Mathematics Advanced • Year 11 • Module 1 • Lesson 5
Odd & Even Functions
Practise HSC-style writing on odd, even and "neither" functions — including a structured proof.
1. Short-answer questions
1.1 Determine, with full algebraic justification, whether the function f(x) = 3x⁴ − x² is even, odd, or neither. 2 marks Band 3
1.2 The graph of y = f(x), where f is an odd function, passes through the point (4, −7). State two other points the graph must pass through, justifying each from the definition or geometric property of an odd function. 3 marks Band 3-4
1.3 Consider f(x) = x³ − 2x + 1.
(a) By substituting x = 1 and x = −1, show that f is neither even nor odd.
(b) Modify the constant term to obtain a new function that is odd, and justify your choice. 4 marks Band 4
2. Extended response
2.1 Let f be a function defined on a domain symmetric about 0.
(a) Prove that if f is both even and odd, then f(x) = 0 for every x in the domain.
(b) Hence, or otherwise, prove that any function g defined on a domain symmetric about 0 can be written uniquely as the sum g(x) = E(x) + O(x), where E is an even function and O is an odd function. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — writes both defining conditions: f(−x) = f(x) and f(−x) = −f(x), and equates them to obtain f(x) = −f(x).
• 1 mark — concludes 2 f(x) = 0 ⇒ f(x) = 0 for all x in the domain, with a closing sentence.
Part (b) — 5 marks
• 1 mark — existence: defines E(x) = ½[g(x) + g(−x)] and O(x) = ½[g(x) − g(−x)].
• 1 mark — verifies E is even: shows E(−x) = E(x) by substitution.
• 1 mark — verifies O is odd: shows O(−x) = −O(x) by substitution.
• 1 mark — sum recovers g: shows E(x) + O(x) = g(x).
• 1 mark — uniqueness: suppose g = E₁ + O₁ = E₂ + O₂; deduce E₁ − E₂ = O₂ − O₁, conclude this difference is both even and odd, then invoke part (a) to force it to be 0.
Your response:
Stuck on uniqueness? Subtract two supposed decompositions and use part (a).How did this worksheet feel?
What I'll revisit before next class:
1.1 — f(x) = 3x⁴ − x² (2 marks)
Sample response. f(−x) = 3(−x)⁴ − (−x)² = 3x⁴ − x² = f(x). Since f(−x) = f(x) for all x in the domain, f is even.
Marking notes. 1 mark — correctly substitutes −x and simplifies (brackets used; both even-power terms correctly become positive). 1 mark — explicit comparison f(−x) = f(x) and classification. A response that writes only "f is even" without the simplification step scores 1/2.
1.2 — Odd function through (4, −7) (3 marks)
Sample response. An odd function satisfies f(−x) = −f(x), so for any point (a, b) on the graph, the point (−a, −b) is also on the graph (geometric meaning: 180° rotation about the origin).
• Point 1: (−4, 7), because f(−4) = −f(4) = −(−7) = 7.
• Point 2: (0, 0), because every odd function defined at 0 satisfies f(0) = 0. (Proof: f(0) = f(−0) = −f(0) ⇒ 2 f(0) = 0 ⇒ f(0) = 0.)
Marking notes. 1 mark each for the two points with a correct algebraic or geometric justification (3 marks total: state odd-function property + (−4, 7) + (0, 0)). Naming (−4, 7) without justification = 0.5; naming (0, 0) without the f(0) = 0 argument = 0.5. Top responses explicitly invoke "180° rotation about the origin" when discussing the geometric meaning.
1.3 — f(x) = x³ − 2x + 1 (4 marks)
(a) Sample response. f(1) = 1 − 2 + 1 = 0 and f(−1) = −1 + 2 + 1 = 2. For even, we would need f(−1) = f(1), i.e. 2 = 0, which is false. For odd, we would need f(−1) = −f(1), i.e. 2 = 0, which is also false. Hence f is neither even nor odd.
(b) Sample response. Replace the constant term +1 by 0, giving g(x) = x³ − 2x. Then g(−x) = (−x)³ − 2(−x) = −x³ + 2x = −(x³ − 2x) = −g(x). Both surviving terms are odd powers of x, so g is odd. (Any non-zero constant term would put an x⁰ — an even power — into the function and break oddness, because constants force g(0) ≠ 0.)
Marking notes. (a) 1 mark — correct values of f(1) and f(−1); 1 mark — explicit elimination of both "even" and "odd" using the substitution test. (b) 1 mark — chooses constant term 0 (the only correct choice); 1 mark — verifies g(−x) = −g(x) by working or states the principle that a sum of odd-power terms is odd. Common error: students change the leading coefficient instead of the constant term, which does not affect parity.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Suppose f is both even and odd on a domain symmetric about 0. Then for every x in the domain,
f(−x) = f(x) [definition of even], and f(−x) = −f(x) [definition of odd]. [1 mark — both conditions stated and equated.]
Setting the two right-hand sides equal gives f(x) = −f(x), so 2 f(x) = 0, hence f(x) = 0. This holds for every x in the domain, so f is the zero function. [1 mark — algebraic conclusion and closing sentence.]
Part (b). Existence. Given g defined on a domain symmetric about 0, define
E(x) = ½ [ g(x) + g(−x) ] and O(x) = ½ [ g(x) − g(−x) ]. [1 mark — explicit construction.]
E is even, because
E(−x) = ½ [ g(−x) + g(−(−x)) ] = ½ [ g(−x) + g(x) ] = E(x). [1 mark — verifies E even.]
O is odd, because
O(−x) = ½ [ g(−x) − g(−(−x)) ] = ½ [ g(−x) − g(x) ] = −½ [ g(x) − g(−x) ] = −O(x). [1 mark — verifies O odd.]
The decomposition works, because E(x) + O(x) = ½ [g(x) + g(−x) + g(x) − g(−x)] = ½ · 2 g(x) = g(x). [1 mark — sum recovers g.]
Uniqueness. Suppose g = E₁ + O₁ = E₂ + O₂ for two pairs (E₁, O₁) and (E₂, O₂) of even/odd functions on the symmetric domain. Subtracting,
E₁(x) − E₂(x) = O₂(x) − O₁(x). Call this common value h(x).
The left side, a difference of even functions, is even. The right side, a difference of odd functions, is odd. So h is both even and odd. By part (a), h(x) = 0 for every x. Therefore E₁ = E₂ and O₁ = O₂, so the decomposition is unique. ▮ [1 mark — uniqueness completed via part (a).]
Total: 7/7.
Band descriptors for marker.
Band 3: States definitions and attempts part (a) but does not algebraically combine them; for part (b) may state the formulas for E and O but not verify the parity or recover g. ≈ 2-3 marks.
Band 4: Completes part (a) correctly and constructs E, O for part (b) but verifies only one of "E is even" / "O is odd"; sum identity correct; no uniqueness argument. ≈ 4-5 marks.
Band 5: Both parities verified, sum identity established, existence complete. Attempts uniqueness but does not close the argument with part (a). ≈ 5-6 marks.
Band 6: Full proof in both parts, uses precise quantifiers ("for every x in the domain", "domain symmetric about 0"), explicitly invokes part (a) inside the uniqueness step ("hence, or otherwise"), and concludes with a closing ▮ or "as required". 7/7.