Composite and inverse functions - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Form composite and inverse functions, determining the correct domain and range of each.”

TCESpecialist MathematicsUnit 38 min readUpdated 2 Jun 2026

Quick answer

A composite $f(g(x))$ feeds the output of $g$ into $f$ , and it only exists where the range of $g$ lies inside the domain of $f$ . An inverse $f^{-1}$ exists only when $f$ is one-to-one; you find it by swapping $x$ and $y$ and solving, and its domain is the range of $f$ .

What this dot point is asking

This dot point opens the functions and sketching graphs topic. Before sketching the more exotic graphs in this strand, you need full control of how functions combine and reverse, and a disciplined habit of tracking domain and range at every step.

Composition of functions

The composite function $f \circ g$ is defined by $(f \circ g)(x) = f(g(x))$ : apply $g$ first, then $f$ . Order matters, because in general $f(g(x)) \ne g(f(x))$ .

When a composite exists

The composite $f(g(x))$ is only defined for inputs $x$ in the domain of $g$ whose outputs $g(x)$ land inside the domain of $f$ . So you must check that the range of the inner function is contained in the domain of the outer function.

For example, if $f(x) = \sqrt{x}$ (domain $x \ge 0$ ) and $g(x) = x - 4$ , then $f(g(x)) = \sqrt{x - 4}$ requires $x - 4 \ge 0$ , so the domain of the composite is $x \ge 4$ , narrower than the domain of $g$ alone.

Inverse functions

The inverse $f^{-1}$ undoes $f$ , so $f^{-1}(f(x)) = x$ . An inverse function exists only when $f$ is one-to-one, meaning no horizontal line crosses the graph more than once. If $f$ is many-to-one, you must restrict its domain to a piece on which it is one-to-one before an inverse exists.

Domain and range swap

For an invertible function the domain and range trade places:

\text{domain of } f^{-1} = \text{range of } f, \qquad \text{range of } f^{-1} = \text{domain of } f.

The graph of

f^{-1}

is the reflection of the graph of

f

in the line

y = x

Finding an inverse

To find an inverse: write $y = f(x)$ , swap $x$ and $y$ , then solve for $y$ . State the domain of the inverse, which equals the range of the original.

Inverse of a restricted quadratic

Find the inverse of $f(x) = x^2 - 4$ for the restricted domain $x \ge 0$ , and state its domain and range.

Step 1: Confirm the function is one-to-one on the restricted domain

A function must pass the horizontal line test before an inverse can exist. On $x \ge 0$ the parabola is strictly increasing, so each output value is produced by exactly one input, and an inverse is guaranteed. The range of $f$ on this domain is $y \ge -4$ , since the minimum output occurs at $x = 0$ where $f(0) = -4$ .

Step 2: Swap $x$ and $y$ and solve for $y$

The standard procedure for finding an inverse is to write $y = f(x)$ , then swap the roles of $x$ and $y$ to get $x = f(y)$ , and finally solve for $y$ . This isolates the inverse formula.

x = y^2 - 4 \implies y^2 = x + 4 \quad\Longrightarrow\quad y = \sqrt{x + 4},

taking the positive square root because the original domain was $x \ge 0$ , so the inverse outputs must be non-negative.

Step 3: State the domain and range of the inverse

The domain of $f^{-1}$ equals the range of $f$ , and the range of $f^{-1}$ equals the restricted domain of $f$ . Both must be stated explicitly.

f^{-1}(x) = \sqrt{x + 4}, \qquad \text{domain } x \ge -4, \qquad \text{range } y \ge 0.

Final answer: $f^{-1}(x) = \sqrt{x + 4}$ , with domain $x \ge -4$ and range $y \ge 0$ . The domain of $f^{-1}$ matches the range of $f$ , and the range of $f^{-1}$ matches the restricted domain of $f$ , as expected.

Checking with composition

A reliable check is to confirm $f(f^{-1}(x)) = x$ . Here $f(f^{-1}(x)) = (\sqrt{x + 4})^2 - 4 = x + 4 - 4 = x$ on the stated domain. If the composition does not return $x$ , an algebra error or a domain mistake has crept in.

A composite both ways

Worked example

Let $f(x) = 2x + 1$ and $g(x) = x^2$ . Find $f(g(x))$ and $g(f(x))$ and confirm they differ.

f(g(x)) = 2x^2 + 1, \qquad g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1.

These are clearly different functions, illustrating that composition is not commutative. Both have domain all real numbers because each inner function outputs values the outer function accepts.

