Maxima and Minima

The first derivative test

A critical point is a value $c$ where $f'(c) = 0$ (or $f'(c)$ is undefined). To classify it, look at how the sign of $f'$ changes as $x$ passes through $c$:

The second derivative test

At a point where $f'(x) = 0$: if $f''(x) < 0$, it's a local maximum; if $f''(x) > 0$, it's a local minimum.

Worked example: local maximum of a cubic

Find the local maximum value of $g(x) = x^3 - 3x$.

Solution: $g'(x) = 3x^2 - 3 = 0$ at $x = \pm 1$. Checking the second derivative, $g''(x) = 6x$:

At $x=-1$: $g''(-1) = -6 < 0$, a local maximum, with value $g(-1) = -1 - (-3) = 2$.

At $x=1$: $g''(1) = 6 > 0$, a local minimum. So the local maximum value is 2.

The same cubic, by the first derivative test

$g'(x) = 3(x - 1)(x + 1)$ is positive on $(-\infty, -1)$, negative on $(-1, 1)$ and positive on $(1, \infty)$. The sign goes $+$ to $-$ at $x = -1$ (a local maximum) and $-$ to $+$ at $x = 1$ (a local minimum), agreeing with the second derivative test.

When $f''(c) = 0$, the second derivative test gives no answer and the first derivative test is the way to decide. For example, $f(x) = x^4$ has $f''(0) = 0$, but $f'$ goes from $-$ to $+$ at $0$, so it is a minimum.

Worked example: absolute extremes on a closed interval

Find the absolute maximum and minimum values of $g(x) = x^3 - 3x$ on $[0, 2]$.

Solution: On a closed interval, the absolute extremes occur either at a critical point inside the interval or at an endpoint. The only critical point in $(0, 2)$ is $x = 1$. Evaluating $g$ at it and at both endpoints:

So the absolute maximum is $2$ (at $x = 2$, an endpoint) and the absolute minimum is $-2$ (at $x = 1$). Note that the endpoint value beats the local maximum here.

Worked example: an optimization problem

Find two positive numbers whose sum is 15 and whose sum of squares is as small as possible.

Solution: Let the numbers be $x$ and $15-x$. Minimize:

$f'(x) = 4x - 30 = 0$ at $x = 7.5$, and $f''(x) = 4 > 0$, confirming a minimum. So the two numbers are $x = 7.5$ and $15 - x = 7.5$, split evenly.

Common mistakes

• Assuming $f'(c) = 0$ means an extreme. It only marks a candidate. $x^3$ has $f'(0) = 0$ and no maximum or minimum there.
• Ignoring the endpoints. On a closed interval the absolute maximum or minimum can sit at an endpoint, as in the example above.
• Giving the location instead of the value. If asked for the maximum value, the answer is $g(-1) = 2$, not $x = -1$.
• Forgetting the constraint in an optimisation problem. Use it to write the quantity in one variable first, as $15 - x$ did in the sum-of-squares example.