Homework 17 - vu(tv2894 – Homework 17(Section 4.5 –...

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Unformatted text preview: vu (tv2894) – Homework 17 (Section 4.5) – miner – (55096) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points P 1 If f is the function whose graph is given by 6 4 2 of f the x-axis and y -axis have been omitted, but the point P = (1, 1) on the graph has been included. Use calculus to determine which of the following f could be. 1. f (x) = 8 1 − 3 x + x 2 + x3 3 3 2 4 6 1 8 2. f (x) = − + 3x + x2 − x3 3 3 3. f (x) = 14 1 − 3 x − x 2 + x3 3 3 which of the following properties does f have? A. critical point at x = 2 , 1 2 4. f (x) = − + 3x − x2 − x3 correct 3 3 5. f (x) = 1 10 − 5 x + 3 x 2 − x3 3 3 B. local minimum at x = 4 . 1. neither of them 2. B only 3. both of them correct 4. A only Explanation: A. True: the graph has a vertical tangent at x = 2 so f ′ (2) does not exist. B. True: by inspection. 002 10.0 points 4 1 6. f (x) = − + 5x − 3x2 + x3 3 3 Explanation: From the graph, f (x) → −∞ as x → ∞. Of the six given choices, therefore, f must have the form 1 f (x) = a + bx + cx2 − x3 3 with {b, c} being one of {3, −1}, Now f ′ (x) = b + 2cx − x2 , Since the graph has a x = 1, it follows that b + 2c − 1 = 0 f ′′ (x) = 2(c − x) . local maximum at c−1 < 0 , {−5, 3}, {3, 1} . In drawing the graph vu (tv2894) – Homework 17 (Section 4.5) – miner – (55096) i.e., b = 3 and c = −1. On the other hand, to determine a we use the fact that f (1) = a + b + c − Consequently, 1 2 f ( x) = − + 3 x − x 2 − x3 . 3 3 1 = 1. 3 5 2 correct 3. 4 3 2 1 0 -1 -2 -3 -4 -5 -6 6 5 4. 4 3 2 1 0 -1 -2 -3 -4 -5 -6 4 2 −4 −2 −2 −4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 003 10.0 points 4 2 −4 −2 −2 −4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 A function y = f (x) is known to have the following properties (i) f (0) = 0, (ii) f changes concavity at x = 0, (iii) f ′′ < 0 on (−∞, −4), f ′′ > 0 on (4, ∞). Which of the following could be the graph of f when dashed lines indicate asymptotes? 5 1. 4 3 2 1 0 -1 -2 -3 -4 -5 -6 5 2 4 5 5. 4 4 2 −4 −2 −2 −4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 3 2 1 0 -1 -2 -3 -4 -5 -6 4 2 −4 −2 −2 −4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2. 4 3 2 1 0 -1 -2 -3 -4 -5 -6 4 2 −4 −2 −2 −4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 Explanation: Since f (0) = 0, the y -intercept of the graph is at the origin, automatically eliminating one of the graphs. As the concavity changes at the origin also, the graph cannot have a local maximum or local minimum at the origin, thus eliminating two more of the graphs. to distinguish between these we look at the concavity of the graphs on (−∞, −4) and on (4, ∞). Hence only vu (tv2894) – Homework 17 (Section 4.5) – miner – (55096) 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 3 4 2 −4 −2 −2 −4 2 4 4. 2 3π 4 3π 2π 5. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 correct can be the graph of f . 004 10.0 points 2 3π 4 3π 2π Which one of the following could be the graph of f ( x) = − 1. 2 3π 2 3π 4 3π 4 3π sin x ? 2 + cos x 6. 2π 2π 2. Explanation: Since f (0) = 0, the graph of f passes through the origin. This already eliminates two of the graphs. On the other hand, by the Quotient Rule, 2 3π 4 3π 2π f ′ ( x) = − cos x(2 + cos x) + (sin x)2 (2 + cos x)2 2 cos x + (cos2 x + sin2 x) (2 + cos x)2 2 cos x + 1 , (2 + cos x)2 =− 3. =− 2 3π 4 3π 2π so f has critical points at the solutions of 4 2 2 cos x = −1, i.e., at x = 3 π and 3 π . This now eliminates two more graphs. To decide which of the remaining two is the graph of f we can look at a sign chart vu (tv2894) – Homework 17 (Section 4.5) – miner – (55096) − 0 ′ 2 3π 4 + 4 3π − 2π for f . Consequently, 2 3π 4 3π 2π is the graph of f . keywords: CurveSketch, ...
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This note was uploaded on 11/07/2010 for the course M 408N taught by Professor Gualdini during the Fall '10 term at University of Texas.

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