Section 4.2 The Mean Value Theorem

Transcription

1 Section 4.2 The Mean Value Theorem Ruipeng Shen October 2nd Ruipeng Shen MATH 1ZA3 October 2nd 1 / 11

2 Rolle s Theorem Theorem (Rolle s Theorem) Let f (x) be a function that satisfies: 1. f is continuous on the closed interval [a, b]. 2. f is differentiable on the open interval (a, b). 3. f (a) = f (b). Then there is a number c in (a, b) such that f (c) = 0. y f'(c 1 )=0 y y=f(x) f'(c 3 )=0 a b x 1 a f'(c 2 )=0 b x y= x -1 1 Ruipeng Shen MATH 1ZA3 October 2nd 2 / 11

3 Proof of Rolle s Theorem Let f (a) = f (b) = k. There are three cases: If f takes on a local maximum value at a number c in (a, b). Then Fermat s Theorem guarantees f (c) = 0. If f takes on a local minimum value at a number c in (a, b). Then Fermat s Theorem guarantees f (c) = 0. Otherwise there is not a local extreme value in (a, b). According to the Extreme Value Theorem, the function f must take on its absolute extreme values at the endpoints. Thus both absolute maximum and minimum value are the same value k. This implies that f (x) = k is a constant. We have f (x) = 0 for each x in (a, b). y y f'(c)=0 y=f(x) y=f(x) k a x b k a f'(c)=0 b x Ruipeng Shen MATH 1ZA3 October 2nd 3 / 11

4 Example Example Show that the only solution to the equation x = arctan x is x = 0. Solution Let f (x) = x arctan x. If there were another solution b to the original equation, we would have f (0) = f (b) = 0. By Rolle s Theorem, there would be a number c between 0 and b, such that f (c) = 0. But the derivative f (x) = d 1 (x arctan x) = 1 dx 1 + x 2 = x x 2 > 0 for any x 0. This is a contradiction. Ruipeng Shen MATH 1ZA3 October 2nd 4 / 11

5 The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be a function that satisfies 1. f is continuous on the closed interval [a, b]. 2. f is differentiable on the open interval (a, b). Then there is a number c in (a, b) such that f (c) = b a = the average rate of change of f or equivalently, = f (c)(b a). Remark 1 If f (x) = kx + q is a linear function, then we have f (x) = k = b a for any x (a, b). Thus the average rate of change is the only choice that guarantees the existence of c in the theorem. Ruipeng Shen MATH 1ZA3 October 2nd 5 / 11

6 Proof of the Mean Value Theorem Step 1 Let us define a linear function It satisfies L(a) = f (a) and L(b) = f (a) + L(x) = f (a) + b a b a (x a). (b a) = f (a) + [] = f (b). Step 2 Consider the function g(x) = f (x) L(x). The function g(x) is still continuous on the closed interval [a, b] and differentiable on the open interval (a, b). In addition, we have g(a) = g(b) = 0. According to Rolle s Theorem, there exists a number c in (a, b) such that g (c) = 0. By the fact g (c) = f (c) L (c), we have g (c) = 0 = f (c) = L (c) =. b a Ruipeng Shen MATH 1ZA3 October 2nd 6 / 11

7 Illustration of the Mean Value Theorem y f'(c)= f(b)-f(a) b-a y=f(x) (f,f(b)) (c,f(c)) g(x) = f(x)-l(x) a (a,f(a)) y=l(x) slope = f(b)-f(a) b-a b x Ruipeng Shen MATH 1ZA3 October 2nd 7 / 11

8 Example Example Prove the inequality x > sin x holds for any 0 < x < π/2. Proof Let f (x) = x sin x. This function is continuous and differentiable everywhere with f (x) = 1 cos x. Given any number b (0, π/2), we can apply the Mean Value Theorem on the function f in the interval [0, b] and conclude that there exists a number c in (0, b), such that f (b) f (0) = f (c)(b 0) = (1 cos c)b > 0 This implies f (b) > f (0) = 0 b > sin b. Therefore we obtain x > sin x for any x (0, π/2). Ruipeng Shen MATH 1ZA3 October 2nd 8 / 11

9 An Application of the Mean Value Theorem Theorem (a) If f (x) = 0 for all x in an interval (a, b), then f is constant on (a, b). (b) If f (x) = g (x) for all x in an interval (a, b), then f g is constant on (a, b); that is, f (x) = g(x) + C where C is a constant. Proof (a) Let x 1 < x 2 be two different numbers in the interval (a, b). According to the Mean Value Theorem, there exists a number c in (x 1, x 2 ), such that f (x 2 ) f (x 1 ) = f (c) (x 2 x 1 ) = 0. Therefore we have f (x 2 ) = f (x 1 ). This argument works for any two numbers x 1 < x 2 in the interval (a, b). Thus the function f is constant in the interval (a, b). (b) If we define h(x) = f (x) g(x), then h (x) = f (x) g (x) = 0 for all x in (a, b). By the part (a), we conclude that h, namely f g, is constant in (a, b). Ruipeng Shen MATH 1ZA3 October 2nd 9 / 11

10 Examples Example Find all functions f (x) that satisfies f (x) = cos x. Solution We have f (x) = cos x = d (sin x). By the theorem, we obtain dx f (x) = sin x + C. Here C is a constant. Example Let f (x) be a differentiable function that satisfies f (x) m for all x. Prove that the inequality f (x 2 ) f (x 1 ) m x 2 x 1 holds for any two numbers x 1 < x 2. Proof Given x 1 < x 2, by the Mean Value Theorem, there exists a number c in (x 1, x 2 ), such that f (x 2 ) f (x 1 ) = f (c)(x 2 x 1 ) Therefore we have f (x 2 ) f (x 1 ) = f (c) x 2 x 1 m x 2 x 1. Ruipeng Shen MATH 1ZA3 October 2nd 10 / 11

11 The Second Mean Value Theorem (Optional) Theorem Let f (x), g(x) be two functions that satisfy: 1. f, g are continuous on the closed interval [a, b]. 2. f, g are differentiable on the open interval (a, b). 3. g (x) 0 for x (a, b). Then there is a number c in (a, b) such that f (c) g = (c) g(b) g(a) Proof Consider the function h(x) = f (x) g(x). It is still continuous g(b) g(a) in the closed interval [a, b] and differentiable in the open interval (a, b). In addition, we have h(b) = h(a). By Rolle s Theorem, there exists a number c in (a, b), such that h (c) = 0 = f (c) g(b) g(a) g (c) = 0 = f (c) g = (c) g(b) g(a). Ruipeng Shen MATH 1ZA3 October 2nd 11 / 11