Maxima and Minima

Finding the largest or smallest value of a function is a main use of the derivative. The method is short: locate the critical points, classify them with a derivative test, and, on a closed interval, compare with the endpoints. The tasks are to find the value or location of a maximum, the number of local extrema, whether a point is a maximum, minimum or neither, the inflection points, or the optimum of a small word problem.

$f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in some open interval around $c$, and a global (absolute) maximum on a set $D$ if $f(c) \ge f(x)$ for all $x \in D$. Minima are defined with $\le$. A global extremum at an interior point is also a local one, but a global extremum on $[a, b]$ can sit at an endpoint, where the local definition (which needs an open interval) does not apply.

Fermat's theorem. If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c) = 0$.

A critical point is an interior point of the domain where $f'(c) = 0$ or $f'(c)$ does not exist. Every interior local extremum is a critical point, but not every critical point is an extremum: $x^3$ has $f'(0) = 0$ and no extremum at 0.

Let $c$ be a critical point and $f$ continuous at $c$.

This test also works where $f'(c)$ does not exist ($|x|$ at 0, $x^{2/3}$ at 0). Relatedly, $f' > 0$ on an interval makes $f$ strictly increasing there, and $f' < 0$ strictly decreasing (a consequence of the mean value theorem).

If $f'(c) = 0$ and $f''(c)$ exists:

Higher-order test. If $f'(c) = \cdots = f^{(n-1)}(c) = 0$ and $f^{(n)}(c) \ne 0$: for even $n$, a minimum when $f^{(n)}(c) > 0$ and a maximum when it is negative; for odd $n$, no extremum.

Sign convention. At a stationary point, $f''(c) < 0$ gives a maximum (the graph bends down); picture $-x^2$.

$f$ is convex (concave up) on an interval if $f'' \ge 0$ there ($f'' > 0$ is sufficient) and concave (concave down) if $f'' \le 0$ ($f'' < 0$ is sufficient). An inflection point is a point of the graph where the concavity changes. If $f''$ exists there it must be 0, but $f''(c) = 0$ is not enough: $x^4$ has $f''(0) = 0$ and is convex on both sides. Check that $f''$ changes sign.

For $f$ continuous on $[a, b]$ (so the extreme value theorem guarantees both extremes):

No classification is needed. On an open or unbounded interval, also examine the limits at the ends, since the extremes may not be attained.

For a $C^2$ function on a neighbourhood of an interior stationary point, first solve $f_x=f_y=0$. Then evaluate $D=f_{xx}f_{yy}-f_{xy}^2$ at that point. Boundary and nondifferentiable points require separate treatment:

$f(x) = x^3 - 3x$, $f'(x) = 3(x - 1)(x + 1)$, $f''(x) = 6x$. Critical points $\pm 1$: $f''(-1) = -6 < 0$ gives a local maximum $f(-1) = 2$; $f''(1) = 6 > 0$ gives a local minimum $f(1) = -2$. $f''$ changes sign at 0, an inflection point. On $[-1.5, 2.5]$: $f(-1.5) = 1.125$ and $f(2.5) = 8.125$, so the global maximum is 8.125 (at the endpoint 2.5) and the global minimum is $-2$ (at 1).

$f(x) = x^4 - 4x^3$, $f'(x) = 4x^2(x - 3)$, $f''(x) = 12x(x - 2)$. At $x = 3$: $f''(3) = 36 > 0$, a local (in fact global) minimum $f(3) = -27$. At $x = 0$: $f''(0) = 0$, and $f' < 0$ on both sides of 0, so there is no extremum. Inflection points where $f''$ changes sign: $x = 0$ and $x = 2$.

$f(x) = x^5 - 5x$: $f'(x) = 5(x^4 - 1) = 0$ at $x = \pm 1$; $f''(x) = 20x^3$. So $x = -1$ is a local maximum ($f = 4$) and $x = 1$ a local minimum ($f = -4$): two local extrema, and no global ones on $\mathbb{R}$ because $f \to \pm\infty$.

Squares of side $x$ are cut from the corners of a $12 \times 12$ sheet and the sides folded up. Volume $V(x) = x(12 - 2x)^2$ for $0 < x < 6$. $V'(x) = (12 - 2x)(12 - 6x) = 0$ gives $x = 2$ in the domain. $V''(2) = -48 < 0$, so the maximum volume is $V(2) = 2 \cdot 8^2 = 128$.

$f(x) = x^{2/3}$ has $f'(x) = \frac{2}{3}x^{-1/3}$, undefined at 0, negative for $x < 0$ and positive for $x > 0$. By the first derivative test, 0 is a minimum (a cusp). The second derivative test cannot be applied.

$f(x) = xe^{-x}$ on $[0, \infty)$: $f'(x) = (1 - x)e^{-x}$, positive before 1 and negative after. $f(0) = 0$ and $f(x) \to 0$ as $x \to \infty$, so the global maximum is $f(1) = 1/e \approx 0.368$.

$f(x) = \sin x + \cos x$: $f'(x) = \cos x - \sin x = 0$ at $x = \pi/4$ (in $[0, 2\pi)$ also $5\pi/4$). $f(\pi/4) = \sqrt{2}$ and $f(5\pi/4) = -\sqrt{2}$. In general $a\sin x + b\cos x$ ranges over $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.

$f(x, y) = x^3 - 3x + y^2$: $f_x = 3x^2 - 3$, $f_y = 2y$, so the critical points are $(1, 0)$ and $(-1, 0)$. $f_{xx} = 6x$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 12x$. At $(1, 0)$: $D = 12 > 0$ and $f_{xx} = 6 > 0$, a local minimum with value $-2$. At $(-1, 0)$: $D = -12 < 0$, a saddle point.