Mathematics Advanced • Year 11 • Module 1 • Lesson 5
Odd & Even Functions
Apply even/odd symmetry to multi-step problems involving graphs, modelling and function decomposition.
Problem 1 — Bridge arch (geometric)
The height (in metres) of a parabolic pedestrian bridge above the deck, as a function of horizontal distance x from the centre of the bridge, is modelled by
h(x) = 9 − 0.04 x², for −15 ≤ x ≤ 15
Set up: What are we solving for?
(i) Show algebraically that h(x) is an even function on its given domain. 2 marks
(ii) A surveyor measures the height at x = 7 m and obtains h(7) = 7.04 m. Without further calculation, state h(−7) and justify your answer using the result from (i). 1 mark
(iii) The engineer wants to add a horizontal handrail at height 6 m. Find the x-values where the rail meets the arch, and explain in one sentence how the evenness of h(x) cut the work in half. 3 marks
Stuck? Revisit lesson § Even Functions (geometric meaning).Problem 2 — Piston displacement (rotational symmetry)
An engineer models the displacement of a piston (in millimetres) about its central rest position, as a function of crank-angle x (in radians) measured from the rest point, by
d(x) = 40 sin(x) + 5 x³, for −1 ≤ x ≤ 1
Set up: What are we solving for?
(i) Show that d(x) is an odd function and state what this means about the piston's motion relative to its rest position. 3 marks
(ii) Given d(0.6) ≈ 23.66 mm (displacement above rest), use the symmetry of d to write down d(−0.6) and interpret the sign physically. 2 marks
(iii) The engineer claims that "for any odd function, d(0) must equal 0". Confirm this for d(x) by direct substitution, then give a one-line algebraic proof that any odd function f with 0 in its domain must satisfy f(0) = 0. 2 marks
Problem 3 — Solar-panel power curve (data)
A north-facing rooftop solar panel produces power P (in watts) as a function of the sun's angle x (in degrees) east of due north, where x = 0 corresponds to solar noon. Data measured at 11 am and 1 pm gave the following symmetric pair:
| x (degrees east of N) | P (watts) |
|---|---|
| −15° (11 am) | 1180 |
| 0° (12 noon) | 1240 |
| +15° (1 pm) | 1180 |
Set up: What are we solving for?
(i) Decide whether the data is consistent with P(x) being even, odd, or neither. Justify in one sentence. 1 mark
(ii) A student proposes the model P(x) = 1240 − 0.267 x². Show that this model is even and that it correctly produces P(±15°) = 1180 W (to the nearest watt). 3 marks
(iii) A different student proposes P(x) = 1240 − 4 x. Test whether this model is even, odd, or neither, and explain in one sentence why the symmetry of the measured data rules it out. 2 marks
Stuck? Revisit lesson § Worked Example 3 — Neither Even nor Odd.Problem 4 — Decomposing a signal into even + odd parts
An audio signal sampled around time t = 0 is described by f(t) = t² + t. A signal-processing technique splits any function f into an even part fe(t) and an odd part fo(t) using
fe(t) = ½ [ f(t) + f(−t) ], fo(t) = ½ [ f(t) − f(−t) ]
Set up: What are we solving for?
(i) For f(t) = t² + t, compute fe(t) and fo(t) explicitly. 3 marks
(ii) Verify by algebraic test that fe(t) is even and fo(t) is odd. 2 marks
(iii) Confirm that fe(t) + fo(t) reproduces the original f(t), and explain in one sentence what this tells us about every function defined on a domain symmetric about 0. 2 marks
Problem 5 — Symmetry shortcut for definite integrals
A property of integrals (you'll meet this formally in Year 12) states that for any function on a symmetric interval [−a, a]:
• If f is odd, then ∫−aa f(x) dx = 0.
• If f is even, then ∫−aa f(x) dx = 2 ∫0a f(x) dx.
Set up: What are we solving for?
(i) Classify each function as even, odd, or neither. 2 marks
A. g(x) = 4x³ − x → ___________ B. g(x) = 3x² + 7 → ___________
C. g(x) = 2x² + 5x → ___________ D. g(x) = sin(x) → ___________
(ii) Using the property above, state the value of ∫−22 (4x³ − x) dx and justify in one line. 2 marks
(iii) Explain in 1-2 lines why the property fails for option C and what (if anything) symmetry tells you about ∫−22 (2x² + 5x) dx. 2 marks
Stuck on C? Try splitting it into its even part and its odd part using the rule from Problem 4.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bridge arch
Set up. We are testing whether the arch is symmetric about the centre (x = 0), using h(x), and then using that symmetry to find heights and pair-up roots without doubling our work.
