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Unformatted text preview: Math 115 Final Exam April 25, 2011 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 9 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. Note that problems 69 will be graded giving very little partial credit. Problem Points Score 1 10 2 13 3 10 4 12 5 11 6 10 7 10 8 14 9 10 Total 100 Math 115 / Final (April 25, 2011) page 2 1 . [10 points] Find a formula for a function of the form f ( x ) = 1 a + x + bx 2 which has a local minimum at (2 , 1 / 2). Be sure to show that your function has a minimum at (2 , 1 / 2). Solution: A local minimum will occur when f ( x ) = 0 or is undefined. The former is when 2 bx + 1 ( bx 2 + x + a ) 2 = 0 . The numerator is zero when x = 1 2 b , and the denominator is zero on either side of this value, when x = 1 2 b 1 2 b 1 4 ab . The first of these gives the local minimum we want: if we let 1 2 b = 2, we have b = 1 4 . This gives f ( x ) = 1 2 x +1 ( 1 4 x 2 + x + a ) 2 , so that if x < 2 we have f ( x ) < 0, and if x > 2, f ( x ) > 0. Thus if b = 1 4 , x = 2 is a local minimum. To require that the minimum occur at (2 , 1 / 2), we want f (2) = 1 a +2 1 = 1 / 2, so that a = 1. Math 115 / Final (April 25, 2011) page 3 x h 9 cm 9 cm 2 . [13 points] A rain gutter attaches to the edge of a roof and collects the rain that falls on the roof. A common gutter design is shown in the figure to the right, and has a trapezoidal crosssection (also shown). In this problem we consider a gutter with base and side length 9 cm, as shown....
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This note was uploaded on 01/28/2012 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.
 Fall '08
 BLAKELOCK
 Math

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