shaped curve passing through the points

(0, 1)

and

(1, 4)

with the

x

-axis as a horizontal asymptote. The second curve passes through

(1, 0)

and

(4, 1)

with the

y

-axis as a vertical asymptote. Explain, using two specific features read from the graph, why the second curve is the inverse of the first.

Show worked solution →

Feature 1: the points are reflected in $y = x$ . The first curve passes through $(0, 1)$ and $(1, 4)$ . Reflecting each point in $y = x$ swaps its coordinates, giving $(1, 0)$ and $(4, 1)$ , which are exactly the two points the second curve passes through. Every point $(a, b)$ on the first curve corresponds to $(b, a)$ on the second.

Feature 2: the asymptote swaps axes. The first curve has the $x$ -axis (a horizontal asymptote); under reflection in $y = x$ a horizontal asymptote becomes a vertical one, and indeed the second curve has the $y$ -axis as its asymptote. This is the graphical signature of domain and range swapping.

Because the two curves are mirror images in $y = x$ (matching swapped points and swapped asymptotes), the second is the inverse of the first.

Marker's note: one mark for identifying the swapped points $(0,1)\!\to\!(1,0)$ and $(1,4)\!\to\!(4,1)$ , one for the horizontal-to-vertical asymptote swap, one for the conclusion tying both to reflection in $y = x$ . A vague "they look symmetrical" without naming specific swapped features caps the answer at one mark.

exam5 marksLet

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

for

x \ne 3

. (a) Show that

f

is its own inverse, that is

f(f(x)) = x

. (b) Hence, or otherwise, find the exact coordinates of any point where the graph of

y = f(x)

meets the line

y = x

Show worked solution →

Part (a): compute $f(f(x))$ . The outer rule is $f(u) = \dfrac{3u + 1}{u - 3}$ ; here the input is $u = f(x) = \dfrac{3x + 1}{x - 3}$ . Build the numerator and denominator separately, clearing the inner fraction by writing each over $(x - 3)$ .

Numerator $3u + 1$ :

3u + 1 = \frac{3(3x + 1)}{x - 3} + 1 = \frac{3(3x + 1) + (x - 3)}{x - 3} = \frac{9x + 3 + x - 3}{x - 3} = \frac{10x}{x - 3}.

Denominator $u - 3$ :

u - 3 = \frac{3x + 1}{x - 3} - 3 = \frac{(3x + 1) - 3(x - 3)}{x - 3} = \frac{3x + 1 - 3x + 9}{x - 3} = \frac{10}{x - 3}.

Divide; the common factor $(x - 3)$ cancels:

f(f(x)) = \frac{10x / (x - 3)}{10 / (x - 3)} = \frac{10x}{10} = x. \checkmark

Since $f(f(x)) = x$ , the function undoes itself, so $f = f^{-1}$ .

Part (b): meet the line $y = x$ . Because $f$ is its own inverse, its graph is symmetric in $y = x$ , and any intersection with that line solves $f(x) = x$ :

\frac{3x + 1}{x - 3} = x \quad\Rightarrow\quad 3x + 1 = x(x - 3) = x^2 - 3x \quad\Rightarrow\quad x^2 - 6x - 1 = 0.

Solve with the quadratic formula:

x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10}.

Both values satisfy $x \ne 3$ , so the graph meets $y = x$ at $\left(3 + \sqrt{10}, \, 3 + \sqrt{10}\right)$ and $\left(3 - \sqrt{10}, \, 3 - \sqrt{10}\right)$ .

Marker's note: two marks for correctly building and simplifying $f(f(x))$ to $x$ (writing each of $3u + 1$ and $u - 3$ over $x - 3$ is the key step), one for stating $f = f^{-1}$ ; one for reducing $f(x) = x$ to $x^2 - 6x - 1 = 0$ , one for both exact points $3 \pm \sqrt{10}$ . Leaving the intersection as a decimal without the exact surd form is penalised.

exam4 marksThe function

f(x) = x^2 - 4x + 7

is defined on the restricted domain

x \ge 2

. (a) Explain why this restriction makes

f

one-to-one. (b) Find

f^{-1}(x)

and state its domain.

Show worked solution →

Part (a): show one-to-one on $x \ge 2$ . Complete the square:

f(x) = x^2 - 4x + 7 = (x - 2)^2 + 3.

The vertex is at $x = 2$ . For $x \ge 2$ the factor $(x - 2)^2$ is strictly increasing, so $f$ is strictly increasing on this domain. A strictly increasing function passes the horizontal line test, so it is one-to-one and an inverse exists.

Part (b): invert on the restricted branch. From $y = (x - 2)^2 + 3$ , swap $x$ and $y$ :

x = (y - 2)^2 + 3 \quad\Rightarrow\quad (y - 2)^2 = x - 3.

Take the square root. Because the restriction $x \ge 2$ forces $y \ge 2$ on the inverse (the range endpoint carries over), choose the positive root:

y - 2 = \sqrt{x - 3} \quad\Rightarrow\quad y = 2 + \sqrt{x - 3}.

So $f^{-1}(x) = 2 + \sqrt{x - 3}$ . The domain of $f^{-1}$ is the range of $f$ ; since the minimum value of $f$ on $x \ge 2$ is $f(2) = 3$ , the range is $[3, \infty)$ , so the domain of $f^{-1}$ is $[3, \infty)$ .

Marker's note: one mark for completing the square and arguing strictly increasing (hence one-to-one) in (a); one for reaching $(y - 2)^2 = x - 3$ , one for choosing $+\sqrt{\,}$ with a reason and writing $f^{-1}(x) = 2 + \sqrt{x - 3}$ , one for the domain $[3, \infty)$ in (b). Taking $-\sqrt{\,}$ (which suits the $x \le 2$ branch instead) is the classic sign error.