Functions, domains, composition and inverses: VCE Mathematical Methods Units 3 & 4

Loav · VCE Mathematical Methods · Units 3 & 4

F01 · 2026 curriculum · Original teaching guide

1. A rule is only part of a function

A function assigns exactly one output to each allowed input. Its rule tells you how to calculate; its domain tells you which inputs are allowed. The range is the set of outputs actually attained. The distinction matters because the same formula on two different domains gives two different functions, and those functions can have different ranges and different inverses. Begin every substantial function question by writing down the stated domain before manipulating the rule.

The notation f: D → ℝ, f(x) = … says that inputs come from D and outputs are real numbers. The target set ℝ is the co-domain; it is not automatically the range. For example, f(x) = x² on ℝ has range [0, ∞), even though its target set can be ℝ. Writing f(3) means the output at input 3. It is a number, whereas f is the entire function. A letter inside the brackets is an input, not multiplication by f.

When a real-valued rule has no specified domain, its maximal real domain consists of every input for which the expression makes sense. A denominator cannot equal zero, a square-root radicand must be non-negative, and a logarithm argument must be positive. Apply these conditions together. A stated restricted domain takes precedence: even if a formula can accept other values, those values do not belong to the defined function.

2. Find the range from attainable outputs

To find a range, ask what outputs can actually occur as the allowed input moves. Useful information includes endpoint values, turning points, asymptotes and whether the function increases or decreases on the interval. Do not simply substitute the two endpoints and assume the outputs between them are the whole answer. A turning point inside the domain may produce a lower minimum or higher maximum than either endpoint.

Square brackets include a finite endpoint; round brackets exclude it. Infinity always takes a round bracket because it is not a real number that the function attains. An excluded input does not necessarily produce an excluded output: another allowed input may give the same output. For an unrestricted quadratic, outputs repeat symmetrically about its axis. That is why you must check attainment, rather than mechanically copy the domain’s bracket style into the range.

Worked example 1: an excluded input, an included output

Let f(x) = (x − 1)² + 2 for x ∈ [−1, 3). The vertex occurs at x = 1, which is allowed, so the minimum is f(1) = 2. At the included left endpoint, f(−1) = 6. As x approaches 3 from below, f(x) approaches 6, but that does not remove 6 from the range: the left endpoint already produces it. Thus the range is [2, 6].

This function is not one-to-one because, for example, f(0) = f(2) = 3. Restricting its domain to [1, 3) keeps the increasing branch. On that branch the range becomes [2, 6), and an inverse can be defined. The domain restriction changes both the function and the endpoint logic.

The graph of y = (x − 1)² + 2 for −1 ≤ x < 3. The closed point (−1, 6) makes 6 part of the range even though (3, 6) is open.

3. Composition is an input-output chain

In (f ∘ g)(x) = f(g(x)), g acts first. Work from the inside out. Calculate the inner output and then use that entire output as the input to the outer rule. Brackets are essential: if f(x) = x² and g(x) = x + 4, then f(g(x)) = (x + 4)², not x² + 4. Reversing the order gives g(f(x)) = x² + 4, a different function.

There is also a domain condition. For a composite of the already defined functions to exist on the whole domain of g, every output of g must be an allowed input to f. In symbols, range(g) ⊆ domain(f). If this fails, do not silently pretend that the original composite exists. State the failure, then, if asked, restrict the inner function so its outputs fit the outer domain. The restricted composition may have a perfectly useful rule.

Worked example 2: restrict the inner function explicitly

Suppose f(x) = √(x − 1) has domain [1, ∞), and g(x) = 5 − 2x has domain ℝ. The range of g is ℝ, so it is not contained in the domain of f. The composite of these stated functions is therefore not defined on all of g’s domain.

For a valid restricted composition, require g(x) ≥ 1. Solving 5 − 2x ≥ 1 gives x ≤ 2; the inequality reverses when dividing by −2. Restrict g to (−∞, 2]. The resulting rule is f(g(x)) = √(4 − 2x), with domain (−∞, 2] and range [0, ∞). At x = 2 the output is zero; decreasing x makes the radicand arbitrarily large.

4. An inverse reverses a one-to-one mapping

A function has an inverse function precisely when every output corresponds to only one input. This is the one-to-one condition. On a graph, a horizontal line must meet the graph at most once. The vertical-line test answers a different question: whether a relation is a function in the first place. A parabola passes the vertical-line test but fails the horizontal-line test unless an appropriate branch is selected.

To find an inverse, write y = f(x), solve for x in terms of y, and then exchange the variable labels. Use the original domain to choose any branch created during rearrangement. Finally state the new domain and range: the inverse’s domain is the original range, and the inverse’s range is the original domain. The graph reflects in y = x because every original point (a, b) becomes (b, a).

The superscript in f⁻¹(x) means the inverse function. It does not mean 1/f(x). A reciprocal reverses multiplication, while an inverse reverses the whole input-output process. To verify an inverse, check that f⁻¹(f(x)) = x on the original domain and f(f⁻¹(x)) = x on the original range. Those domain qualifications prevent false checks caused by an inappropriate square-root branch.

