171 It follows that f is differentiable on Rwith f ′(x) = 2 x3 e− 1 x2 , if x̸ =0; 0, if x =0. In a similar way, we can show that f is twice differentiable onRwith f ′′(x) = 6 x4 + 2 x6 e− 1 x2 , if x̸ =0; 0, if x =0. Based on these calculations, we predict that f is ntimes differentiable for every n∈Nwith f (n)(x) = P 1 x e− 1 x2 , if x̸ =0; 0, if x =0, where Pis a polynomial. Now we proceed to prove this conclusion by induction. The conclusion is true for n=1 as shown above. Given that the conclusion is true for some n∈N, for x̸ =0 we have f (n+1)(x) =−x−2P′ 1 x + 2 x3 P 1 x e− 1 x2 =Q 1 x e− 1 x2 , where Qis also a polynomial. It is an easy exercise to write the explicit formula of Qbased onP. Moreover, successive applications of l’Hôpital’s rule give lim x→0+ f (n)(x)−f (n)(0) x−0 = lim x→0+ 1 x P 1 x e− 1 x2 =lim t→∞ tP(t) et2 =0. In a similar way, we can show that lim x→0− f (n)(x)−f (n)(0) x−0 =0. Therefore, f (n+1)(0) =0. We have proved that for everyn∈N, f is ntimes differentiable and, so, f ∈Cn(R). Here we do not need to prove the continuity of f (n) because the differentiability of f (n) implies its continuity. In a similar way, we can also show that the function f (x) =( e−1 x , if x >0; 0, if x ≤0 is ntimes differentiable for every n∈N. SECTION 4.5 Exercise 4.5.1. Let f (x) =ex. By Taylor’s theorem, for any x >0, there exists c ∈(0,x) such that f (x) =ex = m ∑ k=0 f (k)(0) k! xk + f (m+1)(c) (m+1)! cm+1 = m ∑ k=0 xk k! + ec (m+1)! cm+1 > m ∑ k=0 xk k! .
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