Math 115 — Second Midterm — November 12, 2018EXAM SOLUTIONS1. Do not open this exam until you are told to do so.2. Do not write your name anywhere on this exam.3. This exam has 13 pages including this cover. There are 11 problems.Note that the problems are not of equal difficulty, so you may want to skip over and return to aproblem on which you are stuck.4. Do not separate the pages of this exam. If they do become separated, write your UMID (not name)on every page and point this out to your instructor when you hand in the exam.5. Note that the back of every page of the exam is blank, and, if needed, you may use this space forscratchwork. Clearly identify any of this work that you would like to have graded.6. Please read the instructions for each individual problem carefully. One of the skills being testedon this exam is your ability to interpret mathematical questions, so instructors will not answerquestions about exam problems during the exam.7. Show an appropriate amount of work (including appropriate explanation) for each problem, so thatgraders can see not only your answer but how you obtained it.8. The use of any networked device while working on this exam is not permitted.9. You may use any one calculator that does not have an internet or data connection except a TI-92(or other calculator with a “qwerty” keypad). However, you must show work for any calculationwhich we have learned how to do in this course.You are also allowed two sides of a single 300× 500notecard.10. For any graph or table that you use to find an answer, be sure to sketch the graph or write out theentries of the table. In either case, include an explanation of how you used the graph or table tofind the answer.11. Include units in your answer where that is appropriate.12. Problems may ask for answers in exact form. Recall that x =√2 is a solution in exact form to theequation x2= 2, but x = 1.41421356237 is not.13. Turn off all cell phones, smartphones, and other electronic devices, and remove all head-phones, earbuds, and smartwatches. Put all of these items away.14. You must use the methods learned in this course to solve all problems.Problem Points Score1 172 63 124 125 96 4Problem Points Score7 58 79 1310 711 8Total 100Math 115 / Exam 2 (November 12, 2018) page 21. [17 points] Let g(x) and h(x) be two functions. The graphs of g0(x) and h00(x) are shown below.At right is the graph of y = g0(x), thederivative of g(x).Assume that g(x) is a continuousfunction.Use the graph to answer the questions be-low. Circle all correct answers.1 2 3 4 5 6 7 8−2−1123y = g0(x)xya. [2 points] At which of the following values of x is g(x) not differentiable?Solution:x = 2 x = 4 x = 5 x = 6 x = 7 none of theseb. [2 points] For which of the following values of x does g(x) have a local maximum?Solution:x = 2 x = 4 x = 5 x = 6 x = 7.5 none of thesec. [2 points] For which of the following values of x does g(x) have an inflection point?Solution:x = 2 x = 3 x = 4 x = 5 x = 7.5 none of thesed. [2 points] On which of the following intervals is g(x) linear?Solution:(0, 2) (4, 6) (6, 7) (6, 8) (7, 8) none of thesee. [2 points] For which of the following values of x does g(x) attain a global maximum on theinterval [1, 7]?Solution:x = 2 x = 4 x = 5 x = 6 x = 7 none of theseMath 115 / Exam 2 (November 12, 2018) page 3Use the graph of y = h00(x), the secondderivative of h(x), to answer the ques-tions below. Circle all correct answers.1 2 3 4 5−2−1123y = h00(x)xyf. [2 points] Over which of the following intervals is h(x) concave up on the entire interval?Solution:(0, 1) (1, 3) (2, 4) (4, 5) none of theseg. [2 points] On which of the following intervals is the function h0(x) (the derivative of h(x))decreasing over the entire interval?Solution:(0, 1) (1, 3) (2, 3) (4, 5) none of theseh. [3 points] If h0(4) = 0, which of the following statements must be true?Solution:A. x = 4 is a local maximum of h(x).B. x = 4 is a local minimum of h(x).C. x = 4 is an inflection point of h0(x).D. x = 4 is a critical point of h(x).E. x = 4 is an inflection point of h(x).F. none of theseMath 115 / Exam 2 (November 12, 2018) page 42. [6 points] Let A and B be constants andk(x) =3x +Bxfor 0 < x < 1Bx2+ Ax3for 1 ≤ x.Find the values of A and B that make the function k(x) differentiable on (0, ∞). Show all your workto justify your answers. If there are no such values of A and B, write none.Solution: k(x) will only be differentiable at x = 1 if it is also continuous at x = 1. In order for thisto happen, we plug x = 1 in to both formulas of the original function and set them equal as well:3 + B = B + AFrom this second equation, we can subtract B from both sides to find A = 3.The function 3x +Bxis differentiable on (0, 1), and the function Bx2+ Ax3is differentiable on(1, ∞), so we just need values of A and B that will make k(x) differentiable at x = 1.We can compute the derivative:k0(x) =3 −Bx2for 0 < x < 12Bx + 3Ax2for 1 < x.In order for k(x) to be differentiable at x = 1, we must havelimh→0−k(1 + h) −k(1)h= limh→0+k(1 + h) −k(1)hddx(3x +Bx)x=1=ddx(Bx2+ Ax3)x=13 −Bx2x=1= 2Bx + 3Ax2x=13 − B = 2B + 3AIn addition, Now we plug this value in for A in the earlier equation, giving us3 − B = 2B + 9Solving for B, we get 3B = −6, so B = −2.Answer: A = 3 B = −2Math 115 / Exam 2 (November 12, 2018) page 53. [12 points] Assume the function h(t) is invertible and h0(t) is differentiable. Some of the values ofthe function y = h(t) and its derivatives are shown in the table belowt 0 1 2 3 4h(t) -2 2 3 4 8h0(t) 3.5 0.5 2.5 1.5 5h00(t) 6 0.25 0.3 -0.4 0.6Use the values in the table to compute the exact value of the following mathematical expressions.If there is not enough information provided to find the value, write NI. If the value does not exist,write dne. Show all your work.a. [3 points] Let a(t) = h(t2− 5). Find a0(3).Solution: Since a0(t) = 2th0(t2− 5), then a0(3) = 6h0(4) = 6(5) = 30Answer: 30b. [3 points] Let b(t) =h(t)t2. Find b0(4).Solution: Sinceb0(t) =h0(t)t2− 2th(t)t4then b0(4) =16h0(4) − 8h(4)256=16(5) − 8(8)256=16256=116.Answer:116c. [3 points] Let c(y) = h−1(y). Find c0(2).Solution: Since c0(y) =1h0(h−1(y))then c0(2) =1h0(h−1(2))=1h0(1)= 2. Answer: 2d. [3 points] Let g(t) = ln(1 + 2h0(t)). Find g0(0).Solution: Since g0(t) =2h00(t)1 + 2h0(t)then g0(0) =2h00(0)1 + 2h0(0)=2(6)1 + 2(3.5)=128= 1.5Answer: 1.5Math 115 / Exam 2 (November 12,