Global maximum and minimum

The difference between local and global maxima and minima, and finding the global maximum and minimum of a function on a closed interval by checking stationary points and endpoints.

Local and global

The global maximum is the largest value the function takes anywhere in its domain. A local maximum is only the largest compared with nearby points. On a restricted domain, the global maximum may be at an endpoint, where the curve is not stationary at all.

Global max and min on L ≤ x ≤ U

The blue part is the curve on the interval. Compare the endpoint values with the stationary values.

Try this

Set $U = 1.5$. Is the global maximum now at a stationary point or at an endpoint?

Example 1: a quadratic●

Find the global maximum and minimum of $f(x) = x^2 - 4x + 1$ for $-1 \le x \le 5$.

$f'(x) = 2x - 4 = 0$ at $x = 2$, and $f(2) = -3$.

Endpoints: $f(-1) = 6$ and $f(5) = 6$.

Global maximum $6$ (at both endpoints), global minimum $-3$.

Example 2: a cubic●●

Find the global maximum and minimum of $f(x) = x^3 - 3x$ for $-2 \le x \le 3$.

Stationary points at $x = \pm 1$: $\;f(-1) = 2$ and $f(1) = -2$.

Endpoints: $f(-2) = -2$ and $f(3) = 18$.

Global maximum $18$ at $x = 3$. Global minimum $-2$, reached at both $x = -2$ and $x = 1$.

Flashcards

Global maximum on $a le x le b$

Largest of $f(a)$, $f(b)$ and the stationary values.

Local vs global

Local: largest nearby. Global: largest anywhere in the domain.

Practice

Find the global maximum of $f(x) = x^2 - 4x + 1$ for $-1 \le x \le 5$.
Find its global minimum on the same interval.
Find the global maximum of $f(x) = x^3 - 3x$ for $-2 \le x \le 3$.
Find its global minimum on the same interval.
Find the global maximum of $f(x) = x^3$ for $-1 \le x \le 2$.
Find the global maximum of $f(x) = 6x - x^2$ for $0 \le x \le 4$.
Find the global maximum of $f(x) = x^3 - 3x$ for $0 \le x \le 2$.
On the whole real line, $y = x^2$ has:
On a closed interval, the global maximum can be: