MAC 2311 Midterm: MAC 2311 FIU Exam 109pmk

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turquoisegnu128

15 Feb 2019

School

FIU

Department

Mathematics - Calculus and Pre

Course

MAC 2311

Professor

All

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Document Summary

Exam i: [30 points] compute each limit: 8 x 2 = tan 1(x) : limx 2 q 2x2, limx + , limx + . X2 1 x : [20pt] short answer problems, find all values of (in radians) that satisfy the equation cos( ) = 1. 2: solve x2 4x + 3 < 0, (20pts) answer true or false. 1: [15pts] choose one of the problems below to do. Remarks and answers: the average was approx 55/100, which is fairly low. The worst results were on the limits on page 1, and the proof (#6). 1a) 2 2, or you can leave it as 8. Each part of problem 1 was worth 10 points. For full credit, your method had to be fairly clear, and valid. I accepted a graph as a valid method, since that"s how we did this one in class.

Related Questions

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

MAC 2311 Midterm: MAC 2311 FIU Exam 109pmk

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