MAT134Y5 Midterm: Solutions-Term2-2012-2013

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silvermouse444

20 Jan 2019

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Mat134y5y - calculus for life sciences - utm - 2012-2013. Term test 2 - final answers - version 1264. Part a: (note: equivalent answers are also acceptable. : 9/2 or 4. 5, absolute max at x = 2 , 1, 1, x = 3, 0, 2. No absolute minimum: 2/5 or 0. 4, y = 4x+8, g (cid:16) . 2(cid:17) = 0: x = 3, a = 1/2. 2(cid:17) = 3: f (cid:16) . 4: at that time, the distance between the bicycle and the balloon is increasing at a rate of. 26 m/sec : a. domain: x > 0 (or (0, ) )

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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.

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.

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)

MAT134Y5 Midterm: Solutions-Term2-2012-2013

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