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Unformatted text preview: Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points If f is a continuous function on (5, 3) whose graph is Explanation: 1 A. False: f (x) = 0 at x = 3, 1, while f (x) does not exist at x = 2; in addition, the graph of f has a vertical tangent at x = 1. All of these are critical points. B. False: the graph changes concavity at the point x = 1 in the interval (2, 1); in fact, f (x) > 0 on (2, 1), while f (x) < 0 on (1, 1). C. False: f has a local minimum at x = 2 and a local maximum at x = 1; the graph of f does have a horizontal tangent at (3, 1), but this is an inflection point, so f does not have a local extremum at x = 3. keywords: True/False critical point, local extreme, 4 2 4 2 2 002 (part 1 of 1) 10 points In drawing the graph which of the following properties are satisfied? A. B. C. f has exactly 3 critical points, f (x) > 0 on (2, 1), f has exactly 3 local extrema. P 1. A only 2. none of them correct 3. B and C only 4. A and C only 5. all of them 6. B only 7. C only 8. A and B only of f the xaxis and yaxis 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) = 2. f (x) = 10 1  5x + 3x2  x3 3 3 1 8  3x + x2 + x3 3 3 4 1 3. f (x) =  + 5x  3x2 + x3 3 3 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann 8 1 4. f (x) =  + 3x + x2  x3 correct 3 3 2 1 5. f (x) =  + 3x  x2  x3 3 3 6. f (x) = 14 1  3x  x2 + x3 3 3 4 2 2 4 2 decide which of the following could be the graph of f . 1. 4 2 2 4 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 , f (x) = 2(c  x) . {5, 3}, {3, 1} . 2. 4 2 4 2 2 4 2 4 Since the graph has an inflection point at x = 1, it follows that c1 = 0 , i.e., b = 3 and c = 1. On the other hand, to determine a we use the fact that 1 f (1) = a + b + c  = 1 . 3 Consequently, 8 1 f (x) =  + 3x + x2  x3 . 3 3 keywords: polynomial, local maximum 003 (part 1 of 1) 10 points 4. If f is a function on (4, 4) having exactly three critical points and the sign of f , f are given in 4 f <0 f >0 f >0 f >0 2 0 2 f >0 f <0 f >0 2 4 2 3. 4 2 2 2 4 4 4 2 2 2 4 4 correct Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann 4 2 2 2 4 4 6. 4 2 4 2 2 4 Explanation: For the given sign chart f <0 f >0 f >0 f >0 2 0 2 f >0 f <0 f >0 an inspection of the graphs shows that two of them fail to have exactly three critical points, 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 increasing 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 4 3 5. 4 2 4 2 4 2 2 2 4 keywords: graph, slope, first derivative, second derivative, sign chart 004 (part 1 of 1) 10 points 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 (, 3), f > 0 on (3, ). Which of the following could be the graph of f?
6 5 1.4 3 2 1 0 1 4 2 3 4 5 6 6 6 5 4 5 2.4 3 2 1 0 1 4 2 3 4 5 6 6 5 4 4 2 2 2 4
3 2 1 0 1 2 3 4 5 6 2 4 4 2 2 2 4 2 4 correct 3 2 1 0 1 2 3 4 5 6 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann
6 5 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 6 5 5.4 3 2 1 0 1 2 3 4 5 6 6 5 4 3 2 1 0 1 2 3 4 5 6 4 4 2 4 2 2 4
6 5 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 6 2 4 2 4 can be the graph of f . 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 keywords: concavity, graph 005 (part 1 of 1) 10 points 2 4 Which of the following is the graph of x2 ? x2  9 Dashed lines indicate asymptotes. f (x) =
6 5 1.4 3 2 1 0 1 2 3 4 5 6 4 2 4 2 2 4 2 4 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 2 4 Explanation: Since f (0) = 0, the yintercept 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 (, 3) and on (3, ). Hence only 6 6 5 4 3 2 1 0 1 2 3 4 5 6 5 2. 4 cor4 3 2 2 1 0 1 4 2 2 4 2 2 3 4 4 5 6 rect 6 5 4 3 2 1 0 1 2 3 4 5 6 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann
6 5 3.4 3 2 1 0 1 2 3 4 5 6 6 5 6.4 3 2 1 0 1 2 3 4 5 6 5 4 2 4 2 2 4
6 5 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 6 2 4 2 4 6 5 4.4 3 2 1 0 1 2 3 4 5 6 Explanation: Since x2  9 = 0 when x = 3, the graph of f will have vertical asymptotes at x = 3; on the other hand, since lim x2 = 1, x2  9 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 x 2 4 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) = 2x(x2  9)  2x3 (x2  9)2 =  Thus (x2 18x .  9)2 6 5 5.4 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 6 2 4 while f (x) > 0, x < 0, f (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 properties is 6 5 4 3 2 1 0 1 2 3 4 5 6 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann
6 5 4 4 3. 3 2 2 1 0 1 4 2 2 4 2 2 3 4 4 5 6 6 6 5 4 3 2 1 0 1 2 3 4 5 6 5 4 4 4. 3 2 2 1 0 1 4 2 2 4 2 2 3 4 4 5 6 6 6 5 4 3 2 1 0 1 2 3 4 5 6 5 4 4 5. 3 2 2 1 0 1 4 2 2 4 2 2 3 4 4 5 6 6 5 4 3 2 1 0 1 2 3 4 5 6 6 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 2 4 Consequently, this must be the graph of f . keywords: graph, rational function, asymptote, increasing, decreasing 006 (part 1 of 1) 10 points Which of the following is the graph of f (x) = x2 x 9 if the dashed lines indicate asymptotes?
