Functions is the opening chapter of H2 Mathematics (9758). The language it introduces (domain, range and interval notation) is reused in every later topic, from graphing transformations to calculus. Almost every Promo paper has at least one “state the range” mark inside a functions question, so the definitions and the routine for reading a range off a graph must be exactly right.
The chapter then covers composite functions, inverse functions with domain restriction, the graphs of f and f⁻¹, and self-inverse functions. Finding f⁻¹ is the most heavily examined skill. Almost every promo and prelim question on Functions asks you to justify that an inverse exists, find its rule and state its domain.
What these notes cover
- •Functions, domain and range
- •The vertical line test
- •Function notation and interval notation
- •Finding the range from a graph
- •Composite functions and the condition for existence
- •Range of a composite: the two-stage technique
- •Inverse functions and the horizontal line test
- •Finding f⁻¹ and stating its domain
- •Domain restriction for quadratics
- •Graphs of f and f⁻¹ and solving f(x) = f⁻¹(x)
- •Self-inverse functions
- •Common mistakes and exam technique
Key definitions
- Function
- A rule that assigns to each input x in a set A exactly one output f(x) in a set B. A many-one rule is allowed, but one input with many outputs is not. A function is fully defined only once its domain is stated.
- Domain and range
- The domain Df is the set of all allowed inputs of f. The range Rf is the set of all outputs actually produced as x runs through Df. The range depends on the stated domain, not on the formula alone.
- Composite function fg
- The function formed by applying g first and then f to the result: fg(x) = f(g(x)). Read it from right to left. f²(x) means f(f(x)), not (f(x))².
- One-to-one (injective) function
- A function in which no two different inputs share the same output. A function has an inverse if and only if it is one-to-one.
- Self-inverse function
- A function with f(f(x)) = x for all x in Df, i.e. f² = I, so that f⁻¹ = f. Its graph is symmetric in the line y = x.
Key formulas
Existence of a composite
Rg is the range of the inner function g and Df is the domain of the outer function f. State both sets explicitly as intervals, then compare them.
Domain of a composite
The domain of fg is inherited from the inner function g. Do not read it off the simplified formula for fg(x).
Domain and range of the inverse
f⁻¹ exists only if f is one-to-one. Its domain is the range of f, not the domain of f.
Intersection shortcut (increasing f only)
Valid only when f is increasing. Before using it, write: “Since f is increasing, the graphs of y = f(x) and y = f⁻¹(x) intersect only on the line y = x.”
Inverting a completed square
Keep the + root if the restricted domain is x ≥ a, and the − root if it is x ≤ a. Write the rejection as a line of working, and never leave ± in the answer.
Powers of a self-inverse function
I is the identity function I(x) = x. fⁿ means f applied n times, not a power of f(x).
Worked example
The function g is defined by g: x ↦ x² − 6x + 11, x ∈ ℝ. (a) Show that g⁻¹ does not exist. (b) State the least value of k such that the function g, restricted to the domain x ≥ k, has an inverse. (c) Using this restricted domain, find g⁻¹(x) and state the domain of g⁻¹.
- 1(a) g(2) = 4 − 12 + 11 = 3 and g(4) = 16 − 24 + 11 = 3, so the horizontal line y = 3 cuts the graph of y = g(x) twice, since g(2) = g(4) = 3 with 2 ≠ 4; hence g is not one-to-one and g⁻¹ does not exist.
- 2(b) Completing the square gives x² − 6x + 11 = (x − 3)² + 2, so the vertex of the parabola is (3, 2).
- 3For x ≥ 3 the graph is strictly increasing, so the least value is k = 3.
- 4(c) With domain x ≥ 3, the minimum value is 2 at the vertex and g increases without bound, so Rg = [2, ∞).
- 5Let y = (x − 3)² + 2, so (x − 3)² = y − 2, x − 3 = ±√(y − 2) and x = 3 ± √(y − 2).
- 6Since x ≥ 3, we have x − 3 ≥ 0, so reject x = 3 − √(y − 2) and keep x = 3 + √(y − 2).
- 7Swapping letters, with Dg⁻¹ = Rg = [2, ∞), the inverse is g⁻¹(x) = 3 + √(x − 2), x ≥ 2.
Watch out: The restricted domain decides which sign to keep. Show the rejection as an explicit line of working, because that line carries a mark. Never leave ± in the final answer: a rule with two outputs for one input is not a function.
Common mistakes
Writing the range in x-values or ignoring the restricted domain
Wrong: For f: x ↦ x² + 1, x ≥ 2, writing “range: x ≥ 1”.
Right: Sketch y = x² + 1 on [2, ∞) only. The least output is f(2) = 5 and f increases without bound, so Rf = [5, ∞), written in interval notation and never as an inequality in x.
Claiming fg exists because both functions are defined
Wrong: Writing “fg exists because f and g are both well-defined functions”, or computing fg(x) without justifying existence.
Right: State both sets in interval notation and compare them with ⊆, e.g. Rg = [1, ∞) ⊆ (0, ∞) = Df, so fg exists. If Rg ⊈ Df, quote one value of Rg that lies outside Df.
Finding a range by substituting domain endpoints
Wrong: For g: x ↦ (x − 2)² on [0, 5], taking g(0) = 4 and g(5) = 9 as the ends of the range and giving [4, 9].
Right: The minimum g(2) = 0 lies inside the domain, so Rg = [0, 9]. For a composite, use the two-stage method: sketch g on Dg to read off Rg, then sketch f on Rg to read off Rfg.
Questions students ask
How do you find the range of a function in H2 Math?
Sketch y = f(x) on the GC over the given domain only. Mark any turning points inside the domain and any asymptotes, and decide whether each endpoint is included or excluded. Read the complete set of y-values off the y-axis and write it in interval notation; a turning point inside the domain means the range is not simply from one endpoint value to the other.
How do you show that an inverse function exists?
Show that f is one-to-one using the horizontal line test: “Any horizontal line y = k, k ∈ Rf, cuts the graph of y = f(x) at most once. Hence f is one-to-one and f⁻¹ exists.” To show f⁻¹ does not exist, give a concrete counterexample f(a) = f(b) = k with a ≠ b, using actual numbers from the question.
When can you solve f(x) = f⁻¹(x) by solving f(x) = x?
Only when f is increasing, because the graphs of y = f(x) and y = f⁻¹(x) can then intersect only on the line y = x. You must write this justification before using the shortcut, or the method mark is lost. A decreasing function can meet its inverse off the line y = x; for example, f(x) = −x³ meets its inverse at (1, −1).
Is (fg)⁻¹ = g⁻¹f⁻¹ in the H2 Math syllabus?
No. The 9758 syllabus excludes the relation (fg)⁻¹ = g⁻¹f⁻¹, so do not quote or use it. If a question needs the inverse of a composite, form fg(x) explicitly first and then invert it. Restricting a domain to force a composite to exist is also excluded.
In the full PDF
- •Further worked examples on ranges with asymptotes and turning points, composite functions, inverses of rational functions and solving f(x) = f⁻¹(x)
- •Three promo/prelim-standard worked examples with exam notes on where each mark sits
- •Graphs for each case, including f and f⁻¹ reflected in y = x and two-branch self-inverse hyperbolas
- •The remaining mark-losers: omitting the domain of f⁻¹, keeping both branches of ±, and using f(x) = x without stating that f is increasing
- •Exam-standard sentences for the line tests, set comparisons and reflection in y = x, plus the full interval-notation table
- •A key-results table and the five non-negotiables for H2 Functions