Global extrema on a closed interval

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = x^2$ on $[-1, 3]$ . Its only critical point is $x = 0$ . Is $f(0) = 0$ the largest value on the interval?

No, $f(3) = 9$ is larger
Yes

Key idea

On a closed interval, compare a short list of candidates.

Extreme Value Theorem: a function continuous on

[a, b]

has a maximum and a minimum value there. Closed interval method: evaluate

f

at the critical numbers in

(a, b)

and at

a

and

b

; the largest value is the maximum, the smallest the minimum.

Worked example

Extremes of $f(x) = x^3 - 3x$ on $[0, 3]$ .

Critical number

1

f'(x) = 3x² − 3 = 0 at x = 1 (−1 is outside).

f(0) = 0

,

f(1) = -2

,

f(3) = 18

Evaluate the candidates.

Minimum

-2

(at

x = 1

), maximum

18

(at

x = 3

)

Compare.

Your turn, guided

Your turn: $f(x) = x^2 - 4x$ on $[0, 5]$ . Candidates: $f(0) = 0$ , $f(2) = -4$ , $f(5) = 5$ .

Maximum value

5

The largest candidate.

Practice ★

Find the maximum value of $f(x) = x^2 - 6x$ on $[0, 8]$ .
Find the maximum value of $f(x) = x^2 - 2x$ on $[0, 7]$ .

Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 2]$ .
Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 5]$ .

Watch out!

Practice ★★

Which functions are guaranteed by the Extreme Value Theorem to have a maximum value on the interval? Tick all that are.
1. $x^2$ on $[-7, 7]$
2. $\sin x$ on $[0, 7]$
3. $x$ on $(0, 7)$
4. $\dfrac{1}{x}$ on $[-1, 1]$

Exit ticket

Three questions to check you have got it.

Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 2]$ .
Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 5]$ .
Find the minimum value of $f(x) = x^3 - 27x$ on $[0, 6]$ .

Summary card

Closed interval method

Critical numbers inside $(a, b)$ .
Evaluate $f$ there and at $a$ and $b$ .
Largest = maximum, smallest = minimum.

Watch out: the EVT needs continuity on a closed interval. Give the value, not just where it occurs, when asked.

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