(i) h(−x) = 9 − 0.04(−x)² = 9 − 0.04 x² = h(x). The domain −15 ≤ x ≤ 15 is symmetric about 0, so the algebraic test is valid. Hence h is even.
(ii) Because h is even, h(−7) = h(7) = 7.04 m. No recalculation needed: f(−a) = f(a) for any even function.
(iii) Solve 9 − 0.04 x² = 6 ⇒ 0.04 x² = 3 ⇒ x² = 75 ⇒ x = ±√75 = ±5√3 ≈ ±8.66 m. The rail meets the arch at x ≈ ±8.66 m from centre. Symmetry shortcut: evenness guarantees that any positive solution x = a is paired with x = −a, so we only had to find one root then negate it.
Problem 2 — Piston displacement
Set up. We are testing whether the piston motion is rotationally symmetric (odd) about the rest position, and using that to predict displacement on the opposite side and the value at rest.
(i) d(−x) = 40 sin(−x) + 5(−x)³ = −40 sin(x) − 5 x³ = −(40 sin(x) + 5 x³) = −d(x). Both sin(x) and x³ are odd; a sum of odd functions is odd. So d is odd: a positive displacement at crank-angle +x is matched by an equal-magnitude displacement in the opposite direction at −x.
(ii) d(−0.6) = −d(0.6) ≈ −23.66 mm. The minus sign means the piston is displaced 23.66 mm below rest, i.e. in the opposite direction to its position at x = +0.6.
(iii) Direct: d(0) = 40 sin(0) + 5·(0)³ = 0 + 0 = 0 mm. ✓ Proof: If f is odd then f(−0) = −f(0). But −0 = 0, so f(0) = −f(0), giving 2 f(0) = 0, hence f(0) = 0.
Problem 3 — Solar-panel power curve
Set up. We are matching a measured dataset to candidate models by checking each model's symmetry against the symmetry of the data.
(i) The data shows P(−15°) = P(+15°) = 1180 W, which is consistent with an even function (symmetric about x = 0). It is inconsistent with odd, because odd requires P(−15°) = −P(+15°).
(ii) P(−x) = 1240 − 0.267(−x)² = 1240 − 0.267 x² = P(x). ✓ Even. At x = 15: P(15) = 1240 − 0.267 × 225 = 1240 − 60.075 = 1179.925 ≈ 1180 W. ✓ By evenness, P(−15) = 1180 W also.
(iii) P(−x) = 1240 − 4(−x) = 1240 + 4x, which equals neither P(x) (= 1240 − 4x) nor −P(x). So the model is neither even nor odd. This is incompatible with the data because the data is symmetric: the 11 am and 1 pm readings are equal, but 1240 − 4(15) = 1180 while 1240 − 4(−15) = 1300, contradicting the measured 1180 W at −15°.
Problem 4 — Even + odd decomposition
Set up. We are splitting an arbitrary function into an even component and an odd component, then verifying each, and using the result to recover f.
(i) First compute f(−t) = (−t)² + (−t) = t² − t.
fe(t) = ½ [(t² + t) + (t² − t)] = ½ (2t²) = t².
fo(t) = ½ [(t² + t) − (t² − t)] = ½ (2t) = t.
(ii) fe(−t) = (−t)² = t² = fe(t). ✓ even. fo(−t) = −t = −fo(t). ✓ odd.
(iii) fe(t) + fo(t) = t² + t = f(t). ✓ Any function defined on a domain symmetric about 0 can be uniquely written as the sum of an even function and an odd function — even one that, like f, is itself neither even nor odd.
Problem 5 — Integral shortcut
Set up. We are classifying each integrand by symmetry to decide whether the integral over a symmetric interval vanishes, doubles, or requires direct calculation.
(i) A. g(−x) = 4(−x)³ − (−x) = −4x³ + x = −(4x³ − x) = −g(x) → odd.
B. g(−x) = 3(−x)² + 7 = 3x² + 7 = g(x) → even.
C. g(−x) = 2(−x)² + 5(−x) = 2x² − 5x. Not g(x), not −g(x) → neither.
D. sin(−x) = −sin(x) → odd.
(ii) 4x³ − x is odd, so ∫−22 (4x³ − x) dx = 0, by the odd-on-symmetric-interval rule.
(iii) The rule applies only to functions that are purely even or purely odd; option C is neither, so neither shortcut applies directly. However, using the decomposition from Problem 4 we can split 2x² + 5x into its even part (2x²) and its odd part (5x). The odd part integrates to 0 on [−2, 2], so ∫−22 (2x² + 5x) dx = ∫−22 2x² dx = 2 ∫02 2x² dx, which is a strictly easier calculation than tackling the original integrand whole.