167 and ( 2 2e−1/x x , if x ̸ = 0; f (x)= 0, if x = 0. 2. Suppose that ϕ is bounded and differentiable on R. Defne the function ( xnϕ(1/x), if x ≠ 0; f (x)= 0, if x = 0. Show that if n ≥ 2, the function is differentiable on R and fnd its derivative. Show that if n = 1 and limx →∞ ϕ(x) does not exists, then f is not differentiable at 0. (b) Hint: Observe that 1 1 ′ ′ f = −1 + c < 0 and f = 1+ c > 0. 2nπ (2n+ 1)π SECTION 4.2 Exercise 4.2.1. Defne the function h(x)= f (x) − g(x). Then h has an absolute maximum at x0. Thus, h ′ (x0)= f ′ (x0) − g ′ (x0)= 0, which implies f ′ (x0)= g ′ (x0). Exercise 4.2.3. The inequality holds obviously if a = b. In the case where a ≠ b, the equality can be rewritten as sin(b) − sin(a) ≤ 1. b− a sin(b) − sin(a) The quotient is the slope of the line connecting (a, f (a)) and (b, f (b)). We need b− a to show that the absolute value of the slope is always bounded by 1, which can also be seen from the fgure. The quotient also reminds us of applying the Mean Value Theorem for the function f (x)= sin(x). Consider the case where a < b and defne the function f : [a,b] → R by f (x)= sin(x). Clearly, the function satisfes all assumptions of the Mean Value Theorem on this interval with f ′ (x)= cos(x) for all x ∈ (a,b). By the Mean Value Theorem, there exists c ∈ (a,b) such that f (b) − f (a) = f ′ (c)= cos(c), b − a
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