6 5 4 1. 3 2 1 0 1 4 2 3 4 5 6 6 6 5 4 5 4 2. 3 2 1 0 1 4 2 3 4 5 6 6 5 4 4 2 2 2 4
3 2 1 0 1 2 3 4 5 6 2 4 Explanation: When f (x) = x2 x 9 4 2 2 2 4 2 4 correct the graph of f will have vertical asymptotes at x = 3 and x = 3 as indicated by the dashed vertical lines. In addition the graph will pass through the origin. On the other hand, f (x) =  x2 + 9 < 0 (x2  9)2 3 2 1 0 1 2 3 4 5 6 and so the graph must also be decreasing on any interval on which f is defined. Consequently, the only possible graph for f is 6 5 4 3 2 1 0 1 2 3 4 5 6 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann 2. 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 7 4 2 2 4 4 2 2 4 3. 4 2 keywords: graph, vertical asymptote, horizontal asymptote, rational function 007 (part 1 of 1) 10 points 4. 4 2 2 4 4 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 ? 1. 4 2 4 2 2 4 4 2 6. 4 2 2 4 4 2 2 4 5. 4 2 4 2 2 4 2 correct Explanation: Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann 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 8 4 2 4 2 4 2 2 4 4 2 2 4 because f is INCREASING on (2, 4). On the other hand, the only remaining graph because the graph has 4 critical points, while properties (i) and (ii) fail in 4 2 4 2 2 4 4 2 4 2 2 4 does have all six of the required properties. Consequently, this is a possible graph for f . because the graph has 4 critical points and 2 local maxima, property (iii) fails in keywords: Stewart5e, critical point, local maximum, inflection point 4 2 4 2 2 4 Use calculus to decide which of the following is the graph of f (x) = 2x  3x2/3 . 008 (part 1 of 1) 10 points because the graph is concave DOWN on (4, 2), property (iv) fails in 4 2 4 2 2 4 1. because the graph is concave UP on (0, 2), and property (vi) fails in Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann After differentiation of f (x) = 2x  3x2/3 we see that 2 2(x1/3  1) f (x) = 2  1/3 = ; x x1/3 in particular, (i) f has critical points at x = 0, 1. Differentiating again we next see that 2 , f (x) = 3x4/3 from which it follows that (ii) f > 0, so f concave up, on (, 0), 3. corand again (iii) f > 0, so f concave up, on (0, ) ; 9 2. in particular, f has a local minimum at x = 1. Of the five graphs only rect 4. has these properties. keywords: Stewart5e, fractional point, critical point, local maximum, concavity 009 (part 1 of 1) 10 points Which function could have 5. 2 Explanation: Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann as its graph on [ 0, 2]? sin x 1. f (x) = correct 2 + cos x sin x 2. f (x) = 2  cos x 3. f (x) = sin x cos x  2 Thus f (x) = sin x 2  cos x 10 has critical points when cos x = 1/2, i.e., at x = /3, 5/3, while f (x) = sin x 2 + cos x 4. f (x) = sin x 5. f (x) =  sin x sin x 6. f (x) =  2 + cos x Explanation: As f (/2) > 0, this already eliminates the three choices for f in which f (/2) < 0, leaving only the possibilities sin x , 2  cos x sin x , sin x . 2 + cos x has critical points when cos x = 1/2, i.e., at x = 2/3, 4/3. Consequently, the graph can only be that of f (x) = sin x . 2 + cos x keywords: graph, trig function, local extrema, yintercept, 010 (part 1 of 1) 10 points A function f is continuous and twicedifferentiable for all x except x = 2, 3. If its graph has exactly one critical point and one inflection point, which of the following could be the graph of f ?
