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Unformatted text preview: Math 115 Final Exam December 17, 2009 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 10 pages including this cover. There are 8 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck. 3. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation) for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI92 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3 5 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers , and remove all headphones. 9. Use the techniques of calculus to solve the problems on this exam. Problem Points Score 1 13 2 14 3 12 4 12 5 13 6 12 7 12 8 12 Total 100 Math 115 / Final (December 17, 2009) page 2 1 . [13 points] The graph of the derivative , h ( x ), of a continuous function h is shown below: y = h ( x ) 2 4 2 4 2 4 6 2 a . [3 points] Approximate the xcoordinates of all critical points of h in the interval ( 5 , 5), and classify each as either a local maximum, a local minimum, or neither. Solution: Since h ( x ) is defined at every point in ( 5 , 5), the critical points of h on ( 5 , 5) are the zeros of h . These are x = 4, x = 1, and x = 4. At x = 4, h changes from positive to negative, so h has a local maximum at x = 4. At x = 1, h changes from negative to positive, so h has a local minimum at x = 1. Since the sign of h does not change at x = 4, h has neither a local maximum nor local minimum at x = 4. b . [3 points] Approximate the xcoordinate(s) of any inflection point(s) of h in the interval ( 5 , 5). Solution: At (approximately) x = 2 . 8, x = 1, and x = 4, the sign of the derivative of h changes, so the concavity of h changes. Hence h has an inflection point at each of x = 2 . 8, x = 1, and x = 4....
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 Winter '09
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