Math 295: Exam 3 Name: Caleb M c Whorter Solutions Fall /16/ Minutes

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1 Math 295: Eam 3 Name: Caleb M c Whorter Solutions Fall /16/ Minutes Write your name on the appropriate line on the eam cover sheet. This eam contains 10 pages (including this cover page) and 6 questions. Check that you have every page of the eam. Answer the questions in the spaces provided on the question sheets. Be sure to answer every part of each question and show all your work. If you run out of room for an answer, continue on the back of the page being sure to indicate the problem number. Question Points Score Total: 100

2 Math 295: Eam 3 2 of (15 points) Compute the following its. Be sure to to show all your work and justify your answer completely. (a) 0 e 1 2 L.H. 0 e 1 2 L.H. e 0 2 = 1 2 ln(1 + e 6 ) L.H. (b) 5 1 6e6 1 + e6 5 6e 6 5(1 + e 6 ) L.H. 36e 6 5(6e 6 ) 36 5(6) = 6 5 (c) ln 0 + ln 1/3 L.H /3 1 3 / /3 = 0

3 Math 295: Eam 3 3 of 10 ( 1 (d) 0 1 ) ( e e 0 e 1 ) e 1 e 0 e L.H. e 0 e + e = = (e) 0 (1 + 2) 3/ Let L (1 + 2) 3/. Then ln L ln (1 + 2) 3/ ln (1 + 2) ln (1 + 2) 0 3 L.H ln (1 + 2) = 6. Therefore, ln L = 6 so that L = e 6.

4 Math 295: Eam 3 of 10 f() 2. (15 points) l Hôpital s Rule is a useful tool but can fail to compute a it g() for the following reasons: f () A. does not eist. g () f() B. is not an indeterminate form. g() f () C. The problem loops, i.e. is essentially the same it as the initial g () it. D. f() or g() are not differentiable. For each of the following its, indicate one of the reasons above why l Hôpital s Rule does not apply to the it. C B C A sin 2 + cos e e e + e sin Choose one of the its above and compute it below. Be sure to justify your answer completely / 1/ sin 2 + cos 1/2 1/ sin 1 + cos e e 1 1 1/e e + e 1/e e = e sin 1/ 1/ sin = 2 = = = 1 = = =

5 Math 295: Eam 3 5 of (15 points) Complete each of the following parts. Be sure to show all your work, and justify your answers completely. (a) Verify that f() = satisfies the hypotheses of the Mean Value Theorem on [0, 3]. Find all numbers c satisfying the conclusions of the Mean Value Theorem on this interval. Observe f() = is continuous on [0, 3]. Because f () = is defined on [0, 3], f() is differentiable on (0, 3). By the Mean Value Theorem, there eists c (0, 3) such that f(3) f(0) = f (c)(3 0). But then we have f(3) f(0) = f (c)(3 0) 29 ( 1) = 3(3c 2 + 1) 30 = 3(3c 2 + 1) 10 = 3c = 3c 2 c 2 = 3 c = ± 3 Therefore, c = 3 satisfies the hypothesis of the Mean Value Theorem on [0, 3]. (b) Use the Mean Value Theorem to prove that if f () = 0 for all [a, b], then f() is constant on [a, b]. [Hint: Show f( 0 ) = f(a) for all a 0 b.] Since f () eists on [a, b], f() is differentiable (hence continuous) on [a, b]. Therefore, f() satisfies the Mean Value Theorem on [a, 0 ] for any 0 b. Then by the Mean Value Theorem, f( 0 ) f(a) = f (c)( 0 a) for some c [a, 0 ]. But f () = 0 for all [a, b]. Therefore, f( 0 ) f(a) = f (c)( 0 a) f( 0 ) f(a) = 0 f( 0 ) = f(a) This shows f() = f(a) for all [a, b]. But then f() is constant on [a, b]. Note that this also shows that if f () = g () on [a, b], then f() = g()+c for some constant C. The function H() := f() g() has derivative 0 because H () = f () g () = 0. But then H () = 0 so that H() is constant, i.e. H() = C. But then f() g() = C so that f() = g() + C.

6 Math 295: Eam 3 6 of 10. (20 points) A function f() and its derivatives f () and f () are given below: ( f() = 7 ) /3 f () = 7 3 1/3 ( 1) f 7( 1) () = 9 2/3 Answer the following questions. You may show your work on the net page. Be sure to indicate the part. (a) What are the intervals on which f() is increasing? (, 0) (1, ) (b) What are the intervals on which f() is decreasing? (0, 1) (c) Find all critical values for f(). Classify these critical values as local maima, local minima, or neither. Be sure to use the First or Second Derivative Test to justify your answer. = 0 is a local maima and = 1 is a local minima (d) What are the intervals on which f() is concave? (, 0) (0, 1 ) (e) What are the intervals on which f() is conve? ( 1, ) (f) Find the -values of any points of inflection on f(). = 1 (g) Find the absolute minimum and absolute maimum values of f() on [ 1, 7 ]. 11 is the absolute minimum while 0 is the absolute maimum.

7 Math 295: Eam 3 7 of 10 f() = ( 7 ) /3 f () = 7 3 1/3 ( 1) f () = 7( 1) 9 2/3 Setting f () = 0, we obtain = 0 and = 1. Therefore, = 0 is a maimum and = 1 is a minimum. Setting f () = 0, we obtain = 1. Notice also that f () is undefined at = 0. Since f () switches sign at = 1 and f() is defined at = 1, f() has a point of inflection at = 1. f( 1) = 11 f(0) = 0 f(1) = 3 ( ) 7 f = 0

8 Math 295: Eam 3 8 of (15 points) A rectangular bo has a square bottom and an open top. If only 2,700 cm 2 of material is available to construct the bo, what dimensions maimize the volume of the bo? Be sure to draw a picture and justify completely that these dimensions are optimal. We want to optimize V = lwh = s s h = s 2 h. But we know that But then we have Surface Area = 2, 700 cm 2 Surface Area = sh + s 2 sh + s 2 = 2700 sh = 2700 s 2 h = 2700 s2 s ( ) 2700 s 2 But then V = s 2 h = s s s3 =. Clearly, s [0, 2700], where s in each case the bo has V = 0. We have V s2 =. Now setting V = 0, we have s 2 = s 2 = 0 3s 2 = 2700 s 2 = 900 s = ± 900 = ± s = ±30 But then we must have s = 30 cm Then h = = dimensions then are We confirm this is a maimum, = 15. The

9 Math 295: Eam 3 9 of (20 points) The graph of f () for some function f() is plotted in the figure below. Based on this graph, complete the questions below. 10 y 8 6 f () (a) Find all the intervals on which f() is increasing. (, ) (5, ) (b) Find all the intervals on which f() is decreasing. (, 2) (2, 5)

10 Math 295: Eam 3 10 of 10 (c) Find all critical points for f(). Using the First or Second Derivative Test, classify these -values as locations of maimums, minimums, or neither. Therefore, = is a maima, = is a minimum, and = 2 is neither. (d) Find the intervals on which f() is concave. (, 1.5) (2,.2) (e) Find the intervals on which f() is conve. ( 1.5, 2) (.2, ) (f) Find the -values of any points of inflection on f(). = 1.5, = 2, and =.2

MA 131 Lecture Notes Chapter 4 Calculus by Stewart

MA 131 Lecture Notes Chapter 4 Calculus by Stewart 4.1) Maimum and Minimum Values 4.3) How Derivatives Affect the Shape of a Graph A function is increasing if its graph moves up as moves to the right and