10 min readPathway sheet3-mark and 5-mark questions; the quickest test in most long answers; also a JEE favourite

Second derivative test

If f′(c) = 0 and f″(c) < 0, c is a local maximum; if f″(c) > 0, a local minimum. If f″(c) = 0, the test fails.

Local maximum at cf′(c) = 0 and f″(c) < 0; the local maximum value is f(c)

Local minimum at cf′(c) = 0 and f″(c) > 0; the local minimum value is f(c)

Test failsf′(c) = 0 and f″(c) = 0; use the first derivative test instead

What the test needsf twice differentiable at c

The idea

f″(x) is the rate of change of the slope f′(x). At a hilltop the slope goes from + through 0 to −, so the slope is falling: f″(c) < 0, and the curve bends like a cap (∩). At a valley the slope goes from − through 0 to +, so it is rising: f″(c) > 0, and the curve bends like a cup (∪). Your textbook gives this test as a tool to use; you do not have to prove it.

The test

f′(c)	f″(c)	Conclusion
0	negative	Local maximum at c
0	positive	Local minimum at c
0	0	Test fails: use the first derivative test

Example 1: a cubic

Problem

Find the local maximum and local minimum values of f(x) = x³ − 6x² + 9x + 4.

Working

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3)
f′(x) = 0 gives x = 1 and x = 3.
f″(x) = 6x − 12
f″(1) = −6 < 0, so x = 1 is a point of local maxima.
Local maximum value f(1) = 1 − 6 + 9 + 4 = 8
f″(3) = 6 > 0, so x = 3 is a point of local minima.
Local minimum value f(3) = 27 − 54 + 27 + 4 = 4

Answer

Local maximum value 8 at x = 1; local minimum value 4 at x = 3.

Example 2: a trigonometric function

Problem

Find the local maximum value of f(x) = sin x + √3 cos x, 0 < x < π/2.

Working

f′(x) = cos x − √3 sin x
f′(x) = 0 gives cos x = √3 sin x, so tan x = 1/√3 and x = π/6.
f″(x) = −sin x − √3 cos x
f″(π/6) = −1/2 − √3 × (√3/2) = −1/2 − 3/2 = −2 < 0
So x = π/6 is a point of local maxima.
f(π/6) = 1/2 + √3 × (√3/2) = 1/2 + 3/2 = 2

Answer

Local maximum value 2 at x = π/6.

Example 3: when the test fails

Problem

Find the points of local maxima or minima of f(x) = x⁴.

Working

f′(x) = 4x³ = 0 gives x = 0.
f″(x) = 12x², so f″(0) = 0. The second derivative test fails.
Use the first derivative test: f′(x) = 4x³ < 0 for x < 0 and > 0 for x > 0.
The sign changes from − to +.

Answer

x = 0 is a point of local minima, with local minimum value f(0) = 0.

First derivative test vs second derivative test

First derivative testLooks at the sign of f′ on both sides of c. Works whenever f is continuous at c, even at corners. Slower, as it needs a sign chart.

Second derivative testLooks only at the sign of f″(c). Quick for polynomials and trigonometric functions. Needs f′(c) = 0, and gives no answer when f″(c) = 0.

In a long answer, try the second derivative test first. Switch to the first derivative test if f″(c) = 0.

In real life

Engineers choosing the tilt of a solar panel on a village roof look for the angle that gives the most power. The second derivative test confirms that the angle gives a maximum, not a minimum.

Exam angle

1 marks: “f′(2) = 0 and f″(2) = 5. What kind of point is x = 2?” A point of local minima, since f″(2) > 0.

2 marks: “Using the second derivative test, find the local maximum value of f(x) = 12x − x³.” f′(x) = 12 − 3x² = 0 gives x = ±2. f″(x) = −6x, and f″(2) = −12 < 0, so x = 2 gives a local maximum. Value f(2) = 24 − 8 = 16.

3 marks: “Find the local maximum and local minimum values of f(x) = x³ − 27x + 10.” f′(x) = 3x² − 27 = 0 gives x = ±3. f″(x) = 6x. f″(−3) = −18 < 0: local maximum value f(−3) = −27 + 81 + 10 = 64. f″(3) = 18 > 0: local minimum value f(3) = 27 − 81 + 10 = −44.

Remember it

f″ negative: a frown, so a maximum. f″ positive: a smile, so a minimum.