5. Maximum & Minimum Values

Video coming soon2 worked examples

Theory

On a closed interval $[a, b]$ , the greatest and least values of a continuous function occur either at a stationary point inside the interval or at an endpoint. The method is: find the stationary points in $(a, b)$ , then evaluate the function at those points and at both endpoints, and compare.

The Golden Rule: on a closed interval, always evaluate the endpoints as well as the interior stationary points — the extreme value is often at an end.

⚠️ Common Examiner Traps

Forgetting endpoints: checking only stationary points can miss the true maximum or minimum.
Out-of-range stationary points: discard any that fall outside the interval.
Compare values, not $x$ : the answer is the largest/smallest function value.

Worked examples

Example 1

Find the maximum and minimum values of $f(x) = x^3 - 12x$ on the closed interval $[-3, 3]$ .

Step 1: Stationary points: $f'(x) = 3x^2 - 12 = 0 \implies x = \pm 2$ . Both lie inside $[-3, 3]$ , so both must be tested.

Step 2: Evaluate at the two interior points and both endpoints:

f(-3) = 9, \qquad f(-2) = 16, \qquad f(2) = -16, \qquad f(3) = -9

Step 3: The maximum value is $16$ (at $x = -2$ ) and the minimum value is $-16$ (at $x = 2$ ). Note that here both extremes occur at interior stationary points rather than at the endpoints.

Example 2

Find the maximum and minimum values of $f(x) = 2x^3 - 3x^2$ on $[-1, 2]$ .

Step 1: Stationary points: $f'(x) = 6x^2 - 6x = 6x(x-1) = 0 \implies x = 0, 1$ . Both lie in $[-1, 2]$ .

Step 2: Evaluate at the stationary points and endpoints:

f(-1) = -5, \quad f(0) = 0, \quad f(1) = -1, \quad f(2) = 4

Step 3: The maximum value is $4$ (at $x = 2$ ) and the minimum value is $-5$ (at $x = -1$ ).