Math 115 Practice for Exam 3

Transcription

1 Math 115 Practice for Exam 3 Generated November 17, 2017 Name: SOLUTIONS Instructor: Section Number: 1. This exam has 6 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. 2. 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-92 (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 Winter Winter Fall Winter Winter Winter Total 66 Recommended time (based on points): 72 minutes

2 Math 115 / Final (April 24, 2017) page 9 9. [9 points] A Math 115 coordinator is trying to create functions with certain properties in order to test students understanding of various calculus concepts. a. [5 points] He wants a function f(x) of the form { ax 2 +ax+be x for x < 0 f(x) = a+2cos(x) for x 0 where a and b are constants. Find all value(s) of a and b for which f(x) be differentiable at x = 0. Show enough work to justify your answer. Solution: In order for f(x) to be differentiable at x = 0, f(x) has to be continuous. Then lim +ax+be x = b = lim = a+2 = f(0). x 0 ax2 x 0 +a+2cos(x) Then b = a+2. If f(x) is differentiable then f(0+h) f(0) lim = 2a(0)+a+be 0 = a+b. h 0 h f(0+h) f(0) lim = 2sin(0) = 0. h 0 + h Hence a+b = 0. Using both equation we obtain a = 1 and b = 1. b. [4 points] The coordinator also wants a function g(x) = cx e x, where c is a constant, so that g(x) has at least one critical point. What condition(s) on c will make this true? Find the x-values of all critical points in this case. Your answer may be in terms of c. Solution: The function g(x) has critical points at values of x that satisfy g (x) = c e x = 0. Then the only potential critical point is x = ln(c). This critical point exists only if c > 0. Winter, 2017 Math 115 Exam 3 Problem 9 Solution

3 Math 115 / Final (April 24, 2017) page 7 7. [9 points] Consider the family of functions f(x) = ax 2 e bx where a and b are positive constants. Note that f (x) = ax(2 bx)e bx. a. [4 points] Find the exact values of a and b so that f(x) has a critical point at (4,e 2 ). Solution: Since (4,e 2 ) is a critical point of f(x), then f (4) = 0 or 0 = 4a(2 4b)e 4b. From this equation we get that 2 4b = 0 (since a,e 4b > 0). Then b = 0.5. We also know that the point (4,e 2 ) is in the graph of y = f(x), then e 2 = 16ae 4b. Plugging the value of b, we get e 2 = 16ae 2. This yields 1 = 16a, so a = b. [5 points] Using your values of a and b from the previous part, find and classify the local extrema of f(x). Use calculus to find and justify your answers, and be sure to show enough evidence that you have found them all. For each answer blank, write none if appropriate. Solution: With the values found above f (x) = 1 16 x(2 0.5x)e 0.5x. The critical points are found by solving f (x) = 1 16 x(2 0.5x)e 0.5x = 0 In this case, we have x = 0 or 2 0.5x = 0 (since.e 0.5x > 0). Hence the critical points are x = 0 and x = 4. To classify them we use the first derivative test: OR f ( 1) = 1 16 ( 1)(2 0.5( 1))e 0.5( 1) = f (1) = 1 16 (2 0.5)e 0.5 = f (5) = 1 16 (5)(2 0.5(5))e 0.5(5) = f ( 1) = ( )(+)(+) = f (1) = (+)(+)(+) = + f (5) = (+)( )(+) = Answer: Local max(es) at x = 4 Local min(s) at x = 0 Winter, 2017 Math 115 Exam 3 Problem 7 Solution

