Chapter 6 Differentiation

Section 6.2 The Mean Value Theorem

We begin with a preliminary result that is of interest in its own right. Theorem 6.2.1 If f is differentiable on an open interval (a, b) and if f assumes its maximum or minimum at a point c  (a, b), then f (c) = 0. Proof: Suppose that f assumes its maximum at c. That is, f (x)  f (c) for all x  (a, b). Let (xn) be a sequence converging to c such that a < xn < c for all n. Then, since f is differentiable at c, Theorem 6.1.3 implies that the sequence converges to f (c). But each term in this sequence of quotients is non­negative, since f (xn)  f (c) and xn < c. Thus f (c)  0 by Corollary 4.2.5. Similarly, let ( yn) be a sequence converging to c such that c < yn < b for all n. Then the terms of the sequence are all nonpositive, since f ( yn)  f (c) and yn > c. f But these also converge to f (c), so f (c)  0.       We conclude that f (c) = 0.  a b xn  c  yn

(Rolle’s Theorem) (A special case of the Mean Value Theorem) Let f be a continuous function on [a, b] that is differentiable on (a, b) and such that f (a) = f (b). Then there exists at least one point c in (a, b) such that f (c) = 0. Theorem 6.2.2 Proof: Since f is continuous and [a, b] is compact, Corollary 5.3.3 implies that there exist points x1 and x2 in [a, b] such that f (x1)  f (x)  f (x2) for all x  [a, b]. If x1 and x2 are both endpoints of [a, b], then f (x) = f (a) = f (b) for all x  [a, b]. In this case f is a constant function and f  (x) = 0 for all x  (a, b). Otherwise, f assumes either a maximum or a minimum at some point c  (a, b). But then, by Theorem 6.2.1, f (c) = 0.  f (x) x f (c) = 0 f (a) = f (b) a c b

(Mean Value Theorem) Let f be a continuous function on [a, b] that is differentiable on (a, b). Then there exists at least one point c  (a, b) such that . Theorem 6.2.3 Proof: This ratio represents the slope of the chord through the endpoints of the graph, and the equation of this chord is . Now let h represent the difference between the graph of f and the chord. Then the function h = f – g is continuous on [a, b] and differentiable on (a, b). y x Since h(a) = h(b) = 0, Rolle’s Theorem implies there exists a point c in (a, b) such that h(c) = 0. y = f (x) But then 0 = h(c) = f (c) – g(c) h(x) y = g(x) a x c b

M-V Theorem: f (b) – f (a) = f (c)(b – a) with c  (a, b). Example 6.2.4* We may use the mean value theorem to approximate the value of a function near a point. For example, we can estimate by using the fact that it is close to . Apply the mean value theorem to the function f (x) = on the interval [32, 36] to obtain a point c  (32, 36) such that or Now, 32 < c < 36, so and That is, and 5.600 5.657 5.667 The next several results show how the mean value theorem can be used to relate the properties of a function f and its derivative f .

M-V Theorem: f (b) – f (a) = f (c)(b – a) with c  (a, b). Let f be continuous on [a, b] and differentiable on (a, b). If f (x) = 0 for all x  (a, b), then f is constant on [a, b]. Proof: We prove the contrapositive. That is, suppose that f were not constant on [a, b]. Then there would exist x1 and x2 such that a  x1 < x2  b and f (x1)  f (x2). But then, by the mean value theorem, for some c  (x1, x2) we would have f (c) = [ f (x2) – f (x1)]/(x2 – x1)  0, a contradiction.  Corollary 6.2.7 Let f and g be continuous on [a, b] and differentiable on (a, b). Suppose that f (x) = g (x) for all x  (a, b). Then there exists a constant C such that f = g + C on [a, b]. Proof: Apply Theorem 6.2.6 to the function f – g.  We know that continuous functions satisfy the intermediate value property. The next theorem says that derivatives also have this property, even though they are not necessarily continuous.

(Intermediate Value Theorem for Derivatives) Theorem 6.2.9 Let f be differentiable on [a, b] and suppose that k is a number between f (a) and f (b). Then there exists a point c  (a, b) such that f (c) = k. Theorem 6.2.9 Proof: We assume that f (a) < k < f (b), the proof of the other case being similar. For x  [a, b], let g (x) = f (x) – k x. Then g is differentiable on [a, b] and g(a) = f (a) – k < 0 and g(b) = f (b) – k > 0. Since g is continuous on the compact set [a, b], Corollary 5.3.3 implies that g assumes its minimum at some point c  [a, b]. We claim that c  (a, b). Since , the ratio is positive for all x  U  [a, b], where U is some deleted neighborhood of b. Thus, for x < b with x  U, we must have g (x) < g (b). Hence g (b) is not the minimum of g on [a, b], so c  b. Since g (a) < 0, a similar argument shows that c  a. We conclude that c  (a, b), so Theorem 6.2.1 implies that g (c) = 0. But then f (c) = g (c) + k = k. 

(Inverse Function Theorem) Suppose that f is differentiable on an interval I and f (x)  0 for all x  I. Then f is injective, f – 1 is differentiable on f (I ), and where y = f (x). 2 5 y x For example, let f (x) = x2 + 1 for x  [0, ) and g(x) = for x  [1, ), so that g = f – 1. y = f (x) Note that f (2) = 5 and f (x) = 2x, so that f (2) = 4.  f (2) = 4 We also have so that  We see that y = g(x)