HW 11 - Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels This print-out should have 12 questions. Multiple-choice 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. True: the graph of f is concave UP on (-5, -3). B. 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. C. True: 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. keywords: True/False critical point, local extreme, 4 2 002 (part 1 of 1) 10 points In drawing the graph -4 -2 2 which of the following properties are satisfied? A. B. C. f (x) > 0 on (-5, -3), f has exactly 3 critical points, f has exactly 2 local extrema. P 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. 2 1 1. f (x) = - + 3x - x2 - x3 3 3 2. f (x) = 10 1 - 5x + 3x2 - x3 3 3 1. A only 2. all of them 3. B only 4. C only 5. none of them 6. B and C only 7. A and B only 8. A and C only correct 8 1 3. f (x) = - + 3x + x2 - x3 3 3 4. f (x) = 5. f (x) = 1 8 - 3x + x2 + x3 correct 3 3 14 1 - 3x - x2 + x3 3 3 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 , f (x) = 2(c + x) . {5, -3}, {-3, 1} . 2. 4 2 -4 -2 -2 -4 2 4 2 1. 4 2 -4 -2 -2 -4 2 4 Since the graph has a local minimum at x = 1, it follows that b + 2c + 1 = 0, c+1 > 0 , 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 8 f (x) = - 3x + x2 + x3 . 3 3 -4 keywords: polynomial, local maximum 003 (part 1 of 1) 10 points If f is a function on (-4, 4) having exactly one critical point and the sign of f , f are given in f >0 f >0 f >0 f <0 -2 0 2 f <0 f >0 f <0 decide which of the following could be the graph of f . -2 1 = 1. 3 3. 4 2 2 -2 -4 4 4. 4 2 -4 -2 -2 -4 2 4 correct Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 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 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 (-, -2), f > 0 on (2, ). 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 2 4 correct -3 -2 -1 0 1 2 3 4 5 6 4 2 -2 -2 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 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 keywords: concavity, graph 005 (part 1 of 1) 10 points 2 4 Which of the following is the graph of f (x) = x2 ? x2 - 4 Dashed lines indicate asymptotes. 4 2 -2 -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 6 2 4 2 4 4 2 -2 -2 -4 -3 -2 -1 0 1 2 3 4 5 6 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 (-, -2) and on (2, ). Hence only 2 4 6 5 3.4 3 2 1 0 -1 -2 -3 -4 -5 -6 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 - 4 = 0 when x = 2, the graph of f will have vertical asymptotes at x = 2; on the other hand, since 4 2 -4 -2 -2 -4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x lim x2 = 1, x2 - 4 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 - 4) - 2x3 (x2 - 4)2 = - Thus f (x) > 0, while f (x) < 0, x > 0, x < 0, (x2 8x . - 4)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 cor- 2 4 rect 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 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 -4 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 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. correct 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 Explanation: When f (x) = x2 x -4 4 2 -2 -2 -4 -3 -2 -1 0 1 2 3 4 5 6 the graph of f will have vertical asymptotes at x = 2 and x = -2 as indicated by the dashed vertical lines. In addition the graph will pass through the origin. On the other hand, f (x) = - x2 + 4 < 0 (x2 - 4)2 2 4 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 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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: Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 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 only 2 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 (iv) 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) = 3x2/3 - 2x . 008 (part 1 of 1) 10 points because the graph is concave UP on (0, 2), property (v) fails in 4 2 -4 -2 2 4 1. because (0, 1) is not an inflection point, and property (vi) fails in Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels After differentiation of 2. corwe see that f (x) = in particular, rect (i) f has critical points at x = 0, 1. Differentiating again we next see that 3. f (x) = - 2 , 3x4/3 2 x1/3 -2 = 2(1 - x1/3 ) ; x1/3 f (x) = 3x2/3 - 2x 9 from which it follows that (ii) f < 0, so f concave down, on (-, 0), and again (iii) f < 0, so f concave down, on (0, ) ; 4. in particular, f has a local maximum at x = 1. Of the five graphs only 5. has these properties. keywords: Stewart5e, fractional point, critical point, local maximum, concavity 009 (part 1 of 1) 10 points Explanation: Which function could have Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels while d cos x - dx 2 + sin x 2 = sin x(2 + sin x) + cos2 x (2 + sin x)2 = Thus as its graph on [ 0, 2]? 