168 Solutions and Hints for Selected Exercises Figure 5.10: The function f (x) =sin(x). which implies f (b)−f (a) b−a =|cos(c)| ≤1. It follows that | f (a)−f (b)| ≤ |a−b|. The solution is similar for the case where a>b. It is essential to realize that the most important property required in solving this problem is the boundedness of the derivative of the function. Thus, it is possible to solve the following problems with a similar strategy. 1. Prove that |cos(a)−cos(b)| ≤ |a−b| for all a,b∈R. 2. Prove that |ln(1+e2a)−ln(1+e2b)| ≤2|a−b| for all a,b∈R. Exercise 4.2.4. Let us define f : [−π,π] →Rby f (x) =x+ n ∑ k=1 (ak sinkx+bk coskx). We want to find c ∈(−π,π) such that f (c) =0. Now, consider the function g(x) = x2 2 + n ∑ k=1 − ak cos(kx) k +bk sin(kx) k . Observe that g(−π) =g(π) andg′ =f . The conclusion follows from Rolle’s Theorem. Exercise 4.2.5. Use the identity 1 g(b)−g(a) f (a) f (b) g(a) g(b) = f (a)g(b)−f (b)g(a) g(b)−g(a) = f (a) g(a) − f (b) g(b) 1 g(a) − 1 g(b) . Then apply the Cauchy mean value theorem for two functions φ(x) = f (x) g(x) andψ(x) = 1 g(x) on the interval [a,b].
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