4 Math 115 / Exam 2 (November 12, 2013) page 8 9. [13 points] Consider the function f(x) = axlnx bx with domain x > 0, where a and b are positive constants. Note that this function has exactly one critical point. a. [3 points] Find f (x). Solution: ( f (x) = a (1)lnx+x 1 ) b = alnx+a b x b. [4 points] For which values of a and b does f(x) have a critical point at (e, 2)? Solution: We want f (e) = 0 and f(e) = 2. We plug x = e into f (x) to get f (e) = alne+a b = 2a b = 0 which tells us that 2a = b. Plugging x = e into f(x) gives f(e) = aelne be = (a b)e = 2. Using 2a = b, we get (a (2a))e = 2 so that a = 2/e. Since b = 2a, we get b = 4/e. c. [3 points] Using your values of a and b from part (b), is the critical point from (b) a local maximum, local minimum, or neither? Justify your answer. Solution: The second derivative of f(x) is f (x) = a/x = (4/e)/x. Then f (e) = 4/e 2 > 0 so the graph of f(x) is concave up at x = e. This means that f(x) has a local minimum at x = e. d. [3 points] Using your values of a and b from part (b), find the x-coordinates of any inflection points of f(x) or show that f(x) has no inflection points. Solution: The second derivative of f(x) is f (x) = a/x = (4/e)/x. This is continuous and positive for all values of x in the domain x > 0 of f(x). Since the second derivative never changes sign, f(x) has no inflection points. Fall, 2013 Math 115 Exam 2 Problem 9 Solution

5 Math 115 / Final (April 19, 2012) page [13 points] Use the family of functions of the form f(x) = ax ln(1+e bx ) to answer the following questions. The constants a and b are both positive. a. [4 points] Use the given formula for f(x) to give an explicit expression for the limit definition of f (x). Check your expression carefully, as no partial credit will be given on this part of the problem. Do not evaluate your expression. Solution: f a(x+h) ln(1+e b(x+h) ) ax+ln(1+e bx ) (x) = lim. h 0 h b. [4 points] Compute f (x) using the rules of differentiation. Do not try to evaluate your expression from (a). Solution: f (x) = a bebx 1+e bx. c. [5 points] When a < b, the function f(x) has a critical point at x = 1 ( ) a b ln. b a Using the second-derivative test, determine whether this critical point is a local maximum, local minimum, or neither. Solution: The second derivative is f (x) = (1+ebx )b 2 e bx be bx be bx (1+e bx ) 2 = b2 e bx (1+e bx ) 2. This is always negative, so the critical point is a local maximum. Winter, 2012 Math 115 Exam 3 Problem 8 Solution

6 Math 115 / Exam 2 (March 20, 2012) page 5 4. [12 points] Consider the family of functions where a and b are positive constants. f(x) = ax e bx, a. [4 points] Any function f(x) in this family has only one critical point. In terms of a and b, what are the x- and y-coordinates of that critical point? Solution: We find f (x) = a be bx, so setting this equal to zero and solving for x shows that there is a critical point at x = 1 b ln ( a b The y-coordinate of the critical point is ( 1 ( b)) a y = f b ln = a ( a ) b ln a b b. ). b. [4 points] Is the critical point a local maximum or a local minimum? Justify your answer with either the first-derivative test or the second-derivative test. Solution: The second derivative of f(x) is f (x) = b 2 e bx, so ( 1 f b ln a ) = b 2 a b b = ab, which is negative since a and b are both positive. Therefore, the second derivative test tells us that the critical point is a local maximum. c. [4 points] For which values of a and b will f(x) have a critical point at (1,0)? Solution: We need 1 b ln a b = 1 and a b ln a b a b = 0. The first equation rearranges to ln( a b ) = b, and if we plug this into the second equation, we obtain a a b = b = 0 b = 1. Plugging this into either equation and solving for a gives a = e. Winter, 2012 Math 115 Exam 2 Problem 4 Solution

7 Math 115 / Final (April 25, 2011) page 2 1. [10 points] Find a formula for a function of the form f(x) = 1 a + x + bx 2 which has a local minimum at (2,1/2). Be sure to show that your function has a minimum at (2,1/2). Solution: A local minimum will occur when f (x) = 0 or is undefined. The former is when 2bx + 1 (bx 2 + x + a) 2 = 0. The numerator is zero when x = 1 2b, and the denominator is zero on either side of this value, when x = 1 2b ± 2b 1 1 4ab. The first of these gives the local minimum we want: if we let 1 2b = 2, we have b = 1 4. This gives f (x) = 1 2 x+1 ( 1 4 x2 +x+a) 2, so that if x < 2 we have f (x) < 0, and if x > 2, f (x) > 0. Thus if b = 1 4, x = 2 is a local minimum. To require that the minimum occur at (2,1/2), we want f(2) = 1 a+2 1 = 1/2, so that a = 1. Winter, 2011 Math 115 Exam 3 Problem 1 Solution