exam E 4

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Michigan State University *

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Course

103B

Subject

Mathematics

Date

Apr 3, 2024

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pdf

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7

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MTH 103B Practice Exam for E1-E5 Version 4 Scoring Rubric: Points | Description 4 I own this! I can explain how to do this and why it works. 3 I know how to do this and can do the problems independently, but I am not quite sure why it works. 2 T’'m starting to get it, but I made a couple of mistakes. 1 I have to look at examples to finish this problem. 0 I have no idea what you are talking about! Solutions: 1. Given the functions f(z) = %, g(z) = ILH' and h(z) = (i)(z) find the value of = where h(z) = 1. Solution: Let’s start by finding the domain of h(z). To do this, we must find the values for which f(z), g(z), and h(z) are undefined: fz):a?—1=0=>(z+1)(z-1)=0=>z=1,-1 g@)z+1=0=>z=-1 L 0= z=0 h(z): Therefore, the excluded values are 0, 1 and -1. Page 7
MTH 103B Practice Exam for E1-E5 Version 4 z—2=0andz+1=0 The solutions are z = 2 and z = —1 Because —1 is an excluded value, it is an extraneous solution. Checking = 2 in the simplified function h(z) we see that it does not result in a demoninator that is equal to zero so the solution to this equation sz=2. 2. Find the asymptotes and sketch the graph of the function g(z) = + 1. Show your work/explain . —z+3 your reasoning. Solution: One possible strategy would be to factor out the negative in the denominator as shown below: 2 = +1 9(=) —z+3 2 E 1 —1(z —3) + -2 =——+1 -3 Vertical asymptote: x —3=0=>z=3 Horizontal asymptote: Given the function f(z) = % + ¢, c is the horizontal asymptote so y = 1 is x the h.a. Because a = —2, the graph is reflected over the horizontal asymptote and is steeper than the parent function. Page 8
MTH 103B Practice Exam for E1-E5 Version 4 8 6 4 4.—_—_9_—:::%’___.._‘: ________ | = -8 —6 —4 *272 6 8 —4 —6 -8 ? i / maaR 5 3 / /N -~ —6-5-4-3-2-1 53456738 / e -3 A1 Solution: () Hm ()= 1 (@) Jim f(z) =2 (b) tim f(a) =2 ® lim 1) = (c) zli»]{ll J(z) = does not exist (g) 513% flz)=2 (d) f(=1) =1 () f3)=2 Page 9
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