Calculus B Exam III (Page 1) May 11, 2012

Transcription

1 Calculus B Eam III (Page ) May, 0 Name: Instructions: Provide all steps necessary to solve the problem. Unless otherwise stated, your answer must be eact and reasonably simplified. Additionally, clearly indicate the value or epression that is your final answer. Please put your name on all THREE pages! EXTREMELY IMPORTANT!!! Calculators are NOT allowed for Page but allowed for Pages and 3. However, you MUST turn in this page BEFORE you even pull out your calculator!. Suppose f () = (a) Find all critical numbers of f. (b) Use the Second Derivative Test to determine if each critical point is a local minimum or a local maimum. (c) Find all critical numbers of f. That is, find all values c in the domain of f where f (c) = 0 or does not eist. (Note: These values are called the second-order critical points of f.) (d) Determine whether each second-order critical point is an inflection point.. Find f () if f () = 4 +, where f () = and f (0) =.

2 3. Evaluate the following limits, if they eist. Be sure to indicate if your answer is + or. ln (a) lim (b) lim 0 3sin cos tan (c) lim e 4. Find the most general antiderivative of each of the following. (a) f( ) (b) g ( ) sin 4sec tan 3 e ( ) /

3 Calculus B Eam III (Page ) May, 0 Name: Instructions: Provide all steps necessary to solve the problem. Unless otherwise stated, your answer must be eact and reasonably simplified. Additionally, clearly indicate the value or epression that is your final answer. For optimization problems, you must demonstrate that your answer is correct by using the CIM, the SDT, or the FDT (and don t forget to include the domain of the function to be optimized). EXTREMELY IMPORTANT!!! Calculators are allowed for Pages and 3 (but really only needed for the Newton s method problem). However, you MUST turn in Page BEFORE you even pull out your calculator! 5. For a differentiable function f, what is the Newton s method formula to find n based on the previous approimation? 6. Use Newton s method to find where the curves y = and y = cos intersect. Your initial approimation must be an integer. Provide the values of all subsequent approimations. Give your answer correct to 4 decimal places. (Important!! Make sure your calculator is in radians! Provide ALL steps necessary to verify the solution.)

4 7. Find a positive number in the interval [0., 5] for which the sum of its reciprocal and its square is (a) as large as possible? (b) as small as possible?

5 Calculus B Eam III (Page 3) May, 0 Name: 8. Evaluate lim 4 / 0, if it eists. Be sure to indicate if your answer is + or.

6 9. A poster is to have an area of 80 in with -inch margins at the bottom and sides and a -inch margin at the top. What dimensions will give the largest printed area? [Instructor s Note: Do NOT use decimal approimations!] Etra Credit: Suppose f and g are both concave upward on (,). Under what condition on f will the h ( ) f g ( ) be concave upward? Eplain. composite function

7 Calculus B Solutions to Eam III. (a) Since f () = 9 + = 3( ) = 3(3 )( + ), then setting f (c) = 0 implies that the two critical numbers are c = / 3 and c =. (b) The second derivative is f () = 8 + = 6(3 + ). Since f (c ) = 6(3 / 3 + ) = 4 is positive, then, by the SDT, f is concave upward at c implying that c is a local minimum. Since f (c ) = 6(3( ) + ) = 4 is negative, then, by the SDT, f is concave downward at c implying that c is a local maimum. (c) Since f () = 6(3 + ), then setting f (c) = 0 implies that the second-order critical point is c 3 = / 3. (d) Since f () > 0 if > / 3 and f () < 0 if < / 3, then c 3 is an inflection point.. The antiderivative of is 4 f ( ) f ( ) C. 5 5 Since f (), then f() C and so C = 3 / 0. The antiderivative of 5 f( ) is Since f (0), then f(0) D. Thus, f ( ) f ( ) D (a) ln lim (IF of type 0 / 0 ) LHR lim = = (b) lim 0 3sin cos tan 3cos cos ( sin ) sec 3cos0 cos0sin0 sec LHR lim 0 = = = 3 / (IF of type 0 / 0 ) (c) lim e (IF of type 0) = lim e (IF of type / )

8 LHR LHR lim e lim ( e ) = lime = 0 (IF of type / ) (simplify) 4. (a) Since then the antiderivative is / f( ) / / /, / F( ) ln C ln. C (b) The antiderivative is G ( ) cos 4sec 3e sin C. 5. The formula presented in class (and in the tet) was in terms of n+. This one is slightly different because it is asking you to find n in terms of n. The answer is f( n ) n n. f ( ) 6. n Consider the function f () = cos. We need to find such that f () = 0. Since f () = + sin, then the Newton s Method formula is f ( n) n cosn n n n. f ( n) sinn The closest integer to the -coordinate to the root is. So that will be our initial approimation. Subsequent iterations are given in the following table. (An initial approimation of = 0 would give =, and the same values would follow, incremented by.) n n+ =

