The second derivative and concavity

Higher Level

Three graphs, one x. The curve on top is coloured by the sign of f″: green where it bends upwards like a cup, orange where it bends downwards like a cap. Move the line and watch all three agree.

x = −0.80

–f(x)

–f′(x)

–f″(x)

concave upthe curve is

Stop at x = 1 and watch the colour change at the exact moment f″ crosses zero.

What f″ tells you that f′ cannot

f″(x) > 0: concave up. The gradient is increasing, so the curve holds water.
f″(x) < 0: concave down. The gradient is decreasing, so the curve spills it.
f″(x) = 0 and the sign changes: a point of inflexion. The bend swaps direction.

f′ tells you whether the curve is going up or down. f″ tells you whether it is speeding up or easing off. A curve can be rising the whole time and still change its mind about how fast.

The second derivative test

At a stationary point, where f′(x) = 0, the sign of f″ settles the question in one step.

f(x)= x³ − 3x² + 1 f′(x)= 3x² − 6x = 3x(x − 2), zero at x = 0 and x = 2 f″(x)= 6x − 6 f″(0)= −6, negative, so (0, 1) is a maximum concave down, like a cap f″(2)= 6, positive, so (2, −3) is a minimum concave up, like a cup

If f″ is zero at a stationary point the test tells you nothing, and you must fall back on checking the sign of f′ either side. For y = x⁴ at the origin, f′ and f″ are both zero and it is still a minimum. The test is faster, not stronger.

Points of inflexion

6x − 6= 0, so x = 1 f(1)= 1 − 3 + 1 = −1 so the point of inflexion is (1, −1)

f″ = 0 is not enough. The concavity has to actually change. Here f″(0.5) = −3 and f″(1.5) = 3, so it does. For y = x⁴ at the origin f″ = 0 and the curve stays concave up on both sides, so that is not a point of inflexion at all. It is the same trap as f′ = 0 not guaranteeing a turning point.

Your turn

1. For f(x) = x³ − 3x² + 1, find f″(2).

2. At a stationary point, f″ is negative. The point is:

a local minimum. a local maximum. a point of inflexion. impossible to classify.

3. Give the y coordinate of the point of inflexion of f(x) = x³ − 3x² + 1.

Where the marks go

Using the second derivative test is usually quicker than a sign table and earns the same mark, as long as you state the sign and the conclusion.

For a point of inflexion, solving f″ = 0 gives the x value; you still have to substitute into f for the coordinate, and ideally confirm the concavity actually changes.

Watch the notation. f″(x) and d²y/dx² mean the same thing, and a question may use either.

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