113 for suchk. It follows that f (xk)−f (x0)= f (n)(c k) n! (xk −x0) n ≥0. This implies f (n)(c k) ≥0 for suchk. Since {ck} converges to x0, f (n)(x 0)=limk →∞ f (n)(c k) ≥0. The proof of (ii) is similar. □ ■ Example 4.5.2 Consider the function f (x)=x 2cosxdefinedonR. Then f ′(x)=2xcosx−x2sinx and f ′′(x)=2cosx−4xsinx−x2cosx. Then f (0)=f ′(0)=0 and f ′′(0)=2>0. It follows from the previous theorem that f has a local minimum at 0. Notice, by the way, that since f (0)=0 and f (π) <0, 0 is not a global minimum. ■ Example 4.5.3 Consider the function f (x)=−x 6 +2x5 +x4 −4x3 +x2 +2x−3 defined on R. A direct calculations shows f ′(1)=f ′′(1)=f ′′′(1)=f (4)(1)=0 and f (5)(1) <0. It follows from the previous theorem that f has a local maximum at 1. Exercises 4.5.1 ▷Use Taylor’s theorem to prove that for all x >0 and m∈N, ex > m ∑k=0 xk k! . 4.5.2 Find the 5th Taylor polynomial, P5(x), at x0 =0 for cosx. Determine an upper bound for the error |P5(x)−cosx| for x ∈[−π/2,π/2]. 4.5.3 Use Theorem 4.5.2 to determine if the following functions have a local minimum or a local maximum at the indicated points. (a) f (x)=x3sinx at x 0 =0. (b) f (x)=(1−x)lnx at x0 =1. 4.5.4 Suppose f is twice differentiable on(a,b). Show that for everyx ∈(a,b), lim h→0 f (x+h)+f (x−h)−2f (x) h2 =f ′′(x). 4.5.5 ▶(a) Suppose f is three times differentiable on(a,b) andx0 ∈(a,b). Prove that lim h→0 f (x0 +h)−f (x0)−f ′(x0) h 1! −f ′′(x0) h2 2! h3 = f ′′′(x0) 3! . (b) State and prove a more general result for the case where f is ntimes differentiable on(a,b). 4.5.6 Suppose f is n times differentiable on(a,b) andx0 ∈(a,b). Define Pn(h)= n ∑k=0 f (n)(x 0) hn n! for h∈R.
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