m408k_practicetest3_solutions

# m408k_practicetest3_solutions - Create assignment, 59515,...

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Unformatted text preview: Create assignment, 59515, Practice 3, Dec 02 at 3:44 am 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. CalC4e12b 50:06, calculus3, multiple choice, &gt; 1 min, normal. 001 Which of the following is the graph of f ( x ) = x 2 x 2- 16 ? Dashed lines indicate asymptotes. 1. 2 4- 2- 4 2 4- 2- 4 cor- rect 2. 2 4- 2- 4 2 4- 2- 4 3. 2 4- 2- 4 2 4- 2- 4 4. 2 4- 2- 4 2 4- 2- 4 5. 2 4- 2- 4 2 4- 2- 4 6. 2 4- 2- 4 2 4- 2- 4 Create assignment, 59515, Practice 3, Dec 02 at 3:44 am 2 Explanation: Since x 2- 16 = 0 when x = 4, the graph of f will have vertical asymptotes at x = 4; on the other hand, since lim x x 2 x 2- 16 = 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 to determine where f is increasing or decreasing. Now, by the Quotient Rule, f ( x ) = 2 x ( x 2- 16)- 2 x 3 ( x 2- 16) 2 =- 32 x ( x 2- 16) 2 . Thus f ( x ) &gt; , x &lt; , while f ( x ) &lt; , x &gt; , 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 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 CalC4e20a 50:06, calculus3, multiple choice, &lt; 1 min, wording-variable. 002 If f is a continuous function on (- 4 , 4) such that (i) f has 3 critical points, (ii) f has 1 local maximum, (iii) f 00 ( x ) &gt; 0 on (- 4 ,- 2), (iv) f 00 ( x ) &lt; 0 on (0 , 2), (v) (0 , 1) is an inflection point, (vi) f ( x ) &lt; 0 on (2 , 4), which one of the following could be the graph of f ? 1. 2 4- 2- 4 2 4 correct 2. 2 4- 2- 4 2 4 Create assignment, 59515, Practice 3, Dec 02 at 3:44 am 3 3. 2 4- 2- 4 2 4 4. 2 4- 2- 4 2 4 5. 2 4- 2- 4 2 4 6. 2 4- 2- 4 2 4 Explanation: Five of the graphs fail to have one or more of the 6 properties the graph of f has to have. Indeed, property (i) fails in 2 4- 2- 4 2 4 because the graph has only 2 critical points, while properties (i) and (ii) fail in 2 4- 2- 4 2 4 because the graph has 4 critical points and 2 local maxima, property (iii) fails in 2 4- 2- 4 2 4 because the graph is concave DOWN on (- 4 ,- 2), property (iv) fails in 2 4- 2- 4 2 4 because the graph is concave UP on (0 , 2), and property (v) fails in Create assignment, 59515, Practice 3, Dec 02 at 3:44 am 4 2 4- 2- 4 2 4 because (0 , 1) is not an inflection point. On the other hand, the only remaining graph 2 4- 2- 4 2 4 does have all six of the required properties....
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## m408k_practicetest3_solutions - Create assignment, 59515,...

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