MATH113 Study Guide - Midterm Guide: Asymptote, Squeeze Theorem

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turquoisegnu128

24 Oct 2018

School

University of Alberta

Department

Mathematics

Course

MATH113

Professor

All

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Document Summary

Math 113g1/114e1 midterm solutions (v1) (1) (a) find the domain of the function (cid:114) x + 1 x 2 f (x) = Solution. f (x) is de ned for x (cid:54)= 2 and. 0 x > 2 or x 1. So the domain is ( , 1] (2, ). x + 1 x 2 (cid:3) (b) let g(x) = f (cos x). Find g(cid:48)( /3) if f (1/2) = 2 and f(cid:48)(1/2) = 3 . Solution. g(cid:48)(x) = (f (cos x))(cid:48) = f(cid:48)(cos x)( sin x) and hence g(cid:48)( /3) = f(cid:48)(cos( /3))( sin( /3)) = f(cid:48)(1/2)( . 2x 1 (c) find horizontal and vertical asymptotes of f (x) = 2x 1 it has a vertical asymptote x = 1/2. = lim x limx (1/x) + limx 2 limx 2 limx (1/x) = 1 it has a horizontal asymptote y = 1.

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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.

1. (25 pts) The graph of a function f is given in the figure below. The domain of fis (-, +.). Use the graph of fto answer the questions below. yax) y=f(x) -3 - 2 3 4 (I) (1 pt) Find the range of f. (II) Find the following values. (Note: Possible answers include + , -, or "does not exist".) (a) (1 pt) (x) = 5 (b) (1 pt) lim f (x) = 5 (0) (1 pt) Tim f (x) = 5 (d) (1 pt) f(3) = 2 (e) (2 pts) lim f (x) = + 0 x-2- (8) (2 pts) 11m = (x) = 0 (g) (2 pts) lim f (x) = DOES NOT EXIST x+-2 (h) (2 pts) Lim f (x) = 0 (i) (2 pts) lim f (x) = + X +00 (III) (2 pts) Find all vertical asymptotes. x = -2 (IV) (2 pts) Find all horizontal asymptotes. y = 0 (v) (2 pts) Find all slant asymptotes. | y = x + 2 (VI) (4 pts) Determine the intervals of continuity for f. (-00, -2), [-2, 3), and (3, +c)

Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

MATH113 Study Guide - Midterm Guide: Asymptote, Squeeze Theorem

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