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Let 8 < 0 if x 0 x + 2 if 0 < x < 1 g(x) = : 1 if x 1. g is defined for all x, is not continuous, however g(x) < 3 for all x, but there is no number d for which g(d) = 3. Consequently, g has no global maximum value. On the other hand, g(x) g(0) = 0 for all x, so g(0) = 0 is the global minimum value. Theorem 4.1.2. Suppose that c is an interior point of an interval I, and f (c) is an extreme value of f on I. If f 0 (c) exists, then f 0 (c) = 0. Definition 4.1.2. A number c is called critical number of a function f if f 0 (c) = 0 or if f is not di↵erentiable at c. Remark 4.1.2. To find the extreme values of a function f on an interval [a, b] we proceed as follows: (a) Find all critical numbers of f in the open interval (a, b). These are the numbers in (a, b) where either f is not di↵erentiable, or all numbers c in (a, b) where f is di↵erentiable and such that f 0 (c) = 0. (b) Compute the value of f at all critical numbers and at the end points a and b. (c) The largest of the numbers computed in (b) is the maximum value of f on [a, b] and the smallest of these numbers is the minimum value of f on [a, b]. Warning 4.1.2. Let f (x) = x3 for 1 x 1. f 0 (x) = 3x2 and f 0 (0) = 0 so 0 is a critical number, but f (0) = 0 is not an extreme value of f on [ 1, 1]. For g(x) = x2 , for 0 x 1. g 0 (x) = 2x and g 0 (0) = 0. However, 0 is not a critical point, since it is not an interior point of [0, 1]. Nonetheless, g(0) = 0 g(x) for all 0 x 1. So g(0) = 0 is the minimum value of g on [0, 1]. Let h(x) = x3/4 for x in [0, 1]. h0 (x) = 34 x 1/4 for x 6= 0, and h0 (0) does not exist. But h(0) = 0 is the minimum value for h. 4.2 Mean Value Theorem Theorem 4.2.1 (Rolle’s Theorem). Let f be a continuous on [a, b] and di↵erentiable on (a, b). If f (a) = f (b), then there is a number c in (a, b) such that f 0 (c) = 0. Theorem 4.2.2 (The Mean Value Theorem (MVT)). Let f be continuous on [a, b] and di↵erentiable on (a, b). Then there exists a number c in (a, b) such that f 0 (c) = f (b) f (a) . b a Application 4.2.1. (1) Use Rolle’s theorem to prove that the equation cot x = x has a solution in (0, ⇡/2). 17 (2) Suppose f is di↵erentiable at every number in [ 1, 1] and f 0 (1) = f 0 ( 1) = 1 and assume that f ( 1) = f (1) = 0. Explain why f has at least one zero in ( 1, 1). Explain why f 0 must have at least two zeros in ( 1, 1). 4.3 Consequence of the Mean Value Theorem Definition 4.3.1. If f is a function defined on an interval I, then any di↵erentiable function F on I such that F 0 (x) = f (x) for every interior point x of I is called an antiderivative of f . Theorem 4.3.1. (a) Let f be continuous on an interval I. If f 0 (x) exists and equals 0 for all interior points x, then f is constant on I. (b) Let f and g be continuous on an interval I. If f 0 (x) and g 0 (x) exist and are equal for each interior point x of I, then f g is constant on I. In other words, there is a number C such that f (x) = g(x) + C for all x in I. Warning 4.3.1. Let f and g be defined on I = [ 1, 0] [ [2, 3] by ⇢ 2 if 1x0 f (x) = x + 2 if 2 x < 3, g(x) = ⇢ 3 if 1x0 x + 4 if 2 x < 3. Then f 0 (x) = g 0 (x) for each interior point x of I, but there is no C such that f (x) = g(x) + C for all x in I. The reason for this is that I is not an interval! Definition 4.3.2. A function f is increasing on an interval I if f (x) < f (y) for all x, y in I with x < y. A function f is decreasing on an interval I if f (x) > f (y) for all x, y in I with x < y. Theorem 4.3.2. Let f be a continuous function on an interval I and di↵erentiable at each interior point of I. (a) If f 0 (x) > 0 at each interior point of I, then f is increasing on I. Moreover, f is increasing on I if f 0 (x) > 0 except for a finite number of numbers x in I. (b) If f 0 (x) < 0 at each interior point of I, then f is decreasing on I. Moreover, f is decreasing on I if f 0 (x) < 0 except for a finite number of numbers x in I. 18