Chapter Differentiation Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide Section 6.2 The Mean Value Theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide 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) = Proof: That is, f (x) ≤ f (c) for all x ∈ (a, b) Suppose that f assumes its maximum at c Let (x ) be a sequence converging to c such that a < x < c for all n n n Then, since f is differentiable at c, Theorem 6.1.3 implies that the sequence  f ( xn ) − f (c)   ÷ x − c n   But each term in this sequence of quotients converges to f ′(c) is nonnegative, since f (x ) ≤ f (c) and x < c n n Thus f ′(c) ≥ by Corollary 4.2.5 Similarly, let ( y ) be a sequence converging to c such that c < y < b for all n n n    f ( yn ) − f ( c )   ÷ yn − c   Then the terms of the sequence f are all nonpositive, since f ( y ) ≤ f (c) and y > c n n   But these also converge to f ′(c), so f ′(c) ≤ • a xn → • • • c • ← yn • b Copyright © 2013, 2005, 2001 Pearson Education, Inc We conclude that f ′(c) = ♦ Section 6.2, Slide Theorem 6.2.2 (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) = Proof: Since f is continuous and [a, b] is compact, Corollary 5.3.3 implies that there exist points x and x in [a, b] such that f (x ) ≤ f (x) ≤ f (x ) for all x ∈ [a, b] 2 If x and x 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) = 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) = ♦ f (x) f ′(c) = f (a) = f (b) a Copyright © 2013, 2005, 2001 Pearson Education, Inc c b x Section 6.2, Slide Theorem 6.2.3 (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 f ′(c) = f (b) − f (a ) b−a Proof: This ratio represents the slope of the chord through the endpoints of the graph, and the equation of this chord isy = g ( x ) = f (b) − f (a ) ( x − a ) + f ( a ) b−a 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 Since h(a) = h(b) = 0, Rolle’s Theorem implies y = f (x) there exists a point c in (a, b) such that h′(c) = h(x) But then = h′(c) = f ′(c) – g′(c) = f ′(c) − y = g(x) a x c b Copyright © 2013, 2005, 2001 Pearson Education, Inc f (b) − f (a ) ♦ b−a x Section 6.2, Slide 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 32 For example, we can estimate by using the fact that it is close to Apply the mean value theorem to the function f (x) = 36 = x on the interval [32, 36] to obtain a point c ∈ (32, 36) such that 36 − 32 = Now, 32 < c < 36, so That is, 1 c − (36 − 32) < 32 < c < 6, − 32 = or c < < , c and < < c < − 32 < 25 , − 52 < 32 − < − 13 , and 5.600 53 < 32 < 23 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 ′ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide M-V Theorem: f (b) – f (a) = f ′(c)(b – a) with c ∈ (a, b) Theorem 6.2.6 Let f be continuous on [a, b] and differentiable on (a, b) If f ′(x) = 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 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide Theorem 6.2.9 (Intermediate Value Theorem for Derivatives) 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 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 < and g′(b) = f ′(b) – k > 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) g ( x) − g (b) , the ratio is positive for all x ∈ U ∩ [a, b], x−b g ( x) − g (b) >0 Since x→b x−b g ′(b) = lim   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) =   But then f ′(c) = g ′(c) + k = k ♦   Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide Theorem 6.2.10 (Inverse Function Theorem) Suppose that f is differentiable on an interval I and f ′(x) ≠ for all x ∈ I Then f is injective, f –1 is differentiable on f (I ), and ( f −1)′( y ) =   f ′( x) where y = f (x) y For example, let f (x) = x + for x ∈ [0, ∞) x −1 for x ∈ [1, ∞), so that g = f and g(x) = Note that f (2) = and f ′(x) = 2x, so that –1 y = f (x) • f ′(2) = f ′(2) = We also have so that g ′( x) = g ′(5) = 14 We see that (f ) −1 ′ 1 ( x − 1)− 2 (5) = f ′(2) • g ′(5) = y = g(x) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.2, Slide x ... 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 32 For example, we can estimate by using the fact that it is close to Apply... − 32 = or c < < , c and < < c < − 32 < 25 , − 52 < 32 − < − 13 , and 5.600 53 < 32 < 23 5.657 5.667 The next several results show how the mean value theorem can be used to relate the properties... sequence of quotients converges to f ′(c) is nonnegative, since f (x ) ≤ f (c) and x < c n n Thus f ′(c) ≥ by Corollary 4.2.5 Similarly, let ( y ) be a sequence converging to c such that c < y < b for