Exam3 2 - properties is 2 4-2-4 2 4-2-4 Consequently, this...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 2 6. 2 4 - 2 - 4 2 4 - 2 - 4 Explanation: Since x 2 - 4 = 0 when x = ± 2, the graph of f will have vertical asymptotes at x = ± 2; on the other hand, since lim x →±∞ x 2 x 2 - 4 = 1 , the graph will have a horizontal asymptote at y = 1. This already eliminates some of the possible graphs. On the other hand, f (0) = 0, so the graph of f must also pass through the origin. This eliminates another graph. To decide which of the remaining graphs is that of f we look at the sign of f 0 to determine where f is increasing or decreasing. Now, by the Quotient Rule, f 0 ( x ) = 2 x ( x 2 - 4) - 2 x 3 ( x 2 - 4) 2 = - 8 x ( x 2 - 4) 2 . Thus f 0 ( x ) > 0 , x < 0 , while f 0 ( x ) < 0 , x > 0 , so the graph of f is increasing to the left of the origin and decreasing to the right of the origin. The only graph having all these
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Unformatted text preview: properties is 2 4-2-4 2 4-2-4 Consequently, this must be the graph of f . keywords: graph, rational function, asymp-tote, increasing, decreasing 002 (part 1 of 1) 10 points A rectangle is inscribed in a semi-circle of diameter d as shown in d What is the largest area the rectangle can have when d = 4? 1. Area = 2 sq. units 2. Area = 4 sq. units correct 3. Area = 1 sq. units 4. Area = 3 sq. units 5. Area = 5 sq. units Explanation: Lets take the semi-circle to be the upper half of the circle x 2 + y 2 = 4 having radius 2 and center at the origin. For the rectangle we take one side on the x-axis...
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