Exam2_2 7 - x 2-9 3. f ( x ) = x 2 + 3 x-8 x-3 4. f ( x ) =...

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Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 7 Since only - 3 lies in the interval [ - 6 , 3], we see that c = - 3 . keywords: Rolle’s theorem, rational function 014 (part 1 of 1) 10 points Determine lim x →∞ 6 x 2 - 3 x + 8 7 + 2 x - 7 x 2 . 1. limit = 3 7 2. limit = 0 3. none of the other answers 4. limit = - 6 7 correct 5. limit = Explanation: Dividing the numerator and denominator by x 2 we see that 6 x 2 - 3 x + 8 7 + 2 x - 7 x 2 = 6 - 3 x + 8 x 2 7 x 2 + 2 x - 7 . On the other hand, lim x →∞ 1 x = lim x →∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = - 6 7 . keywords: limit, limit at inFnity, rational function 015 (part 1 of 1) 10 points ±ind the rational function f so the Fgure below is the graph of y = f ( x ) on [ - 8 , 8]. 2 4 6 - 2 - 4 - 6 2 4 6 - 2 - 4 - 6 1. f ( x ) = x 2 + 3 x - 8 x 2 - 9 correct 2. f ( x ) = x 2 - 3 x + 8
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Unformatted text preview: x 2-9 3. f ( x ) = x 2 + 3 x-8 x-3 4. f ( x ) = x 2 + 3 x + 8 x 2-9 5. f ( x ) = x 2 + 3 x-8 x 2 + 9 6. f ( x ) = x 2-3 x + 8 9-x 2 Explanation: The graph has vertical asymptotes at x = 3, so a term x 2-9 must appear in the denominator of f ( x ); in particular, therefore, f ( x ) 6 = x 2 + 3 x-8 x 2 + 9 and f ( x ) 6 = x 2 + 3 x-8 x-3 , eliminating two of the possible answers. The graph also has a horizontal asymptote y = 1, so the ratio of coecients of degree 2 terms in the numerator and denominator of f ( x ) must be 1. Thus f ( x ) 6 = x 2-3 x + 8 9-x 2 ,...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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