The Second Derivative Test

A trough bends upwards

On $y=x^2$y = x² the gradient is $\frac{dy}{dx}=2x$dy/dx = 2x. At x = −1 it is −2, at x = 0 it is 0, and at x = 1 it is 2. Walking left to right through the trough, the gradient climbs from negative, through zero, to positive: it is growing.

The rate at which the gradient grows is the derivative of the derivative, the second derivative, written $\frac{d^{2}y}{d{x}^2}$d²y/dx². For $y=x^2$y = x² it is 2, a positive number. A curve whose gradient is growing bends upwards, like the inside of a bowl.

A peak bends downwards

On $y=-x^2$y = −x² the gradient is −2x: 2 at x = −1, 0 at x = 0, and −2 at x = 1. Now the gradient falls from positive, through zero, to negative: it is shrinking, and $\frac{d^{2}y}{d{x}^2}=-2$d²y/dx² = −2 is negative. The curve bends downwards, like an upturned bowl.

The gold curve $y=x^2$y = x², second derivative 2, bends upwards from its trough at the origin. The dashed curve $y=-x^2$y = −x², second derivative −2, bends downwards from its peak at the same point.

The test

Let c be a stationary point, so the gradient there is 0. Work out the second derivative at c.

If it is positive, the curve bends upwards at c, and the stationary point is a local minimum. If it is negative, the curve bends downwards, and the stationary point is a local maximum. If it is 0, the test gives no answer.

The reason is the gradient. A positive second derivative means the gradient is growing as x passes c, and it is 0 at c, so it goes from negative to positive: the curve falls into c and climbs out, a trough. A negative second derivative means the gradient is shrinking through 0: the curve climbs to c and falls away, a peak.

Only the sign counts, not the size. A second derivative of 2 decides as firmly as one of 200.

A cubic with both

Take $y=x^3-6x^2+9x$y = x³ − 6x² + 9x. Its derivative is $\frac{dy}{dx}=3x^2-12x+9=3\left(x-1\right)\left(x-3\right)$dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3), which is zero at x = 1 and x = 3.

The second derivative is $\frac{d^{2}y}{d{x}^2}=6x-12$d²y/dx² = 6x − 12. At x = 1 it is 6 − 12 = −6, which is negative, so x = 1 is a local maximum, at height 1 − 6 + 9 = 4. At x = 3 it is 18 − 12 = 6, which is positive, so x = 3 is a local minimum, at height 27 − 54 + 27 = 0.

So the curve has a peak at (1, 4) and a trough at (3, 0), where it touches the x-axis.

The gold curve $y=x^3-6x^2+9x$y = x³ − 6x² + 9x. At (1, 4), under the level gold tangent, the second derivative is −6 and the curve bends down; at (3, 0) it is 6 and the curve bends up.

A curve in two pieces

Take $y=x+\frac{4}{x}$y = x + 4/x, which has no value at x = 0. Its derivative is $\frac{dy}{dx}=1-\frac{4}{x^2}$dy/dx = 1 − 4/x², which is zero when $x^2=4$x² = 4, at x = 2 and x = −2.

The second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{8}{x^3}$d²y/dx² = 8/x³. At x = 2 it is $\frac{8}{8}=1$8/8 = 1, which is positive, so (2, 4) is a local minimum, since $2+\frac{4}{2}=4$2 + 4/2 = 4. At x = −2 it is $\frac{8}{-8}=-1$8/(−8) = −1, which is negative, so (−2, −4) is a local maximum.

Here the local maximum, −4, is lower than the local minimum, 4. Each is a statement about its own neighborhood: near x = 2 every height is at least 4, and near x = −2 every height is at most −4. The heights at x = 1 and x = 4 are both 5, above the minimum; at x = −1 and x = −4 they are both −5, below the maximum.

The gold curve $y=x+\frac{4}{x}$y = x + 4/x, with dashed asymptotes x = 0 and y = x. Its right-hand branch bends upwards around the minimum (2, 4); its left-hand branch bends downwards around the maximum (−2, −4), which is lower.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, with second derivative 2x. Drag the point to x = 1: the circle that fits the curve sits above it, like a bowl, and the second derivative is 2. At x = −1 the circle hangs below and the second derivative is −2.

When the second derivative is zero

At the origin $y=x^4$y = x⁴ has gradient $4x^3=0$4x³ = 0 and second derivative $12x^2=0$12x² = 0. The origin is a minimum: $x^4$x⁴ is 0.0001 at both x = −0.1 and x = 0.1, and positive for every x except 0.

$y=x^3$y = x³ also has gradient $3x^2=0$3x² = 0 and second derivative 6x = 0 at the origin. But it is no turning point: $x^3$x³ is −0.001 at x = −0.1 and 0.001 at x = 0.1, so the curve climbs straight through.

And $y=-x^4$y = −x⁴ has gradient and second derivative both 0 there too, with a maximum at the origin.

So a second derivative of 0 fits a minimum, a maximum, and a point that is neither. The test is silent: it does not say the point is a point of inflection, it says nothing at all. Look at the curve, or use the first derivative test, the next row, which settles every one of these cases.

Three curves with gradient 0 and second derivative 0 at the origin: the gold curve $y=x^4$y = x⁴ has a minimum there, the dashed curve $y=-x^4$y = −x⁴ has a maximum, and the dashed curve $y=x^3$y = x³ passes straight through.

The usual mistakes

Reading the sign the wrong way round. Positive means bending upwards, a trough; a second derivative of 6 at a stationary point is a minimum, not a maximum.

Treating a small value as zero. A second derivative of 0.2 is positive and decides just as 6 does; only exactly 0 is silent.

Calling a zero second derivative a point of inflection. On $x^4$x⁴ it is 0 at the origin, which is a minimum.

Using the test where the gradient is not 0. On $y=x^3-3x$y = x³ − 3x at x = 2 the second derivative is 12, but the gradient is 3 × 4 − 3 = 9, so the curve is climbing there: bending upwards, with no trough. The test classifies stationary points only.

Three beds beside a canal

In the application below, 240 m of fencing makes a plot against a canal, divided into three beds. The area is a quadratic in one length, its stationary point gives the dimensions, and the second derivative, −8, confirms a maximum.