Stationary Points

Where the tangent is level

The derivative of $y=x^2$y = x² is 2x. To the left of the origin, at x = −1, the gradient is −2, so the curve is falling. To the right, at x = 1, the gradient is 2, so the curve is rising. At the very bottom, x = 0, the gradient is 2 × 0 = 0, and the tangent there is horizontal: it is the x-axis itself.

A point where the gradient is zero is a stationary point. For an instant the curve is neither rising nor falling.

The curve $y=x^2$y = x² with its tangents at (−1, 1), falling with gradient −2, and at (1, 1), rising with gradient 2. Between them, at the origin, the tangent is the x-axis, with gradient 0.

Set the derivative to zero

So to find a stationary point, set the derivative equal to zero and solve. For $y=x^2$y = x² that is 2x = 0, so x = 0. Put x = 0 back into the equation of the curve for the y-coordinate: y = 0, so the stationary point is (0, 0).

Take $y=x^2-6x$y = x² − 6x. Its derivative is $\frac{dy}{dx}=2x-6$dy/dx = 2x − 6. Set 2x − 6 = 0, so 2x = 6 and x = 3. The height there is $3^2-6\times 3=9-18=-9$3² − 6 × 3 = 9 − 18 = −9, so the stationary point is (3, −9).

Check it with a chord that straddles x = 3. At x = 2.99 the height is 8.9401 − 17.94 = −8.9999, and at x = 3.01 it is 9.0601 − 18.06 = −8.9999. The two ends are level, so the chord gradient is 0.

The curve crosses the x-axis where $x^2-6x=x\left(x-6\right)=0$x² − 6x = x(x − 6) = 0, at x = 0 and x = 6. Those are the points where the height is zero, not the gradient. The stationary point sits halfway between them, at x = 3, because the curve is symmetric about that line.

The curve $y=x^2-6x$y = x² − 6x crosses the x-axis at x = 0 and x = 6. Its level tangent, y = −9, touches it halfway between, at the stationary point (3, −9).

A peak or a trough

The tangent is horizontal at the top of a peak as well. On $y=-x^2$y = −x² the derivative is −2x, which is 0 at x = 0. At x = −1 the gradient is 2 and at x = 1 it is −2, so the curve rises up to the origin and falls away after it.

So the sign of the gradient on each side tells the two kinds apart. Rising then falling is a peak, a local maximum. Falling then rising is a trough, a local minimum. Both are called turning points.

The cubic $y=x^3-3x$y = x³ − 3x has both. 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. The heights are ${\left(-1\right)}^3-3\left(-1\right)=-1+3=2$(−1)³ − 3(−1) = −1 + 3 = 2 and $1^3-3\times 1=-2$1³ − 3 × 1 = −2. Test the gradient on each side: at x = −2 it is 3 × 4 − 3 = 9, at x = 0 it is −3, and at x = 2 it is 9 again. The curve rises, falls, then rises, so (−1, 2) is a local maximum and (1, −2) is a local minimum.

The curve $y=x^3-3x$y = x³ − 3x, with the level tangent y = 2 on top of its peak at (−1, 2) and the level tangent y = −2 under its trough at (1, −2).

The second derivative

Differentiating the derivative gives the second derivative, written $\frac{d^{2}y}{d{x}^2}$d²y/dx². It is the rate at which the gradient changes. At a maximum the gradient goes from positive, through 0, to negative, so it is decreasing and $\frac{d^{2}y}{d{x}^2}$d²y/dx² is negative. At a minimum the gradient is increasing and $\frac{d^{2}y}{d{x}^2}$d²y/dx² is positive.

For $y=x^3-3x$y = x³ − 3x the second derivative is 6x. At x = −1 it is −6, which is negative, so (−1, 2) is a maximum. At x = 1 it is 6, which is positive, so (1, −2) is a minimum. Both answers agree with the gradient on each side.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, whose derivative is $x^2-1$x² − 1 and second derivative 2x. Drag the point to x = −1, the peak at $\left(-1,\frac{2}{3}\right)$(−1, 2/3): the circle that fits the curve there hangs below it, and the second derivative is −2. At x = 1, the trough at $\left(1,-\frac{2}{3}\right)$(1, −2/3), the circle sits above and the second derivative is 2.

Neither a peak nor a trough

A zero gradient does not always mean a turn. On $y=x^3+2$y = x³ + 2 the derivative is $3x^2$3x², which is 0 at x = 0, so (0, 2) is a stationary point. But at x = −1 and at x = 1 the gradient is 3, positive on both sides. The curve rises, levels off for an instant, and keeps rising. This is a stationary point of inflection.

The second derivative cannot always decide. On $y=x^3+2$y = x³ + 2 it is 6x, which is 0 at x = 0. On $y=x^4$y = x⁴ the second derivative $12x^2$12x² is also 0 at x = 0, yet the gradient $4x^3$4x³ is −4 at x = −1 and 4 at x = 1, so the origin is a minimum. When the second derivative is 0, test the gradient on each side.

The curve $y=x^3+2$y = x³ + 2 and its level tangent y = 2 at (0, 2). The curve rises on both sides of the point and crosses the tangent there.

The largest area

Stationary points answer questions about the largest or smallest value. A rectangle with a perimeter of 40 m has width x and length 20 − x, so its area is $A=x\left(20-x\right)=20x-x^2$A = x(20 − x) = 20x − x². Then dA/dx = 20 − 2x, which is 0 when x = 10. The second derivative is −2, which is negative, so x = 10 gives the maximum area, 10 × 10 = 100 square meters. Either side is smaller: x = 9 and x = 11 both give 99 square meters.

A rectangle of width x and length 20 − x, and its area $A=20x-x^2$A = 20x − x² plotted against x. Drag the width to 10: the area reaches 100 square meters at the top of the curve, where dA/dx = 0.

The usual mistakes

Solving y = 0 instead of $\frac{dy}{dx}=0$dy/dx = 0. On $y=x^2-6x$y = x² − 6x, x = 6 is where the curve crosses the x-axis. The gradient there is 2 × 6 − 6 = 6, not 0.

Dropping a term of the derivative. On $y=x^2-6x$y = x² − 6x the gradient at x = 0 is −6, because the −6x term contributes −6. The stationary point is at x = 3.

Stopping at x. A stationary point is a point: put x = 3 back into the curve to get y = −9, and give (3, −9).

Calling every stationary point a maximum. On $y=x^3-3x$y = x³ − 3x the point (1, −2) is a minimum, and on $y=x^3+2$y = x³ + 2 the point (0, 2) is neither. Check the gradient on each side, or the sign of the second derivative.

A sheep pen against a wall

In the application below, 40 m of fence makes three sides of a pen, and a wall makes the fourth. The area is a quadratic in the width, and its stationary point gives the greatest area.