The First Derivative Test

Where the second derivative goes silent

On $y=x^4$y = x⁴ the gradient is $4x^3$4x³ and the second derivative is $12x^2$12x². At x = 0 both are 0. The origin is a stationary point, and the second derivative test says nothing about it.

The gradient itself still speaks. At x = −0.5 it is 4 × (−0.125) = −0.5, and at x = 0.5 it is 0.5. The curve is falling before the origin and rising after it, so the origin is a trough, a local minimum.

A pause, not a turn

On $y=x^3$y = x³ the gradient is $3x^2$3x². It is 0 at x = 0, and 3 at both x = −1 and x = 1; in fact $3x^2$3x² is positive for every x except 0. Walking left to right, the sign of the gradient runs +, 0, +.

So the curve climbs up to the origin, levels off for an instant, and climbs on. There is no peak and no trough. A stationary point like this, where the curve does not turn, is a stationary point of inflection.

The test

Let c be a critical point of f, with f continuous at c. Look at the sign of f' just before c and just after it.

If f' changes from positive to negative, f has a local maximum at c. If it changes from negative to positive, f has a local minimum at c. If the sign is the same on both sides, f has neither.

The reason is what the sign of f' means. Where f' is positive on an interval, f is increasing there; where it is negative, f is decreasing. If f' is positive just before c and negative just after, f rises all the way up to f(c) and falls all the way after it, so f(c) is the highest value nearby.

A sign table

Take $y=x^3-3x$y = x³ − 3x. Its derivative is $3x^2-3=3\left(x+1\right)\left(x-1\right)$3x² − 3 = 3(x + 1)(x − 1), which is zero at x = −1 and x = 1. These two points cut the number line into three intervals.

Between neighboring critical points f' is continuous and never zero, so it cannot change sign there: one test value per interval is enough. At x = −2, f' = 12 − 3 = 9, positive. At x = 0, f' = −3, negative. At x = 2, f' = 9, positive.

So the signs run +, then −, then +. At x = −1 the sign changes from + to −: a local maximum, at height −1 + 3 = 2. At x = 1 it changes from − to +: a local minimum, at height 1 − 3 = −2.

The gold curve $y=x^3-3x$y = x³ − 3x. It climbs while f' is positive, up to the peak at (−1, 2); falls while f' is negative, down to the trough at (1, −2); then climbs again.

Where both tests are needed, and one is enough

Take $f\left(x\right)=x^4-4x^3$f(x) = x⁴ − 4x³. Its derivative is $f^{\prime }\left(x\right)=4x^3-12x^2=4x^2\left(x-3\right)$f'(x) = 4x³ − 12x² = 4x²(x − 3), zero at x = 0 and x = 3.

The second derivative is $12x^2-24x$12x² − 24x. At x = 3 it is 108 − 72 = 36, positive, so x = 3 is a minimum. At x = 0 it is 0, and that test is silent.

The sign table settles both. At x = −1, f' = 4 × 1 × (−4) = −16. At x = 1, f' = 4 × 1 × (−2) = −8. At x = 4, f' = 4 × 16 × 1 = 64. The factor $4x^2$4x² is never negative, so the sign of f' is the sign of x − 3, and it runs −, −, +.

At x = 0 the sign is − on both sides: the curve falls, levels off at (0, 0), and keeps falling, so x = 0 is neither a maximum nor a minimum. At x = 3 it changes from − to +: a local minimum, at height 81 − 108 = −27, which agrees with the second derivative.

The gold curve $y=x^4-4x^3$y = x⁴ − 4x³, with the vertical scale squeezed. It is level at (0, 0) and keeps falling; it turns only at (3, −27), where f' changes from − to +.

Where f' does not exist

The first derivative test also works at a critical point where f' does not exist, as long as f is continuous there. $f\left(x\right)=x^{\frac{2}{3}}$f(x) = x^(2/3) has $f^{\prime }\left(x\right)=\frac{2}{3\sqrt[3]{x}}$f'(x) = 2/(3∛x), with no value at 0. For x < 0 the cube root is negative, so f' is negative; for x > 0 it is positive. The sign changes from − to +, so x = 0 is a local minimum, the cusp at the origin.

The second derivative test cannot even start here, because f'(0) is not 0: it does not exist.

The gold curve $y=x^4$y = x⁴, where f' runs −, 0, + across the origin, a minimum; and the dashed curve $y=x^3$y = x³, where f' runs +, 0, +, with no turn. Both have second derivative 0 at the origin.

The usual mistakes

Testing past the next critical point. For $x^3-3x$x³ − 3x, a test value for x = −1 must lie between −1 and 1 on the right; x = 2 is beyond x = 1, where f' is positive again, and would make −1 look like a pause.

Reading the sign of f instead of f'. $x^3-3x$x³ − 3x is positive at x = −0.5 and negative at x = 0.5, but what decides is the gradient, which is negative at both.

Swapping the two cases. + then − is a peak: the curve climbs, then falls.

Calling a pause a turn. $x^3$x³ has f' positive on both sides of 0, so the curve only levels off there.

A bottling machine

In the application below, the waste of glass depends on a machine setting through ${\left(x-12\right)}^4$(x − 12)⁴. Its second derivative is 0 at the best setting, and the sign of the first derivative on each side settles it.