HW 11 solutions-1

# HW 11 solutions-1 - mandel(tgm245 – HW11 – Radin...

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Unformatted text preview: mandel (tgm245) – HW11 – Radin – (56470) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If f is a function on (- 4 , 4) having exactly one critical point and the sign of f ′ , f ′′ are given in- 2 2 f ′ < f ′ < f ′ < f ′ > f ′′ > f ′′ < f ′′ > decide which of the following could be the graph of f . 1. 2 4- 2- 4 2 4- 2- 4 2. 2 4- 2- 4 2 4- 2- 4 correct 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 Explanation: For the given sign chart mandel (tgm245) – HW11 – Radin – (56470) 2- 2 2 f ′ < f ′ < f ′ < f ′ > f ′′ > f ′′ < f ′′ > an inspection of the graphs shows that two of them fail to have exactly one critical point, leaving just four possible graphs for f . To distinguish among these we use the fact that (i) if f ′ ( x ) > 0 on ( a, b ), then f ( x ) is in- creasing on ( a, b ), while (ii) if f ′ ( x ) < 0 on ( a, b ), then f ( x ) is decreasing on ( a, b ), and that (iii) if f ′′ ( x ) > 0 on ( a, b ), then the graph is concave UP on ( a, b ), while (iv) if f ′′ ( x ) < 0 on ( a, b ), then the graph is concave DOWN on ( a, b ). Consequently, again by inspection we see that the only possible graph for f is 2 4- 2- 4 2 4- 2- 4 002 10.0 points Which of the following is the graph of f ( x ) = x 2 x 2- 1 ? Dashed lines indicate asymptotes. 1. 2 4- 2- 4 2 4- 2- 4 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 cor- mandel (tgm245) – HW11 – Radin – (56470) 3 rect 5. 2 4- 2- 4 2 4- 2- 4 6. 2 4- 2- 4 2 4- 2- 4 Explanation: Since x 2- 1 = 0 when x = ± 1, the graph of f will have vertical asymptotes at x = ± 1; on the other hand, since lim x →±∞ x 2 x 2- 1 = 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- 1)- 2 x 3 ( x 2- 1) 2 =- 2 x ( x 2- 1) 2 . Thus f ′ ( x ) > , x < , while f ′ ( x ) < , x > , 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 . 003 10.0 points 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 ′′ ( x ) > 0 on (- 4 ,- 2), (iv) f ′′ ( x ) < 0 on (0 , 2), (v) (0 , 1) is an inflection point, (vi) f ′ ( x ) < 0 on (2 , 4), which one of the following could be the graph of f ? mandel (tgm245) – HW11 – Radin – (56470) 4 1....
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HW 11 solutions-1 - mandel(tgm245 – HW11 – Radin...

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