Worked example 3: invert a decreasing square-root function

Let f(x) = 3 − 2√(x + 1), x ≥ −1. The square root increases from zero, so f decreases from 3 without a lower bound. Its range is (−∞, 3], and it is one-to-one. Starting with y = 3 − 2√(x + 1) gives √(x + 1) = (3 − y)/2. The right side is non-negative because y ≤ 3.

Squaring and rearranging gives x = ((3 − y)/2)² − 1. Therefore f⁻¹(x) = ((3 − x)/2)² − 1, with domain (−∞, 3] and range [−1, ∞). For a check, f(8) = −3 and f⁻¹(−3) = 8. Extending this inverse formula to all real x would destroy its role as the inverse of the original function.

5. Piecewise rules still define one function

A piecewise function uses different formulas on different input intervals. First locate the input in the correct interval, then substitute into that interval’s formula. At a joining point, the bracket or inequality determines which rule applies. If two rules both include the same input but give different outputs, the proposed definition is not a function. If neither includes it, that input is absent from the domain.

Find a piecewise range one branch at a time and take the union. Overlap is counted once. For inverse questions, compare outputs across branches as well as within each branch: two separately increasing pieces can still produce repeated outputs. A graph with open and closed endpoint markers makes these issues easier to see, but your algebra should explain the set you report.

Practise, then check

For f(x) = x² − 4 on [−2, 1), find f(−1) and the range.
Find the maximal real domain of h(x) = √(x + 2) + 1/(x − 1). Explain both restrictions.
Let f(x) = 2x − 1 on ℝ and g(x) = x² on [0, 3]. Find f ∘ g and its domain and range.
Find the inverse of f(x) = 1 + 3/(x − 1), x ≠ 1, including its domain and range.
Find the inverse of p(x) = (x − 2)² + 1 on (−∞, 2]. Explain the sign of the root.
Let f(x) = √x on [0, 9] and g(x) = x + 1 on [−1, 10]. Does f ∘ g exist on all of g’s stated domain? Find the largest suitable restriction and the resulting range.
Define q(x) = x + 2 for −2 ≤ x < 0, and q(x) = x² + 1 for 0 ≤ x ≤ 2. Find q(0), its range and whether it has an inverse function.

Full reasoned answers

f(−1) = 1 − 4 = −3. The vertex x = 0 is allowed and gives −4. The included input −2 gives the greatest output, 0. Therefore the range is [−4, 0]. The missing input 1 does not exclude −3 because −1 already produces it.
The radicand requires x ≥ −2, while the denominator requires x ≠ 1. Taking both conditions gives [−2, 1) ∪ (1, ∞). The endpoint −2 is permitted because the root is zero and the denominator is −3.
The range of g is [0, 9], which lies inside f’s domain ℝ. Thus (f ∘ g)(x) = 2x² − 1 for x ∈ [0, 3]. Since x² increases there, the range is [−1, 17]. Both endpoint outputs are attained.
From y(x − 1) = x + 2 obtain x(y − 1) = y + 2, hence x = (y + 2)/(y − 1). No original input gives output 1, and every other real output is possible. Thus f⁻¹(x) = (x + 2)/(x − 1), with domain and range both ℝ \ {1}. This function is its own inverse.
The original range is [1, ∞). Since the original input satisfies x − 2 ≤ 0, solving y − 1 = (x − 2)² requires x − 2 = −√(y − 1). Thus p⁻¹(x) = 2 − √(x − 1), domain [1, ∞), range (−∞, 2].
No: g has range [0, 11], exceeding f’s allowed inputs. Require 0 ≤ x + 1 ≤ 9, giving −1 ≤ x ≤ 8. On this restricted inner domain, the composite is √(x + 1) with range [0, 3].
The second rule applies at zero, so q(0) = 1. The first range is [0, 2), and the second is [1, 5]; their union is [0, 5]. There is no inverse function on the full domain because q(−1) = q(0) = 1.

Exam habits and technology checks

Write the domain before using a solve command or drawing a graph.
For a composite, show the inner-range/outer-domain check, not just substitution.
For an inverse, provide the rule and domain together. Show why the branch is valid.
Use technology to test a suspected range or intersection, then justify endpoint inclusion algebraically. A viewing window can hide an excluded point or a far-away branch.

Curriculum and sources

Scope: VCE Mathematical Methods Units 3 and 4, current 2023–2027 study design, Functions, relations and graphs and Algebra, number and structure. See the official VCAA Mathematical Methods curriculum and VCAA Mathematical Methods FAQs, updated February 2026. This guide covers domain, range, composition, inverse conditions and inverse graphs.

Further reading: Michael Evans, Kay Lipson and David Greenwood, Cambridge Senior Mathematics VCE Mathematical Methods Units 3 & 4, second edition (2023), ISBN 9781009110495. Chapter 1, sections 1A–1F, pp. 2–42; Chapter 7, section 7B, pp. 303–307. Explanations, diagrams, worked examples and practice questions in this guide are original. This is an independent Loav learning resource, not a VCAA publication.