FIGURE N4–7
Verify the graph on your calculator.F. GLOBAL MAXIMUM OR MINIMUM
CASE I. DIFFERENTIABLE FUNCTIONS.
If a function f is differentiable on a closed interval a ≤ x ≤ b, then f is also continuous on the closed
interval [a, b] and we know from the Extreme Value Theorem that f attains both a (global) maximum
and a (global) minimum on [a, b]. To find these, we solve the equation f ′(x) = 0 for critical points on
the interval [a, b], then evaluate f at each of those and also at x = a and x = b. The largest value of f
obtained is the global max, and the smallest the global min.
EXAMPLE 16
Find the global max and global min of f on (a) −2 ≤ x ≤ 3, and (b) 0 ≤ x ≤ 3, if f (x) = 2x^3 − 3x^2 −
12 x.
SOLUTION:
(a) f ′(x) = 6x^2 − 6x − 12 = 6(x + 1)(x − 2), which equals zero if x = −1 or 2. Since f (−2) = −4, f
(−1) = 7, f (2) = −20, and f (3) = −9, the global max of f occurs at x = −1 and equals 7, and the
global min of f occurs at x = 2 and equals −20.
(b) Only the critical value 2 lies in [0,3]. We now evaluate f at 0, 2, and 3. Since f (0) = 0, f (2)
= −20, and f (3) = −9, the global max of f equals 0 and the global min equals −20.CASE II. FUNCTIONS THAT ARE NOT EVERYWHERE DIFFERENTIABLE.
We proceed as for Case I but now evaluate f also at each point in a given interval for which f is
defined but for which f ′ does not exist.
EXAMPLE 17
The absolute-value function f (x) = |x| is defined for all real x, but f ′(x) does not exist at x = 0.
Since f ′(x) = −1 if x < 0, but f ′(x) = 1 if x > 0, we see that f has a global min at x = 0.EXAMPLE 18