APPM 1350 Final: appm1350spring2014examfinal_sol

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turquoisegnu128

31 Jan 2019

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1. (a) (5 pts. each) find the following derivatives, f (x), for the f given: (a) f (x) = x2 sin 1(x2) (b) f (x) = 1 + x2 (c) f (x) = x ln x (d) f (x) = xx (b) (15 pts) using rules for di erentiation it is easy to show that (3x2 + 1) = 6x. Show that this is true directly from the de nition of di erentiation. No credit will be given for quoting rules of di erentiation. d dx. Solution: (a) f (x) = 2x sin 1(x2) + x2. 3(x + h)2 + 1 (3x2 + 1) h (e) d dx (3x2 + 1) = lim h 0. 3x2 + 6xh + 3h2 + 1 3x2 1 h. = lim h 0: (5 pts each) in answering the following questions, justify each part. 0 (b) chop up into two regions, x > 0 and x < 0.

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4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.