Exam2_Solutions_F10 - Math 115 Second Midterm November 16,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 115 — Second Midterm November 16, 2010 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 9 pages including this cover. There are 8 problems. Note that the problems are not of equal diFculty, 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 TI-92 (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 ±nd 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 of all cell phones and pagers , and remove all headphones. Problem Points Score 1 10 2 14 3 15 4 14 5 12 6 10 7 15 8 10 Total 100
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
page 2 1 . [10 points] Given below is a graph of a function f ( x ) and a table for a function g ( x ). 3 5 7 x M 2 2 f L x R x 0 1 2 3 4 g ( x ) 4 3 1 2 20 3 g ( x ) -2 - 5 2 1 2 3 - 1 3 Give answers for the following or write “Does not exist.” No partial credit will be given. i) d dx f ( g ( x )) at x = 0 By the chain rule d dx f ( g ( x )) = f ( g ( x )) g ( x ). At x = 0 we have f ( g (0)) g (0) = f (4) · ( - 2) = 4 . ii) d dx [ f ( x ) g ( x )] at x = 2 By the product rule d dx [ f ( x ) g ( x )] = f ( x ) g ( x ) + f ( x ) g ( x ). At x = 2 we have f (2) g (2) + f (2) g (2) = (2 / 3)(1) + (4 / 3)(1 / 2) = 4 / 3 . iii) d dx b f ( x ) g ( x ) B at x = 4 By the quotient rule d dx b f ( x ) g ( x ) B = f ( x ) g ( x ) - f ( x ) g ( x ) [ g ( x )] 2 . At x = 4 we have f (4) g (4) - f (4) g (4) [ g (4)] 2 = ( - 2)(20 / 3) - (0)( - 1 / 3) (20 / 3) 2 = - 3 / 10 . iv) d dx [ g ( f ( x ))] at x = 3 We know g ( f (3)) = 1 / 2, so for values of y near f (3), g ( y ) looks like a line with slope 1/2. So g ( f ( x )) “looks like” 1 2 f ( x )+ b for some constant b , for x near 3. Since f ( x ) is “pointy” at x = 3, 1 2 f ( x ) + b looks like a vertically compressed version of this pointy graph (near x = 3), which is still pointy. So g ( f ( x )) is also pointy at x = 3, hence not diFerentiable. v)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 9

Exam2_Solutions_F10 - Math 115 Second Midterm November 16,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online