Absolute maxima and minima - the table method

To find the absolute maxima and minima of a continuous function on a closed interval $[a,b]$, Weierstrass’s theorem guarantees that they exist. The practical method consists of building a table of candidates.

In brief — Table method for absolute max/min

1. Compute $f'(x)$ and find all the stationary points ($f'(x)=0$) interior to $(a,b)$.
2. Compute $f$ at the stationary points and at the endpoints $a, b$.
3. Put everything into a table and compare: the largest value is the absolute maximum, the smallest the absolute minimum.

The idea is that an absolute extremum on $[a,b]$ can fall only at an interior stationary point or at an endpoint of the interval: it is therefore enough to compute $f$ at these few candidate points and compare their values.

$x$	$y=f(x)$	Type
$a$	$f(a)$	Left endpoint
$b$	$f(b)$	Right endpoint
$x_1$	$f(x_1)$	Stationary point
$x_2$	$f(x_2)$	Stationary point

$\to$ absolute max $=$ largest $y$ value    $\to$ absolute min $=$ smallest $y$ value

Example — Absolute max/min on an interval

Let us look for the absolute maximum and minimum of $f(x) = x^3-3x$ on $[0,7]$. The derivative is $f'(x) = 3x^2-3 = 0 \iff x=\pm 1$; only $x=1\in(0,7)$ is interior to the interval (the other is outside).

$x$	$f(x)$	Type
$0$	$0$	Endpoint
$7$	$343-21=322$	Endpoint
$1$	$1-3=-2$	$f'=0$

Comparing the values: absolute maximum $\boxed{322}$ at $x=7$, absolute minimum $\boxed{-2}$ at $x=1$.

Links

Topics: Function study
Concepts: Absolute maxima and minima · Stationary point · Weierstrass’s theorem
Skills: Studying a function