7 6 1.5 4 3 2 1 0 1 2 3 4 5 6 7 To decide among these we check critical points because the graph has a local maximum in (/2, ) and a local minimum in (, 3/2). Now sin x has critical points at x = /2, 3/2, eliminating this choice. On the other hand, by the Quotient Rule, d sin x dx 2  cos x = cos x(2  cos x)  sin2 x (2  cos x)2 2 cos x  1 = , (2  cos x)2 while d sin x dx 2 + cos x cos x(2 + cos x) + sin2 x = (2 + cos x)2 = 2 cos x + 1 . (2 + cos x)2 6 4 2 4 2 2 4 6
6543210 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 7 6 2.5 4 3 2 1 0 1 2 3 4 5 6 7 6 4 2 4 2 2 4 6
6543210 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann
7 6 6 3.5 4 4 3 2 2 1 0 1 2 4 6 2 4 2 2 3 4 4 5 6 6 7 7 6543210 1 2 3 4 5 6 6 6 4.5 4 4 3 2 2 1 0 1 2 4 6 2 4 2 2 3 4 4 5 6 6 7 7 6543210 1 2 3 4 5 6 6 6 5.5 4 4 3 2 2 1 0 1 2 4 6 2 4 2 2 3 4 4 5 6 6 7 7 6543210 1 2 3 4 5 6 6 6 6.5 4 4 3 2 2 1 0 1 2 4 6 2 4 2 2 3 4 4 5 6 6 7 6543210 1 2 3 4 5 6 11 Consequently, if f is to have exactly one critical point and one inflection point its graph can only be
7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 10 6 4 2 4 2 2 4 6
6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1011 7 8 9 10 11 2 4 6 8 10 8 10 keywords: critical point, inflection point, graph, vertical asymptote
7 8 9 10 11 011 (part 1 of 1) 10 points A function f is continuous and twicedifferentiable for all x = 1. Its derivatives have the properties (i) f (1) = 0,
7 8 9 10 11 8 10 (ii) f (iii) f > 0 on (, 2) < 0 on (2, 1). (1, ), correct 8 10 If the lines x = 1 and y = 1 are asymptotes of the graph of f , which of the following could be the graph of f ?
7 6 6 5 1. 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 7 8 9 10 11 Explanation: If f is differentiable everywhere except at x = 2, 3, then already 3 of the graphs can be eliminated since these have vertical asymptotes at x = 3, 2. On the other hand, the critical points of f occur when f (x) = 0, i.e., when the tangent line to the graph is horizontal, while the inflection points occur where the concavity changes sign. Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann
7 6 6 2. 5 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 correct 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 7 6 6 5 5. 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 12 7 6 6 5 3. 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 Explanation: Since y = 1 is a horizontal asymptote, two of the graphs can be eliminated immediately by inspection. On the other hand, all the remaining graphs exhibit the same behaviour to the left of the vertical asymptote at x = 1. But to the right of the vertical aymptote, only one graph is concave up. Consequently,
7 6 6 5 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 7 6 6 5 4. 4 4 3 2 2 1 0 1 8 6 4 2 2 4 6 2 2 3 4 4 5 6 6 7 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 is the only graph having all the stated properties. keywords: Stewart5e, asymptote, graph, concavity 012 (part 1 of 1) 10 points Which of the following is the graph of f (x) = x  1 x2 when dashed lines indicate asymptotes? Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann 6. 1. 13 7. 2. 8. 3. correct 4. 5. Explanation: Note first that the vertical asymptote must be the yaxis. On the other hand, since 1 lim f (x)  x = lim  2 = 0 , x x x the line y = x is a slant asymptote. This already eliminates four of the eight possible graphs. To identify which of the remaining four graphs is that of f we can use concavity because each graph is always concave up or always concave down, on either side of the yaxis. Now 6 2 f (x) =  4 , f (x) = 1 + 3 , x x in which case f (x) < 0 on (, 0) , while f (x) < 0 on (0, ) . Thus the graph of f is concave down both to the left and the right of the yaxis. Jolley, Garrett Homework 11 Due: Nov 6 2007, 3:00 am Inst: R Heitmann Consequently, the graph of f is 14 keywords: graph, vertical asymptote, rational function, slant aymptote, concavity, ...
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