169 Exercise 4.2.6. 1. Apply Rolle’s theorem to the function 2 xn x f (x)= a1x + a2 + ··· + an 2 n on the interval [0,1]. 2. Apply Rolle’s theorem to the function n sin(2k + 1)x f (x)= ∑ 2k + 1 k=0 on the interval [0,π/2]. Exercise 4.2.8. (a) Given ε > 0, frst fnd x0 large enough so that a − ε/2 < f ′ (x) < a + ε/2 for x > x0. Then use the identity f (x)− f (x0) f (x0) f (x) f (x) − f (x0)+ f (x0) + x −x0 x−x0 = = , x 0 x x − x 0 + x0 1 + x −x0 and the mean value theorem to show that, for x large, f (x) a − ε < < a+ ε. x (b) Use the method in part (a). (c) Consider f (x)= sin(x). SECTION 4.3 Exercise 4.3.2. (a) We can prove that f is uniformly continuous on R by defnition. Given any ε > 0, 1 ε α choose δ = and get ℓ + 1 ε | f (u) − f (v)|≤ ℓ|u− v| α <ℓδ α = ℓ < ε ℓ + 1 whenever |u− v| < δ . Note that we use ℓ + 1 here instead of ℓ to avoid the case where ℓ = 0. (b) We will prove that f is a constant function by showing that it is differentiable on R and f ′ (a)= 0 for all a ∈ R. Fix any a ∈ R. Then, for x ̸ = a, f (x) − f (a) ℓ|x − a|α = ℓ|x − a| α−1 ≤ . x− a |x − a| Since α > 1, by the squeeze theorem, f (x) − f (a) lim = 0. x→a x − a
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