Written Assignment #1 - SOLUTIONS

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1 Written Assignment #1 - SOLUTIONS Question 1. Use the definitions of continuity and differentiability to prove that the function { cos(1/) if 0 f() = 0 if = 0 is continuous at 0 but not differentiable at 0. We begin by proving that f() is continuous at 0. We need to show that for every ɛ there eists some δ such that f() f(0) < ɛ whenever 0 < δ. We begin with the epression f() f( 0 ) : f() f(0) = cos(1/) 0 = cos(1/) 1 (since 1 cos(1/) 1 for all ) So, if we choose δ = ɛ, and suppose 0 < δ, we have f() f(0) 1 = 0 1 < δ 1 = ɛ Therefore, f() satisfies the ɛ δ definition of continuity at = 0. Net we will prove that f() is not differentiable at 0. By the definition of derivative, we have that f f(0 + h) f(0) (0) = lim h 0 h h cos(1/h) 0 = lim h 0 h = lim cos(1/h) h 0 To prove that the limit does not eist, we use the ɛ δ definition of limit. Suppose, for a proof by contradiction, that lim h 0 cos(1/h) = L for some real number L. Note that L 1 or L 1. Without loss of generality, we may assume that L 1. We need to show that there eists some ɛ such that for ANY δ, there eists some a ( δ, δ) with cos(1/a) L ɛ. 1

2 Choose ɛ = L 1 / (which is positive, as L 1). Let δ be arbitrary. One can choose a natural number N such that 1/(Nπ) < δ. Set a = 1/(Nπ). Then 0 < a 0 < δ, and as cos(1/a) = 1, we have cos(1/a) L = 1 L > ɛ. We have therefore shown that f() is not differentiable at = 0. Question. Prove that if a nonconstant function f satisfies the hypotheses of Rolle s Theorem on the interval [a, b], then there eist points 1 and in (a, b) such that f ( 1 ) < 0 and f ( ). Since f satisfies the hypothess of Rolle s Theorem, we know that f is continuous on [a, b], that f is differentiable on (a, b) and that f(a) = f(b). Denote by M the absolute maimum of f on [a, b] and by m the absolute minimum of f on [a, b]. If f is nonconstant on [a, b] then we must have M > m. Since f(a) = f(b), at least one of these etrema (M or m) must occur at a point in (a, b). We assume first that f(c) = M for some c (a, b), and hence that M > f(a) = f(b). By considering the interval [a, c] and applying the Mean Value theorem, we obtain M f(a) c a = f (r), for some r (a, c) Since M > f(a) and c > a, the LHS of this equation must be positive. Hence f (r). Now, applying the Mean Value theorem to the interval [c, b], we obtain f(b) M b c = f (s), for some s (c, b) Since M > f(b) and c < b, the LHS of this equation must be negative. Hence f (s) < 0. If however M = f(a) = f(b), then we must have m = f(d) for some d (a, b), and hence m < f(a) = f(b). Applying the same reasoning as above, we obtain 0 > m f(a) d a 0 < f(b) m b d = f (p), for some p (a, d) = f (q), for some q (d, b)

3 Question 3. Suppose that both f and f are differentiable on [0, ), f(0) = f (0) = 0 and that f () for all. Prove that f() for all. Consider a point 0 and apply the Mean Value Theorem to the function f () on the interval [0, 0 ]. We obtain f ( 0 ) f (0) 0 = f (c), for some c (0, 0 ) Since f (0) = 0 and f (c), this means that f ( 0 ) 0 and hence f ( 0 ) (since 0 ). Since 0 was chosen arbitrarily, we have shown that f () for all. Net, we consider another point 1 and apply the Mean Value Theorem to the fuction f() on the interval [0, 1 ]. We obtain f( 1 ) f(0) 1 = f (d), for some d (0, 1 ) Since f(0) = and f (d) by our result above, we must have f( 1) 1 and hence f( 1 ). Since 1 was also chosen arbitrarily, we have shown that f() for all. Question 4. Prove the given inequalities using the Mean Value Theorem: 1. sin() < for all. sin() > 3 for all 3. e > n n! for all and positive integers n. 1. Consider the function f() = sin() and the interval [0, ]. The function is continuous and differentiable on (, ), so we may apply the Mean Value Theorem. We note first that f () = 1 cos(). sin() (0 sin(0)) 0 = 1 cos(c) for some c (0, ) From this, we note that cos(c) 1 and so 1 cos(c) 0. This gives sin() and since this implies sin(). It remains to show that sin() for all. 3 0

4 Suppose 0 < < π. In this interval we have cos() < 1 and so we have sin() Again, since this implies that > sin(). On the other hand, if π, then sin() 1 < and so sin() here as well.. Consider the function f() = sin() + 3 and the interval [0, ]. The function is continuous and differentiable on (, ), so we may apply the Mean Value theorem. We note first that f () = cos() 1 +. sin() = cos(c) 1 + c for some c (0, ) Net we consider the function f () = cos() 1 + and the interval [0, c]. This function is also continuous and differentiable, so we may apply the Mean Value Theorem here as well. We note that f () = sin() +. (cos(c) 1 + c ) (cos(0) 1) c 0 Simplifying this equation gives = sin(d) + d for some d (0, c) cos(c) 1 + c c = d sin(d) From the previous question, we know that > sin() for all, so in particular, d sin(d), which implies cos(c) 1 + c. Thus, we have sin() + 3 = cos(c) 1 + c Finally, since, this implies that sin() + and thus sin() >. 3. We consider the statement P (n) : e < n and will prove that P (n) n! is true for all n N using mathematical induction. Base Case: We must show that e > 1 +. Consider the function f() = e 1 and the interval [0, ]. The function is continuous and differentiable on (, ), so we may apply the Mean Value theorem. We note first that f () = e 1. 4

5 (e 1 ) (e 0 1) 0 = e c 1 for some c (0, ) Now since e > 1 for all, we have e c 1. Thus, e 1 Since this implies e 1 and hence e > 1 +. Inductive Step: We assume that P (k) is true for some k 1. That is, e > k We must show that P (k + 1) is true, i.e. that for some k 1 (Inductive Hypothesis) Consider the function e > k+1 (k + 1)! f() = e 1 k k+1 (k + 1)! The function is continuous and differentiable on (, ), so we may apply the Mean Value theorem. We note first that f () = e 1 Applying the Mean Value Theorem, k 1 (k 1)! k e 1 k+1 (k+1)! (e0 1) = e c 1 c ck for some c (0, ) By the Inductive Hypothesis, we know that e c 1 c ck and so e 1 k+1 (k+1)! Again, since this implies that e > k+1 (k+1)!. Therefore, by the process of mathematical induction, we have shown that e > n n! for all n N 5