MTH 142 Final: MTH 142 University of Rochester 2014 Spring Final solutions

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turquoisegnu128

31 Jan 2019

School

University of Rochester

Department

Mathematics

Course

MTH 142

Professor

All

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Part a: (24 points) (a), (6 points) find the vertical and horizontal asymptotes of f (x) = Answer: (a) the denominator factors into (2x 1)(x + 1). Therefore the function has vertical asymp- totes at x = 1. To nd the horizontal asymptotes, we let x and x . Factoring out the leading term x2 from both numerator and denominator, we nd lim x . 2 is a horizontal asymptote as x . A similar calculation shows that y = 3. Answer: (b) plugging in x, we nd that x3 sin(x) 2 + x2 ex2 f ( x) = ( x)3 sin( x) Note that sin(x) and x3 are odd functions, but x2 is even. 1 and therefore the function is symmetric with an even symmetry. (c), (6 points) find the intervals of increase and decrease for the following function. Nd the points x where the function has a local maximum or local minimum. f (x) =

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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.

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I know its a lot but i would really appreciate the help please!I will give you 5 stars and all the points!

need to find:

a. vertical and horizontal asymptotes

b.x and y intercepts

1. R(x) = x + 1 / x(x+4)

2. R(x) = 3x+3 / 2x+4

3. R(x) = 3 / x^2 - 4

4. R(x) = x^2 / x^2+x-6

5. R(x) = 3 / (x-1)(x^2-4)

6. R(x) = x^2+x -12 / x^2-x-6

Hello I need help with f and g!!! Thank you (:

MTH 142 Final: MTH 142 University of Rochester 2014 Spring Final solutions

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