9 cos sin cos( ) sin( ) cos(0.739) sin(0.739) The root of f (rounded to four decimal places) is The point of intersection to the two curves is (0.739,.0739). 7. Let be a positive number in [0., 5]. Let S be the sum of the reciprocal and the square of. That is, S() = / + = +. The domain of this function is [0., 5]. To optimize S, we differentiate to get 3 3 S( ). Setting equal to zero and solving gives the critical number 3 S () = Is this value a minimum or a maimum? 3 /3 To see, use the SDT. Since S () = 3 + > 0, then = /3 must be a minimum. To determine the maimum, use the CIM. The domain of the function is [0., 5] and the critical number occurs when = /3. Since S(0.) S ( ) ( ) 0 0.0, 3 S ( ) ( ).890, and /3 S (5) 5 5., 5 then = 5 corresponds to the absolute maimum value and = /3 is the location of the absolute minimum value. Instructor s Note: My eplanation above is longer than it needs to be. The CIM will determine both the minimum and maimum, but I wanted to illustrate the minimum using the SDT.. Overall, the numbers = /3 gives the minimum and = 5 a gives the maimum. 8. lim 4 / 0 is an IF of type. (Recall that the alternate form of the eponential function is e = ep().) 0 / lim 4 (use the cancellation eqn: A = ep(ln A)) / = lim epln( 4 ) 0 = lim ep ln( 4 ) 0 = eplim ln( 4 ) 0 (simplify using properties of logarithms) (distribute limit through eponential function) (IF of type 0)

10 = LHR = = = ln( 4 ) eplim 0 d d ln( 4 ) ep lim 0 d d 4 4 eplim 0 4 eplim ep 4 0 = ep(4) = e 4 (IF of type 0 / 0 ) (simplify) 9. Solution #: Let be the width of the poster and y the height. y Since y = 80, then y = 80 /. The area of the printed region is A = ( )(y 3). Using the values determined above for y, then A ( ) ( ) The domain of A() is [, 60]. Instructor s Note: If you took the relation y = 80 and solved for, then your area equation would be A(y) = 86 y 540y with domain [3, 90]. To optimize, the derivative is ( ) Setting A( ) 0 A and solving gives

11 We can disregard the negative solution since it is outside the domain of A(). 3 Use the SDT to show that this value is a maimum. Since A( ) 70 is negative for all values in the domain of A, then 30 in corresponds to the absolute maimum. Thus, the dimensions of the poster that gives the largest printed area is y in in in 3 30 in 0.95 in 6.43 in. Solution #: Let be the width of the printed area of the poster and y the height. y Since ( + )(y + 3) = 80, then 80 y 3. The area of the printed region is A = y. Using the values determined above for y, then 80 A ( ) The domain of this function is [0, 58]. To optimize, the derivative is

12 A( ) 80 3 ( ) ( ) Setting A( ) 0 and solving gives ( ) ( ) ( ) ( ) We can disregard the negative solution since it is outside the domain of A(). Thus the only solution is Use the SDT to show that this value is a maimum. Since A( ) 70( ) is negative for all values in the domain of A, then 0 in corresponds to the absolute maimum. Thus, the dimensions of the poster that gives the largest printed area is 80 y3 0 in in in 3 30 in 0.95 in 6.43 in. Instructor s Note: Even if you got the right answer, there were some automatic deductions for not including information that was emphasized in class. Reason Percentage Not including the domain 5 % Including the incorrect domain 3 % No justification of the maimum value 8 % Not labeling justification as FDT, SDT, or CIM % Providing approimations, not eact values 4 % Suggestions for improvement: Draw a picture Label functions Eplain what you are doing! Etra Credit: If h() is concave upward, then by the Concavity Test, h() > 0. Differentiating h() once using the Chain Rule gives h( ) f g( ) g( ) Differentiating again using the Product Rule and the Chain Rule gives

13 d d h( ) f g( ) g( ) g( ) f g( ) d d ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) f g g g f g g f g g f g g? Since f and g are both concave upward, then f () and g() are both positive. [g()] is automatically positive. The only thing left is f (g()). This must be positive in order for h() to be positive. So if we know that f is increasing on (,), then that would imply that f (g()) is positive. Condition: f is increasing on (,).

14 Calculus B Eam III (Page ) P Answer/Solution A(%) M(%) O a c = / 3 and c = b c is a local minimum c is a local maimum c c 3 = / d c 3 is an inflection point f ( ) a b 3 / c a F( ) ln C b G ( ) cos4sec 3e sin C Overall