Points of Inflection

The bend changes sides

On $y=x^3$y = x³ the gradient is $3x^2$3x² and the second derivative is 6x. Left of the origin, 6x is negative: the gradient is shrinking, and the curve bends downwards. Right of the origin, 6x is positive: the gradient is growing, and the curve bends upwards.

So at the origin the bend changes sides. A point on a curve where the direction of bending changes is a point of inflection.

Concave up and concave down

The two kinds of bend have names. Where the second derivative is positive, the gradient is increasing and the curve is concave up, bending like the inside of a bowl. Where the second derivative is negative, the gradient is decreasing and the curve is concave down.

So $y=x^3$y = x³ is concave down for x < 0 and concave up for x > 0, and a point of inflection is where a curve passes from one to the other.

The gold curve $y=x^3$y = x³, with the vertical scale squeezed: concave down to the left of the origin, where the second derivative 6x is negative, and concave up to the right, where it is positive.

Where the second derivative is zero

Where the bend flips, the second derivative passes from negative to positive, or from positive to negative, and so passes through 0. That gives the candidates: the points where $\frac{d^{2}y}{d{x}^2}=0$d²y/dx² = 0, together with any points of the curve where the second derivative does not exist.

Take $y=x^3-3x^2$y = x³ − 3x². Its gradient is $3x^2-6x$3x² − 6x and its second derivative is 6x − 6, which is 0 at x = 1. The height there is 1 − 3 = −2, so the candidate is (1, −2).

A zero is not enough

On $y=x^4$y = x⁴ the second derivative is $12x^2$12x², which is 0 at x = 0. But $12x^2$12x² is positive on both sides of 0: at x = −1 and at x = 1 it is 12. The curve is concave up on both sides, so its bend never changes, and the origin is not a point of inflection. It is the minimum of $x^4$x⁴.

So $\frac{d^{2}y}{d{x}^2}=0$d²y/dx² = 0 marks a candidate, not an inflection.

The gold curve $y=x^4$y = x⁴, concave up on both sides of the origin. Its second derivative, $12x^2$12x², is 0 at the origin without changing sign.

Check the sign on each side

So test the sign of the second derivative on each side of the candidate. For $y=x^3-3x^2$y = x³ − 3x², 6x − 6 is −6 at x = 0 and 6 at x = 2. It runs −, 0, +, so the curve passes from concave down to concave up, and (1, −2) is a point of inflection.

The gradient there is 3 − 6 = −3, not 0: a point of inflection does not have to be a stationary point. The tangent at (1, −2) is y = −2 − 3(x − 1) = −3x + 1. The curve lies below it on the left, by 0.125 at x = 0.5, and above it on the right, by 0.125 at x = 1.5. At a point of inflection the curve crosses its own tangent.

The gold curve $y=x^3-3x^2$y = x³ − 3x² and the gold tangent y = −3x + 1 at (1, −2). The curve is below the tangent on the left, where it is concave down, and above it on the right, where it is concave up.

The curve $y=x^3-3x$y = x³ − 3x with its tangent at a point you move. The shaded gap shows on which side the curve lies. Slide the point through x = 0: the curve changes sides of the tangent, and there the tangent has gradient −3.

Two inflections

Take $y=x^4-6x^2$y = x⁴ − 6x². Its second derivative is $12x^2-12=12\left(x-1\right)\left(x+1\right)$12x² − 12 = 12(x − 1)(x + 1), which is 0 at x = −1 and x = 1.

At x = −2 it is 36, at x = 0 it is −12, and at x = 2 it is 36. The signs run +, −, +, so the bend changes at both candidates. The heights are 1 − 6 = −5 at each, and the points of inflection are (−1, −5) and (1, −5). Between them the curve is concave down, around the local maximum at the origin; outside them it is concave up.

The gold curve $y=x^4-6x^2$y = x⁴ − 6x², concave down between its two points of inflection, (−1, −5) and (1, −5), and concave up outside them.

Two cases to watch

A curve can inflect where the second derivative does not exist. On $y=\sqrt[3]{x}$y = ∛x the second derivative is $-\frac{2x^{-\frac{5}{3}}}{9}$−2x^(−5/3)/9, with no value at x = 0. At x = −1 it is $\frac{2}{9}$2/9, positive, and at x = 1 it is $-\frac{2}{9}$−2/9, negative. So the bend changes at the origin, which is a point of inflection with a vertical tangent.

A sign change at a point not on the curve is no inflection. On $y=\frac{1}{x}$y = 1/x the second derivative is $\frac{2}{x^3}$2/x³: −2 at x = −1 and 2 at x = 1. The concavity changes across x = 0, but the curve has no point there, so it has no point of inflection.

The usual mistakes

Stopping at $\frac{d^{2}y}{d{x}^2}=0$d²y/dx² = 0. $x^4$x⁴ has that at the origin, and the origin is a minimum.

Testing the sign of f' instead of the second derivative. f' changing sign marks a peak or a trough; at the inflection of $x^3-3x^2$x³ − 3x², f' is negative on both sides.

Thinking a point of inflection must be flat. The tangent at (1, −2) on $x^3-3x^2$x³ − 3x² has gradient −3.

Giving only x = 1. A point of inflection is a point: (1, −2).

An outbreak

In the application below, the total number of cases in an outbreak is a cubic in time. New cases appear fastest on the day the curve of totals changes from concave up to concave down.