Mathematics Advanced • Year 11 • Module 1 • Lesson 5
Odd & Even Functions
Build procedural fluency in the algebraic test for whether a function is even, odd, or neither.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the algebraic definitions:
A function f is even if f(−x) = ____________ for all x in the domain.
A function f is odd if f(−x) = ____________ for all x in the domain.
Q1.2 Describe the geometric symmetry of each in one phrase:
Even function: ________________________________________________________________
Odd function: ________________________________________________________________
Q1.3 Name the only function that is both even and odd. f(x) = ______________
2. Worked example — testing f(x) = x⁴ − 2x²
Follow each line of algebra. Every step has a reason on the right.
Problem. Determine whether f(x) = x⁴ − 2x² is even, odd, or neither.
Step 1 — Substitute −x into every term using brackets.
f(−x) = (−x)⁴ − 2(−x)²
Reason: replace each x by (−x); brackets prevent sign errors.
Step 2 — Simplify each power separately.
(−x)⁴ = x⁴ and (−x)² = x²
∴ f(−x) = x⁴ − 2x²
Reason: an even power of (−x) is positive.
Step 3 — Compare f(−x) with f(x).
f(x) = x⁴ − 2x² ✓ identical to f(−x)
Reason: f(−x) = f(x) is the definition of even.
Step 4 — Verify it is not odd (optional safety check).
−f(x) = −(x⁴ − 2x²) = −x⁴ + 2x²
−f(x) ≠ f(−x), so f is not odd.
Conclusion. f(x) = x⁴ − 2x² is an even function.
3. Faded example — fill in the missing steps
Determine whether f(x) = x⁵ + 3x is even, odd, or neither. Fill in each blank line. 4 marks
Step 1 — Substitute −x:
f(−x) = (−x)⁵ + 3(−x)
Step 2 — Simplify each power: (−x)⁵ = ____________ and 3(−x) = ____________
∴ f(−x) = ________________________________________________
Step 3 — Factorise the result:
f(−x) = ________________ × ( x⁵ + 3x )
Step 4 — Compare: f(−x) = ________ · f(x), so f satisfies the definition of an ____________ function.
Conclusion. f(x) = x⁵ + 3x is ____________________.
4. Graduated practice — classify each function
For each function, state whether it is even, odd, or neither. Show the line of algebra for f(−x) and the comparison with f(x) and −f(x). Assume the largest natural domain unless stated.
Foundation — single-term polynomials (4 questions)
| Q | Function | f(−x) simplified | Classification |
|---|---|---|---|
| 4.1 1 | f(x) = x² | ||
| 4.2 1 | f(x) = x⁷ | ||
| 4.3 1 | f(x) = 5 | ||
| 4.4 1 | f(x) = −3x |
Standard — typical HSC difficulty (6 questions)
Show your working in the space below each part — at least one line for f(−x) and one for the comparison.
4.5 f(x) = 2x⁴ − 5x² + 1 2 marks
4.6 f(x) = x³ − 4x 2 marks
4.7 f(x) = x² + 2x 2 marks
4.8 f(x) = |x| − 3 2 marks
4.9 f(x) = 1 / x², x ≠ 0 2 marks
4.10 f(x) = x / (x² + 1) 2 marks
Extension — combine concepts (2 questions)
4.11 Let f(x) = (x² − 4)(x⁴ + 1). Without expanding, classify f and justify in one line. Then verify by computing f(−x). 3 marks
4.12 Prove that if g(x) is an odd function and h(x) is an even function (both defined on a domain symmetric about 0), then the product p(x) = g(x) · h(x) is an odd function. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Algebraic definitions
Even: f(−x) = f(x). Odd: f(−x) = −f(x).
Q1.2 — Geometric symmetry
Even: reflection (mirror) symmetry in the y-axis. Odd: rotational symmetry of 180° about the origin.
Q1.3 — Both even and odd
f(x) = 0 (the zero function). It is the only function satisfying both f(−x) = f(x) and f(−x) = −f(x), because 0 = −0.
Q3 — Faded example f(x) = x⁵ + 3x
Step 1: f(−x) = (−x)⁵ + 3(−x).
Step 2: (−x)⁵ = −x⁵; 3(−x) = −3x. ∴ f(−x) = −x⁵ − 3x.
Step 3: f(−x) = −1 × (x⁵ + 3x).
Step 4: f(−x) = −1 · f(x), so f satisfies the definition of an odd function.
Conclusion: f(x) = x⁵ + 3x is odd.
Q4.1 — f(x) = x²
f(−x) = (−x)² = x² = f(x). Even.
Q4.2 — f(x) = x⁷
f(−x) = (−x)⁷ = −x⁷ = −f(x). Odd.
Q4.3 — f(x) = 5
f(−x) = 5 = f(x). Even. Every non-zero constant function is even (and not odd, because 5 ≠ −5).
Q4.4 — f(x) = −3x
f(−x) = −3(−x) = 3x = −(−3x) = −f(x). Odd.
Q4.5 — f(x) = 2x⁴ − 5x² + 1
f(−x) = 2(−x)⁴ − 5(−x)² + 1 = 2x⁴ − 5x² + 1 = f(x). Even. All powers of x are even (constant counts as x⁰).
Q4.6 — f(x) = x³ − 4x
f(−x) = (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x). Odd. Both terms are odd powers of x.
Q4.7 — f(x) = x² + 2x
f(−x) = (−x)² + 2(−x) = x² − 2x. This is not equal to f(x) = x² + 2x (the linear term has the wrong sign) and not equal to −f(x) = −x² − 2x (the quadratic term has the wrong sign). Neither. Mixing an even power (x²) with an odd power (2x) breaks both symmetries.
Q4.8 — f(x) = |x| − 3
f(−x) = |−x| − 3 = |x| − 3 = f(x). Even. A vertical shift of an even function (here |x|) remains even.
Q4.9 — f(x) = 1/x², x ≠ 0
f(−x) = 1/(−x)² = 1/x² = f(x). Even. Domain is symmetric about 0 (all x ≠ 0), which is required for evenness.
Q4.10 — f(x) = x/(x² + 1)
f(−x) = (−x)/((−x)² + 1) = −x/(x² + 1) = −f(x). Odd. Numerator is odd, denominator is even, so the ratio is odd.
Q4.11 — f(x) = (x² − 4)(x⁴ + 1)
Without expanding: Both factors are even functions of x (each contains only even powers of x). Even × even = even, so f is even.
Verification: f(−x) = ((−x)² − 4)((−x)⁴ + 1) = (x² − 4)(x⁴ + 1) = f(x). ✓
Q4.12 — Proof: odd × even = odd
Let g be odd and h be even on a domain symmetric about 0. Define p(x) = g(x) · h(x).
Then p(−x) = g(−x) · h(−x) [substitute −x into the product]
= (−g(x)) · (h(x)) [g odd ⇒ g(−x) = −g(x); h even ⇒ h(−x) = h(x)]
= −(g(x) · h(x))
= −p(x).
Hence p(−x) = −p(x) for all x in the domain, so p is odd. ▮