Maxima and Minima

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.

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.