Math 115 Practice for Exam Generated October 30, 017 Name: SOLUTIONS Instructor: Section Number: 1. This exam has 5 questions. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.. Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you hand in the exam. 3. Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam. 4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate. 5. You may use any calculator except a TI-9 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3 5 note card. 6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 7. You must use the methods learned in this course to solve all problems. Semester Exam Problem Name Points Score Fall 016 7 10 Winter 015 13 Fall 01 5 14 Winter 013 9 10 Winter 016 3 7 9 Total 56 Recommended time (based on points): 54 minutes
Math 115 / Exam (November 14, 016) page 8 7. [10 points] Suppose f(x) is a continuous function defined for all real numbers whose derivative and second derivative are given by ( ) (x 3) f (x) = arctan (x+1) 3 and f (x) = (x+1)3 (x 3)(13 x) (x+1)[(x+1) 6 +(x 3) 4 ]. a. [5 points] Find all critical points of f(x) and all values of x at which f(x) has a local extremum. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all local extrema. For each answer blank below, write none if appropriate. Solution: First we find the critical points, which occur when f (x) = 0 or f (x) does not exist. Since f (x) = 0 when x = 3 and f (x) is not defined when x = 1, the two critical points are x = 1 and x = 3. Next we must determine whether there is a local min, local max, or neither at each critical point. The second derivative test is inconclusive, because f ( 1) does not exist and f ( 3 ) = 0. So we must use the first derivative test. Notice that arctan preserves signs (i.e., arctan(x) is positive when x is positive ( and arctan(x) ) is negative when x is (x 3) negative) so we only need to check the sign of (x+1) 3. The factor (x 3) is always positive, while (x+1) 3 is negative when x < 1 and positive when x > 1. This gives us the resulting signs: Interval x < 1 1 < x < 3 x > 3 Sign of f + (x) = + + = + + + = + So f(x) has a local minimum at x = 1 and no local maxima. Answer: Critical point(s) at x = 1, 3 Local max(es) at x = None Local min(s) at x = 1 b. [5 points] Find the x-coordinates of all inflection points of f(x). If there are none, write none. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all inflection points. Solution: First we find the candidate inflection points, which occur when f (x) = 0 or f (x) does not exist. We can see that f (x) = 0 at x = 3 13 and and that f (x) is undefined when x = 1. To determine whether these are actually inflection points (where concativty changes), we must test the sign of the second derivative on either side of each of the points. We find the following: Interval x < 1 1 < x < 3 3 < x < 13 x > 13 Sign of f + (x) + = + + + + = + + + + + = + + + + + = The inflection points of f(x) occur at the points where the sign of the second derivative changes, that is, at x = 3 13 and x =. 3 Answer: Inflection point(s) at x =, 13 Fall, 016 Math 115 Exam Problem 7 Solution
Math 115 / Exam (March 4, 015) page 3. [13 points] A function g(x) and its derivative are given by g(x) = x3 +34x +73x+5400 (x+30) 4 and g (x) = (x 6) (x 10) (x+30) 5. a. [8 points] Find all critical points of g(x) and all values of x at which g(x) has a local extremum. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all local extrema. Solution: To find critical points we (i) solve for x in the equation g (x) = 0 and (ii) look for points in the domain of the function where the derivative is undefined. The critical points occur at x = 6 and x = 10. Note that x = 30 is not a critical point since it is not in the domain of the function. To test whether each of these critical points is a local extremum, we consider the sign of g (x) on the intervals on each side of the critical points and then apply the First Derivative Test. Note that (x 6) is positive for all x 6, and (x+30) 5 is positive for all x > 30. Finally, note that x 10 is positive for x > 10 and negative for x < 10. We summarize the resulting signs of the derivative in the table below. Interval 30 < x < 6 6 < x < 10 x > 10 Sign of g + (x) = + + = + + + = + + + We see that g(x) does not have a local extremum at x = 6 and that, by the First Derivative Test, g(x) has a local maximum at x = 10. (For each answer blank below, write none in the answer blank if appropriate.) Answer: critical point(s) at x = 6, 10 Answer: local min(s) at x = NONE Answer: local max(es) at x = 10 b. [5 points] Find the values of x that minimize and maximize g(x) on the interval [0, ). Use calculus to find your answers, and be sure to show enough evidence that the points you find are indeed global extrema. Solution: The only local extremum is the local maximum at x = 10 so since g is continuous on the interval [0, ),this must also be the global max. (Alternatively, note that from part (a), we know that g(x) is increasing for 0 x 10 and decreasing for x 10, so g(10) must be the global maximum value of g(x) on the interval [0, ).) To check for a global min we need to consider the ends of the interval. Now, g(0) = 1 150 and 1 lim g(x) = 0, which is less than x 150. This implies that there is no global minimum of g(x) on the interval [0, ). (For each answer blank below, write none in the answer blank if appropriate.) Answer: global min(s) at x = NONE Answer: global max(es) at x = 10 Winter, 015 Math 115 Exam Problem Solution
