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PROBLEM 1 Exercise 4.2.4. Let I C R be an open interval, let CE I and let f: I R be a function. Suppose that f(x) < (x-c)² - for all X € I. Prove that f is differentiable at C and fl (c) = 0. (The function f in Example 4.2.5 (1) is a special case of this exercise, where C = 0.) PROBLEM 2 Exercise 4.3.2. Let I C R be an open interval, letce. I, let n € N and let f1,...,fn : I R be functions. Suppose that fi is differentiable at C for all i E {1 n}. Prove that f1,f2 fn is differentiable at C, and find (and prove) a formula for (f1,22 fn)'(c) in terms of fi (c) fn (c) and fi(c) fn(c). PROBLEM 3 Theorem 4.3.3 (Chain Rule). Let I, J C R be open intervals, let c € I and let f : I J and g: J R be functions. Suppose that f is differentiable at C, and that g is differentiable at f(c). Then g o f is differentiable at C and [80 f)'(c) = g' (f(c)) f'(c). Before we give a proof of the Chain Rule, we want to present an attempted proof of this theorem that takes the most straightforward possible approach, though in this case the straightforward approach has a flaw. The attempted proof is "lim (g 30ff(x)-(80f())() = x-+C X-C X-C 8(f(x))-g(f(c))f(x-f(c) - - - lim X-+C f(x) - f(c) X-C = g' (f(c)) . f'(c)." " Proof of Theorem 4.3.3 (Chain Rule). Let k: J IR be defined by g(y)-s(f(c)) if yEJ-{f(c)} k(y) f(c) = Because g is differentiable at f(c), we know that lim g(y)-g(f(c)) y-f(c) exists and equals g' (f(c)). It therefore follows from Lemma 3.3.2 that k is continuous at f(c). Because / is differentiable at c. we know by Theorem 4.2.4 that f is continuous at c. Theorem 3.3.8 (2) then implies that k o / is continuous at C. It follows from Lemma 3.3.2 that lim (ko f)(x) exists and equals (ko f)(c). By the definition of k we see that if J - {(((c)) then (y)ly-f(c)]=g(y)-g(f(c)) - - (4.3.1) Equation 4.3.1 also holds when y - f(c). which is seen by simply substituting y - f(c) into both sides of the equation. Hence Equation 4.3.1 holds for all YE J. Let x € 1 - {c}. Then / "(x) € J. and so we can substitute y = ((x) into Equa- tion 4.3.1 to obtain - - - Dividing both sides of this last equation by x - c (which is not zero) yields (kon()) - x-c Finally, using the continuity of kof atc, Theorem 3.2.10 (4), the definition of * and the fact that f is differentiable at C, we see that lim o x-c = Hence (gofl'(c) exists and equals g'(f(c)).) f'(c). The following result is an immediate consequence of the Chain Rule (Theo- rem 4.3.3). Exercise 4.3.6. [Used in Section 4.3.] Explain the flaw in the attempted proof of the Chain Rule (Theorem 4.3.3) that is given prior to the correct proof. Restate the Chain Rule with modified hypotheses that would make the attempted proof into a valid proof. PROBLEM 4 Exercise 4.3.7. [Used in Lemma 7.3.4 and Theorem 7.3.12.] Let (a.b) C R and [b,c) C R be non-degenerate closed bounded intervals, and let f: [a,b] R and g: (b,c) - R be functions. Suppose that f(b) = g(b). Let h : [a,c] R be defined by f(x), if XE [a,b] h(x) = g(x), if x € (b,c). (1) Suppose that f and g are differentiable, and that f' (b) = g'(b), where f'(b) and g' (b) are one-sided derivatives. Prove that h is differentiable. PROBLEM : 5 Give an example of a function that is defined for all real numbers, is continuous everywhere, but its derivative does not exist at the point X = 3. PROBLEM 6 Consider the function f(x)=|x|. Prove that this function is not differentiable at the point x=0. (HINT: You may want to think about limits from the left and right in the definition of the derivative.) PROBLEM 7 Consider the function (x)=x2+2x Using the limit definition of the derivative, show that for any real number a, the derivative of this function exists at X=a and is equal to 2a+2. PROBLEM 8 Consider the function g(x)=(3x²-5x)(2x-3)(2x³+5x-6)(4x5-3x2+2x-8). Find the equation of the line tangent to the graph of this function at the point where x=-1. Show your work. PROBLEM 9 Consider the function where n is a positive integer. Use induction to prove that for any value x=a, f (a)=n (a) 2 (n-1) for for all positive integers n=1,2,3

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