Reflecting the graph but forgetting to restrict the domain

When a function is many-to-one, such as $y = x^2$ over all real $x$ , reflecting in $y = x$ produces a sideways parabola that fails the vertical line test and is not a function. The fix is to restrict the original domain first (for instance $x \ge 0$ ) so the function becomes one-to-one, and to carry that restriction through as the range of the inverse. Skipping the restriction gives a relation, not a valid inverse function.

Inverses of rational functions

Many TASC questions ask for the inverse of a rational function of the form $f(x) = \dfrac{ax + b}{cx + d}$ . The method is always the same: write $y = f(x)$ , swap $x$ and $y$ , then solve the resulting equation for $y$ by clearing the denominator and collecting the $y$ terms on one side. The domain of the inverse is found from the range of the original, and for these functions the excluded value of the inverse is the horizontal asymptote of $f$ (namely $y = \tfrac{a}{c}$ ).

The horizontal asymptote becomes the excluded input

For $f(x) = \dfrac{ax + b}{cx + d}$ the range excludes the horizontal asymptote value $\tfrac{a}{c}$ , so the domain of $f^{-1}$ excludes $x = \tfrac{a}{c}$ . Stating that excluded value is usually worth a mark on its own.

Why this matters

Composition and inverses underpin the rest of the sketching topic, where reciprocal, absolute value and rational functions are all built by transforming and combining simpler functions. Tracking domain and range carefully here prevents errors that would otherwise propagate into every later graph.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Let

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

for

x \ne 3

. (a) Find the inverse function

f^{-1}(x)

and state its domain. (b) Hence show that

f

is self-inverse, that is

f(f(x)) = x

Show worked answer →

(a) Set $y = \dfrac{2x + 1}{x - 3}$ and swap $x$ and $y$ : $x = \dfrac{2y + 1}{y - 3}$ . Clear the denominator: $x(y - 3) = 2y + 1$ , so $xy - 3x = 2y + 1$ . Collect the $y$ terms: $xy - 2y = 3x + 1$ , so $y(x - 2) = 3x + 1$ and $y = \dfrac{3x + 1}{x - 2}$ . Hence $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ , with domain $x \ne 2$ (the value $x = 2$ is the horizontal asymptote of $f$ , which $f$ never attains). (3 marks)

(b) Since $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ , this is a different expression from $f$ , so $f$ is not literally identical to its inverse; instead verify directly. $f(f(x)) = \dfrac{2\left(\frac{2x+1}{x-3}\right) + 1}{\frac{2x+1}{x-3} - 3}$ . Multiply top and bottom by $x - 3$ : the numerator becomes $2(2x + 1) + (x - 3) = 5x - 1$ and the denominator becomes $(2x + 1) - 3(x - 3) = -x + 10$ . This is not identically $x$ , so $f$ is not self-inverse; the verification shows the composite is $\dfrac{5x - 1}{10 - x}$ , confirming that you must compute $f^{-1}$ by the swap-and-solve method rather than assuming self-inversion. Markers reward the correct swap-and-solve inverse and a stated domain. (2 marks)

Practice4 marks

Given

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

(domain

x \ge 2

) and

g(x) = x^2 + 1

, find

f(g(x))

and state its domain and range.

Show worked answer →

Compute the composite: $f(g(x)) = \sqrt{g(x) - 2} = \sqrt{x^2 + 1 - 2} = \sqrt{x^2 - 1}$ .

The composite requires $g(x)$ to lie in the domain of $f$ , that is $g(x) \ge 2$ . So $x^2 + 1 \ge 2$ , giving $x^2 \ge 1$ , that is $x \le -1$ or $x \ge 1$ . This is the domain of $f(g(x))$ .

For the range, $x^2 - 1 \ge 0$ over this domain and is unbounded above, so $\sqrt{x^2 - 1} \ge 0$ and the range is $y \ge 0$ . Markers reward checking that the range of the inner function lands in the domain of the outer function, the restricted domain $x \le -1$ or $x \ge 1$ , and the range $y \ge 0$ .

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC

What this dot point is asking

Composition of functions

Inverse functions

Finding an inverse

Step 1: Confirm the function is one-to-one on the restricted domain

Step 2: Swap xxx and yyy and solve for yyy

Step 3: State the domain and range of the inverse

Checking with composition

Worked example

Inverses of rational functions

Why this matters

Exam-style questions

Practise this

Sources & how we know this

Step 2: Swap $x$ and $y$ and solve for $y$