Maximum and Minimum Values

Video lesson · from 1:22:441 worked example

One lesson video covers all of Differentiation, so it opens at 1:22:44 for this topic — not from the beginning.

In a closed interval, the maximum and minimum values of a function are either at a stationary point or at an end point of the interval.

Check both ends: this is the classic way to lose marks here — most candidates who drop marks never consider both ends of the closed interval. You must evaluate the function at both endpoints as well as at every stationary point inside the interval.
Compare values, not natures: the greatest value is simply the largest of the numbers you have worked out. A maximum turning point inside the interval is not automatically the greatest value — an endpoint can beat it.
Discard stationary points outside the interval: if solving $\frac{dy}{dx}=0$ gives an $x$ outside the stated range, it plays no part.
Answer with the value: the question asks for the greatest and least values of the function, so give the $y$ values.

Example 1

Find the maximum and minimum values of $y = 4x^3 - x^2 - 4x + 1$ in the closed interval $-2 \leq x \leq 2$ .

Step 1: Check stationary points.

\begin{aligned} \frac{dy}{dx} &= 12x^2 - 2x - 4 \\ 12x^2 - 2x - 4 &= 0 \\ 6x^2 - x - 2 &= 0 \\ (3x - 2)(2x + 1) &= 0 \end{aligned}

Roots: $x = \frac{2}{3}$ and $x = -\frac{1}{2}$ .

Check their $y$ -values:

$y\left(\frac{2}{3}\right) = 4\left(\frac{8}{27}\right) - \frac{4}{9} - 4\left(\frac{2}{3}\right) + 1 = -\frac{23}{27}$
$y\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) - \frac{1}{4} - 4\left(-\frac{1}{2}\right) + 1 = \frac{9}{4}$

Step 2: Check end points ( $-2$ and $2$ ).

Compare all values: $-\frac{23}{27}$ , $2.25$ , $-27$ , $21$ .

Maximum value is 21 (at $x=2$ ).

Minimum value is -27 (at $x=-2$ ).