Math 115 / Exam (November 13, 01) page 6 5. [14 points] The function f is has a continuous second derivative on the interval 10 x 19. Some values of its derivative function f are given in the table below. x 10 11 1 13 14 15 16 17 18 19 f (x) -34-3 -1 - -3 31 6 70 66 37 a. [4 points] f has exactly one inflection point on the interval 15 x 19. Given the information provided, give the smallest x interval on which this inflection point is guaranteed to lie, making it clear whether your endpoints are included. Solution: 16 < x < 18 or (16,18). b. [8 points] f has exactly four critical points, with x-values 11., 11.7, 1.6, and 14., respectively. Classify each point as a local minimum, a local maximum, or neither, given that f has either a local maximum or a local minimum at x = 11.. For each point below, circle only one option. At x = 11., f has a local maximum a local minimum At x = 11.7, f has a local maximum a local minimum neither At x = 1.6, f has a local maximum a local minimum neither At x = 14., f has a local maximum a local minimum neither c. [ points] Is there at least one inflection point on the interval 11 < x < 1? (Circle one.) Yes No Not possible to determine Fall, 01 Math 115 Exam Problem 5 Solution
Math 115 / Exam (March 1, 013) page 10 9. [10 points] The function f(x) is twice-differentiable. Some values of f and f are given in the following table. In addition, it is known that f (x) is positive. x 0 1 3 4 f(x) 7 6 7 9 1 f (x) - 1 1 4 No partial credit will be given on any part of this problem. a. [4 points] Circle any statement which is true, and draw a line through any statement which is false. (i.) For some value of x with 0 < x < 1, f has a critical point. (ii.) For some value of x with 1 < x <, f has a critical point. (iii.) For some value of x with < x < 3, f has a critical point. (iv.) For some value of x with 3 < x < 4, f has a critical point. b. [3 points] If possible, find the global minimum value of f(x) on the closed interval [0,4]. (Give the y-coordinate, not the x-coordinate.) Do not give an approximation. If it is not possible to find it exactly, write IT IS NOT POSSIBLE TO FIND IT EXACTLY. Solution: We know that f(x) is decreasing until some critical point p between 0 and 1 and is increasing after that (because we know that f goes from negative to positive between 0 and 1 and never becomes negative again, since f > 0). The minimum occurs at some point that s not included in the table, so IT IS NOT POSSIBLE TO FIND IT EXACTLY. c. [3 points] If possible, find the global maximum value of f(x) on the closed interval [0,4]. (Give the y-coordinate, not the x-coordinate.) Do not give an approximation. If it is not possible to find it exactly, write IT IS NOT POSSIBLE TO FIND IT EXACTLY. Solution: The only critical point is a local minimum, so the maximum value must be at one of the endpoints. Looking at the table, we see the maximum value is 1 (when x = 4). Winter, 013 Math 115 Exam Problem 9 Solution
Math 115 / Final (April 1, 016) page 8 7. [9 points] Consider the family of functions given by f(x) = e x +Ax+B for constants A and B. a. [6 points] Find and classify all local extrema of f(x) = e x +Ax+B. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all local extrema. For each answer blank, write none if appropriate. Your answers may depend on A and/or B. Solution: First, we find the critical points of f(x). Notice that f(x) is differentiable, so f(x) only has critical points where f (x) = 0. Since f (x) = (x+a)e x +Ax+B, the only critical point of f(x) occurs where x+a = 0, i.e., at x = A. We test whether this critical point is a local maximum, a local minimum, or neither. Applying the First Derivative Test: For x < A : x+a < 0 and ex +Ax+B > 0, so f (x) < 0. For x > A : x+a > 0 and ex +Ax+B > 0, so f (x) > 0. Hence, f (x) changes from negative to positive at x = A, so f(x) has a local minimum at x = A (and no local maxima). (Note that we could instead apply the Second Derivative Test: f (x) = (x+a)(x+ A)(e x +Ax+B +(e x +Ax+B ) = ( (x+a) + ) e x +Ax+B which is always positive (since both factors are always positive). So in particular f ( A/) > 0 so f(x) has a local minimum at x = A/. Answer: Local min(s) at x = A Answer: Local max(es) at x = none b. [3 points] Find exact values of the constants A and B so that the point (3,1) is a critical point of f(x) = e x +Ax+B. Solution: As we showed in part a., f(x) has its only critical point at x = A, so now we must have that A = 3 so A = 6. To find B, we use the fact that (3,1) is a point on the graph of y = f(x) (so f(3) = 1). Hence, A = 6 and B = 9. 1 = f(3) = e 3 +( 6)(3)+B = e B 9 ln(1) = ln ( e B 8) 0 = B 9 B = 9 Answer: A = 6 and B = 9 Winter, 016 Math 115 Exam 3 Problem 7 Solution