Finding Stationary Points

Where the curve levels off

At the lowest point of $y=x^2$y = x² the curve stops falling and starts rising. For that instant the tangent is level, the x-axis itself, and its gradient is 0. The derivative 2x agrees: it is 0 at x = 0.

A point where the derivative is 0 is a stationary point: there the function is, for an instant, neither increasing nor decreasing.

The method

So to find the stationary points, differentiate, set the derivative equal to 0, and solve. Each solution is the x-coordinate of a stationary point; put it back into f, not f', to get the height.

For $f\left(x\right)=x^3-3x$f(x) = x³ − 3x, $f^{\prime }\left(x\right)=3x^2-3$f'(x) = 3x² − 3. Setting $3x^2-3=0$3x² − 3 = 0 gives $x^2=1$x² = 1, so x = −1 or x = 1. The heights are f(−1) = −1 + 3 = 2 and f(1) = 1 − 3 = −2. The stationary points are (−1, 2) and (1, −2). Check by substitution: f'(−1) = 3 − 3 = 0 and f'(1) = 0.

For $f\left(x\right)=x^3-6x^2+9x$f(x) = x³ − 6x² + 9x, $f^{\prime }\left(x\right)=3x^2-12x+9=3\left(x-1\right)\left(x-3\right)$f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), which is 0 at x = 1 and x = 3. The heights are 1 − 6 + 9 = 4 and 27 − 54 + 27 = 0, so the stationary points are (1, 4) and (3, 0).

Every one of them

Solving f'(x) = 0 completely finds every stationary point, so every solution of the equation counts, including x = 0 when a factor of x comes out.

For $f\left(x\right)=x^4-8x^2$f(x) = x⁴ − 8x², $f^{\prime }\left(x\right)=4x^3-16x=4x\left(x^2-4\right)=4x\left(x-2\right)\left(x+2\right)$f'(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2), which is 0 at x = −2, 0 and 2. The heights are f(−2) = 16 − 32 = −16, f(0) = 0 and f(2) = −16. So there are three stationary points: (−2, −16), (0, 0) and (2, −16). Dividing $4x^3=16x$4x³ = 16x by x would lose the one at the origin.

For $f\left(x\right)=x+\frac{4}{x}$f(x) = x + 4/x, $f^{\prime }\left(x\right)=1-\frac{4}{x^2}$f'(x) = 1 − 4/x², which is 0 when $x^2=4$x² = 4, at x = −2 and x = 2. The heights are −2 − 2 = −4 and 2 + 2 = 4, so the stationary points are (−2, −4) and (2, 4). At x = 0 the function is not defined, so there is no point there to find.

The curve $y=x^4-8x^2$y = x⁴ − 8x² and its three stationary points, at x = −2, 0 and 2. The level line y = −16 is the tangent at both (−2, −16) and (2, −16), and at the origin the tangent is the x-axis.

The curve $y=x+\frac{4}{x}$y = x + 4/x, in two pieces either side of x = 0. Its stationary points are (2, 4), with the level tangent y = 4, and (−2, −4), with the level tangent y = −4.

Peaks, troughs and others

The tangent is level at the top of a peak as well as the bottom of a trough. On $y=-x^2$y = −x² the derivative −2x is 0 at x = 0, where the curve is highest. So f' = 0 picks out both kinds, and the equation alone does not say which kind each point is.

Some stationary points are neither. For $f\left(x\right)=x^3$f(x) = x³, $f^{\prime }\left(x\right)=3x^2$f'(x) = 3x² is 0 at x = 0, so (0, 0) is a stationary point, but the curve levels off there for an instant and then carries on upward. Sorting the points into kinds is the job of the next lessons.

And some curves have none. For $f\left(x\right)=x^3+3x$f(x) = x³ + 3x, $f^{\prime }\left(x\right)=3x^2+3$f'(x) = 3x² + 3 is at least 3 for every x, so f'(x) = 0 has no solution and the curve has no stationary point.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, whose derivative is $x^2-1$x² − 1. At x = 0.4 the tangent has gradient m = −0.84. Drag the point P to x = 1 or x = −1, the solutions of $x^2-1=0$x² − 1 = 0: there m = 0 and the tangent is level.

The usual mistakes

Solving y = 0 instead of $\frac{dy}{dx}=0$dy/dx = 0. For $y=x^2-6x$y = x² − 6x, the curve crosses the axis at x = 0 and x = 6, but its stationary point is where 2x − 6 = 0, at x = 3, and its height is 9 − 18 = −9.

Putting the x-value back into f' for the height. f' is 0 there by construction; the height comes from f.

Dividing by x and losing a solution. $4x^3-16x=0$4x³ − 16x = 0 has three solutions, not two.

Stopping at one solution. $x^2=1$x² = 1 gives x = 1 and x = −1.