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Unformatted text preview: Solution: The numerator is zero when x = 0 , 1, while the denominator is zero when x = 1 , 1 / 2. This tells us that the xintercepts (zeros) are x = 0 , 1 and the vertical asymptotes are at x = 1 , 1 / 2. Page 2 Math 171: Exam 2 A rational function can only change sign at zeros and vertical asymptotes, so we can determine when the function is positive or negative by considering the inter vals (1 , 1), ( 1 , 0), (0 , 1 / 2), (1 / 2 , 1), and (1 , 1 ). By checking points in those intervals we find that it is +, , , , and +, respectively. To find the yintercept we just evaluate the function at x = 0, at which point it’s zero (as we have already seen). For the horizontal asymptote, lim x !1 ( x 1) x 2 (2 x 1) 2 ( x + 1) = lim x !1 x 3 + smaller order terms 4 x 3 + smaller order terms = lim x !1 x 3 4 x 3 = lim x !1 1 4 = 1 4 . So the graph has a horizontal asymptote at y = 1 / 4. Similarly, lim x !1 ( x 1) x 2 (2 x 1) 2 ( x +1) = 1 4 . (We also know that rational function can have at most one horizontal asymptote.) We now have everything we need to sketch a graph (see the last page). 5. Let f ( x ) = 1 2 x 1 . (a) [6 points] Express f ( x ) as a limit. Solution: By the definition of the derivative: f ( x ) = lim h ! 1 2( x + h ) 1 1 2 x 1 h . (b) [6 points] Using the limit definition of the derivative, find f ( x ). Solution: f ( x ) = lim h ! 1 2( x + h ) 1 1 2 x 1 h = lim h ! 1 2( x + h ) 1 1 2 x 1 h (2( x + h ) 1)(2 x 1) (2( x + h ) 1)(2 x 1) = lim h ! (2 x 1) (2( x + h ) 1) h (2( x + h ) 1)(2 x 1) = lim h ! 2 x 1 2 x 2 h + 1 h (2( x + h ) 1)(2 x 1) = lim h !...
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 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Derivative, Product Rule, lim

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