← Calculus methods

Proving a maximum or minimum

Finding f'(x) = 0 only LOCATES a stationary point — every year's Excellence mark requires actually proving which kind it is.

Two valid proof methods

• Second derivative test: find f''(x) and substitute the x-value. NEGATIVE means the curve bends downward (a maximum); POSITIVE means it bends upward (a minimum).
• Sign test: evaluate f'(x) at a value slightly BELOW and slightly ABOVE the stationary point. A change from + to − is a maximum; − to + is a minimum.
• Either method is acceptable — but you MUST show the working for whichever one you pick, not just assert "this is a maximum".

Worked example Proving a minimum with the second derivative A = 80r − 2πr² has a stationary point at r = 40/π (2πr found from dA/dr = 80 − 4πr = 0, in a slightly different setup from the previous example). Prove whether it is a maximum or minimum. Find the second derivative: A''(r) = d/dr(80 − 4πr) = −4π. −4π is a NEGATIVE constant, for every value of r (it does not depend on r at all here). A negative second derivative at the stationary point means the curve bends downward — a MAXIMUM. ✓ The second derivative is always negative, so this stationary point is a maximum, confirmed without needing to test specific values either side. ✕

Asserting the answer is a maximum/minimum without showing the test

Finding r = 40/π and simply writing "this is the maximum value" earns no proof credit, however correct the arithmetic. The schedule specifically rewards SHOWING either the second-derivative sign or the sign change of f'(x) either side of the point — the proof is graded as its own, separate piece of evidence from the calculation that found the point.

Practise just this