Maxima & minima

Ride a bike over the crest of a hill: first it goes uphill, then for a brief moment it is level, then downhill. That one moment, where it goes neither up nor down, is the highest point of the route. A function graph does the same thing: where it turns around, it runs flat for an instant. Such turning spots are called maxima and minima, together extreme points, and they usually carry the information you actually care about: the greatest height of a throw, the lowest cost, the biggest profit.

At the turning point it is flat

The derivative $f'(x)$ measures the slope at position $x$. As long as the graph climbs, it is positive, as long as the graph falls, it is negative. So when the graph turns around, the derivative has to switch from positive to negative or the other way round, and in between lies the only value that is neither positive nor negative: 0. If you lay a tangent on the graph there, it runs flat, and its slope is exactly 0. For $f(x) = x^2$ with $f'(x) = 2x$ this happens at $x = 0$, and that is where the lowest point of the parabola sits. For the downward parabola $f(x) = -x^2$ with $f'(x) = -2x$ the derivative is zero at $x = 0$ as well, only there it marks the highest point. The $x$ coordinate of such a turning spot is called an extreme position, and the point together with its height is called an extreme point.

Working out the candidates

An example: $f(x) = x^2 - 6x + 11$. You build the derivative term by term, every power drops by one and its old exponent moves in front as a factor. So $x^2$ becomes $2x$, $-6x$ becomes $-6$, and the plain number $11$ drops out, because a constant does not change the slope. All together: $f'(x) = 2x - 6$. Now you set the derivative to zero, and $2x - 6 = 0$ gives $x = 3$. You get the matching height by substituting $x = 3$ into the original function: $f(3) = 9 - 18 + 11 = 2$. So the candidate is the point $(3,\ 2)$. If the equation $f'(x) = 0$ has several solutions, you get several candidates, and each one is checked on its own.

High or low? The sign decides

A zero of the derivative does not yet tell you whether the graph turns around at the top or at the bottom. The sign of $f'$ just left and just right of it does. If it switches from plus to minus, the graph rises first and then falls, and that is a maximum. If it switches from minus to plus, it falls first and then rises, and that is a minimum. Test it on our example $f'(x) = 2x - 6$: at $x = 2$ we get $f'(2) = -2$, so the graph falls there. At $x = 4$ we get $f'(4) = 2$, so it rises there. Minus before plus, so at $x = 3$ there is a minimum.

There is one case where the derivative is zero and the graph still does not turn around. For $f(x) = x^3$ we have $f'(x) = 3x^2$, and $3x^2 = 0$ gives $x = 0$. But if you plug in test values you get $f'(-1) = 3$ and $f'(1) = 3$, positive both times. At $x = 0$ the graph merely takes a short flat pause and then keeps climbing, and such a spot is called a saddle point. That is why $f'(x) = 0$ is only the entry ticket: only the change of sign decides whether there really is a maximum or a minimum.