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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 < 0 , 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 0 ( x ) = ( x 2 + 1) ° 3 = 2 > 0 ; the curve is increasing everywhere. Because y 00 ( x ) = ° 3 x ( x 2 + 1) ° 5 = 2 ; the curve is concave up for x < 0 and concave down for x > 0 . There is an in±ection point at (0 ; 0) . Here is the graph: 1

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-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1.0 -0.5 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) ² 0 , namely D = ( °1 ; ° 1] [ [0 ; 1 ) . B. The x -intercept and y -intercept occur at (0 ; 0) . C. The function has no apparent symmetry. D. Because lim x !1 ° p x 2 + x ° x ± = lim x !1 x p x 2 + x + x = lim x !1 1 q 1 + 1 x + 1 = 1 2 ; a horizontal asymptote is the line y = 1 = 2 . There are no vertical asymptotes.
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