Calculus AB and Calculus BC

CHAPTER 4 Applications of Differential Calculus

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) = 2x³ − 3x² − 12x.

SOLUTION:

(a) f ′(x) = 6x² − 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

The function has neither a global max nor a global min on any interval that contains zero (see Figure N2–4). However, it does attain both a global max and a global min on every closed interval that does not contain zero. For instance, on [2,5] the global max of f is the global min