APPM 1345 Midterm: appm1345spring2016exam1_sol

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turquoisegnu128

31 Jan 2019

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Instructions: books, notes, and electronic devices are not permitted. Solution: (a) all real numbers except where x2 5 = 0, or all x"s such that x 6= 5 . x x 5. = y = (x2 + 5) (x2 5)2 . y < 0 for all x . = y = (x 5) (x 5)(x + 5) 1 x + 5 (b) y = (c) y = x (cid:0) 4 5(cid:1)2 x2 5. Therefore, x = 5 is a vertical asymptote, and x = 5 is a hole in the graph . (d) 2 = = a2 = 10 = a = 10. 1: consider the function f (x) = x4 4x3 + 10 (a) (6 points) name the relative extrema of f (x). (b) (6 points) name any points of in ection for f (x). Solution: (a) y = cos(cos x) sin x and the only time y = 0 is when one of its two factors is 0.

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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.

4. (21 pts) (I) (6 pts) Determine the value of a constant b for which the function g is continuous at 0. Show your work. Justify your answer! 2 x +b if x<0, X-5 g(x) = if x 20, x#4. We have to show that lim g (x) = g (0). g(0) = 0 +18 = -1; lim X0* 0 + 16 ) = lim x + 16 **0* x2 - 160 - 16- Since lim g (x) has to exist (in order for g to be continuous at x = 0), it has to be true that x-0- x0+ X- 0 lim g(x) - lim g(x) = lim g(x). Therefore, -1, and , so b = 5. so, if b = 5, lim g (x) = g(0), and therefore, g is continuous at x = 0. (II) (4 pts) Evaluate the limit. Show your work! 2. b limx- (a) lim g(x) = lim *+- 2 x + b X - 5 : 1 * +16 10 g (x) = lim x++ - Elim x2 - 16 x++ (x - 10). OR 2x+5. 2; (a) lim g (x) = lim x--00 X- (b) lim x++ 2 + 16 g(x) = lim **** x2 - 16 X-+ (III) (4 pts) Determine whether g has any horizontal asymptotes. If so, write the equation (or equations) of all horizontal asymptotes. y = 2, by part (II) (a) y = 0, by part (II) (b) (IV) (7 pts) Determine whether g has any vertical asymptotes. If so, write the equation (or equations) of all vertical asymptotes. Show your work. Justify your answer! 5 + 16 There is no vertical asymptote at x = 5, because lim g(x) = x+5 52-16' -8 + b (x) = there is no vertical asymptote at x = -4, because lim X -4 - 4 - 5 But, lim g(x) = lim + 16 x+4+ x2 - 16 lim_ (x + 16) -20 x+4+ (x - 4) 60 (x + 4) +8 (or the numerator goes to 20, and the denominator is positive and goes to o, as x + 4*) = +00 So, the function g has only one vertical asymptote, x = 4.

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)

APPM 1345 Midterm: appm1345spring2016exam1_sol

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