1. f (x) = - cos x 2. f (x) = cos x correct sin x - 2 cos x 3. f (x) = 2 - sin x f (x) = 1 + 2 sin x . (2 + sin x)2 cos x sin x - 2 10 has critical points when sin x = 1/2, i.e., at x = /6, 5/6, while f (x) = - cos x 2 + sin x 4. f (x) = cos x 5. f (x) = cos x 2 + sin x cos x 6. f (x) = - 2 + sin x has critical points when sin x = -1/2, i.e., at x = 7/6, 11/6. Consequently, the graph can only be that of f (x) = cos x . sin x - 2 Explanation: Since the graph has negative y-intercept, the inequality f (0) < 0 must hold; this already eliminates the three choices for f in which f (0) > 0, leaving only the possibilities cos x cos x , - cos x , - . sin x - 2 2 + sin x To decide among these we check critical points because the graph has a local maximum in (0, /2) and a local minimum in (/2, ). Now - cos x has critical points at x = 0, 2, eliminating this choice. On the other hand, by the Quotient Rule, d cos x dx sin x - 2 - sin x(sin x - 2) - cos2 x = (2 - sin x)2 = 2 sin x - 1 , (sin x - 2)2 keywords: graph, trig function, local extrema, y-intercept, 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 two critical points 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 6 4 2 -4 -2 -2 -4 -6 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 7 6 2.5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 7 6 3.5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 7 6 4.5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 7 6 5.5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 7 6 6.5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 11 6 4 2 -4 -2 -2 -4 -6 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 6 4 2 -4 -2 -2 -4 -6 2 4 6 8 10 correct 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. Consequently, if f is to have exactly two critical points and one inflection point its graph can only be 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 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 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 6 4 2 -4 -2 -2 -4 -6 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 keywords: critical point, inflection point, graph, vertical asymptote 011 (part 1 of 1) 10 points 6 4 2 -4 -2 -2 -4 -6 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 A function f is continuous and twicedifferentiable for all x = 1. Its derivatives have the properties (i) f (-1) = 0, (ii) f (iii) f > 0 on (-, -2) < 0 on (-2, 1). (1, ), 6 4 2 -4 -2 -2 -4 -6 -6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 If the lines x = 1 and y = 2 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 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels 7 6 5 4. 4 3 2 1 0 -1 -8 -6 -2 -3 -4 -5 -6 -7 7 -9 -8 -7 -6 6 5 5. 4 3 2 1 0 -1 -8 -6 -2 -3 -4 -5 -6 -7 -9 -8 -7 -6 12 6 4 2 -4 -2 -2 -4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 2 4 6 6 4 2 -4 -2 -2 -4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 7 6 6 5 2. 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 2 4 6 7 6 6 3. 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 Explanation: Since y = 2 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 is the only graph having all the stated properties. Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels keywords: Stewart5e, asymptote, graph, concavity 012 (part 1 of 1) 10 points Which of the following is the graph of f (x) = 1 -x x 6. 13 5. when dashed lines indicate asymptotes? 1. 7. correct 2. 8. 3. Explanation: Note first that the vertical asymptote must be the y-axis. On the other hand, since 4. x lim f (x) + x = lim x 1 = 0, 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 Tran, Tony Homework 11 Due: Nov 6 2007, 3:00 am Inst: Samuels always concave down, on either side of the y-axis. Now f (x) = - in which case f (x) < 0 on (-, 0) , while f (x) > 0 on (0, ) . Thus the graph of f is concave down to the left of the y-axis, and concave up to the right. Consequently, the graph of f is 1 - 1, x2 f (x) = 2 , x3 14 keywords: graph, vertical asymptote, rational function, slant aymptote, concavity, ...
View Full Document

This note was uploaded on 04/27/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.

Ask a homework question - tutors are online