MATH 103 Midterm: MATH 103 UPenn m103 sExam 2 Exam

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turquoisegnu128

31 Jan 2019

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Math 103 sample final exam 2: at what value or values of x is the function f (x) = . |x + 1| 1 if x < 0 x2 + x if 0 x < 1 ifx 1. X ln x cos 2x 1 x (c) lim x 0 sin2(x2) x3 (d) lim x 2(cid:5) x. 2 (g) lim x x3 + 2x 3 x4 5x + 1 cos t5 + 1 dt. 1 ln 2 (c) 1 + ln 2 (g) 1 + 1. 2 ln 2 (d) 1 + ln 2 (h) 0. + ln 2: suppose f is an even function that satis es (cid:5) 2. 0 f (x) dx = 6 and (cid:5) 2. 0 (a) h = 2 (e) h = 2 2. 2 (b) h = 2 3 2 (f) h = 2 + 3. Be the angle the camera makes with the horizontal.

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3. (21 pts) Evaluate the limit or say that the limit does not exist. If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L'Hospitals's Rule to justify your answer. (Note: Possible answers include + or -..) (a) (7 pts) x2 - 2x - 3 - lim x+3 Vx+1 - 2 x2 - 2x - 3 Vx+ 1 + 2 lim x+3 Vx+1 -2 Vx+1 +2 (x - 3) (x + 1) (Vx+ = lim x+3 x + 1 - 4 = lim x+3 (x - 3) (x + 1) (Vx+1 +2). -- x-3 = lin (x + 1) (Vx+1 +2) = (3 + 2) (V3+1 +2) = = 4 (V4 +2) = 16 (b) (7 pts) lim xsin (In x) X >0+ NOTE: We cannot apply the Product Law, since lim sin (1n x) DOES NOT EXIST. On the other hand, -1s sin (in x) s 1 and , therefore, -xs x sin (in x) sx, (since x > 0). Since lim X = lim (-X) = 0, x0+ X0 the inequality above implies that lim xsin (in x) = 0, by the Squeeze Theorem . x0+ EXPLANATION : Since lim ln x = -0, x →0+ sin (lnx) oscillates between 1 and -1 as x + 0*, and, therefore lim sin (In x) DOES NOT EXIST. x >0+ (c) (7 pts) in x+2- (lnx) +In 20 -= -00 (x - 2) 70-

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.