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Unformatted text preview: Answer Guide 6 Math 408C: Unique Numbers 56975, 56980, and 56985 Tuesday, October 13, 2009 Homework problems Section 4.4 16. lim x !1 x + 2 p 9 x 2 + 1 = lim x !1 1 + 2 x q 9 + 1 x 2 = 1 + 0 p 9 + 0 = 1 3 : 20. In the calculations below, we assume that x < , so that p x 2 = & x . Thus, lim x !&1 & x + p x 2 + 2 x = lim x !&1 & x + p x 2 + 2 x x & p x 2 + 2 x x & p x 2 + 2 x = lim x !&1 x 2 & ( x 2 + 2 x ) x & p x 2 + 2 x = lim x !&1 & 2 x & x x & x & q 1 + 2 x = 2 & 1 & p 1 + 0 = & 1 : 46. The function y ( x ) = x p x 2 + 1 is de&ned for all x in R , so it has no vertical asymptotes. As x ! 1 , y ! 1 , so the lines y = 1 are horizontal asymptotes. Because y ( x ) = ( x 2 + 1) & 3 = 2 > ; the curve is increasing everywhere. Because y 00 ( x ) = & 3 x ( x 2 + 1) & 5 = 2 ; the curve is concave up for x < and concave down for x > . There is an inection point at (0 ; 0) . Here is the graph: 1654321 1 2 3 4 5 61.00.5 0.5 1.0 x y Section 4.5 22. Consider f ( x ) = p x 2 + x & x = p x ( x + 1) & x: A. The domain of the function consists of all x such that x ( x + 1) , namely D = ( &1 ; & 1] [ [0 ; 1 ) . B. The xintercept and yintercept occur at (0 ; 0) ....
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This note was uploaded on 02/09/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas at Austin.
 Fall '06
 McAdam
 Math, Calculus

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