Absolute maxima and minima on an interval

Problem

Find the absolute maximum and minimum of $f(x) = x^3-3x \quad \text{on } [0,7].$

Solution

Stationary points. We compute the derivative and set it to zero: $f'(x) = 3x^2-3 = 0 \iff x=\pm 1.$ Only $x=1\in(0,7)$ is interior to the interval (the other, $x=-1$, is outside).

Table of candidates. We compare the value of $f$ at the edges and at the stationary point:

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

Comparison. The largest value is $322$, the smallest is $-2$: $\text{absolute max} = \boxed{322} \text{ at } x=7, \qquad \text{absolute min} = \boxed{-2} \text{ at } x=1.$

Links

Topics: Curve sketching
Concepts: First derivative · Absolute maxima and minima · Stationary point
Skills: Differentiating · Sketching a function
Exercise type: Curve sketching