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Unformatted text preview: Student name: Student ID: TAs name and/or section: MATH 31A (Butler) Midterm I, 22 January 2010 This test is closed book and closed notes. No calculator is allowed for this test. For full credit show all of your work (legibly!). Problem 3 is worth 15 pts, the remaining problems are worth 10 points (a total of 55 points). 1. Given y = 5 2 x is tangent to f ( x ) at x = 3, find the line tangent to the function g ( x ) = 3 x 2 f ( x ) 6 at x = 3. First we note that to find the tangent line to g ( x ) at x = 3 we need to compute g (3) and g (3). Using the rules of derivatives we have g ( x ) = 6 xf ( x ) + 3 x 2 f ( x ) . So we have the following. g (3) = 3 3 2 f (3) 6 = 27 f (3) 6 g (3) = 6 3 f ( x ) + 3 3 2 f ( x ) = 18 f ( x ) + 27 f ( x ) So what we really need are f (3) and f (3). Fortunately for us we have the tangent line at x = 3, in particular we have that f (3) = 5 2 3 = 1 (i.e., the same point as the tangent line) and f (3) = 2 (the slope of the tangent line). So we have the following. g (3) = 27( 1) 6 = 33 g (3) = 18( 1) + 27( 2) = 72 Finally, the tangent line will be y = g (3) + g (3)( x 3) = 33 + ( 72)( x 3) = 72 x + 183 . 2. Let ( x ) be the function shown below. 1 2 3 1 2 3 4 5 6 7 Answer the following questions. (The answer might be that it does not exist.) (a) lim x 1 ( x ) = The limit is 5 / 2. Since the function from both sides of 1 approach the values2....
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 Winter '10
 BUTLER
 Math

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