Solution to Exam 1

Name: __Solution___Methods of CalculusSummer 2004Exam1 - June 2, 2004Instructions: Compare the answers in this solution document to your version of the test..SECTION A - Multiple ChoiceChoose the best answer choice.1. Find the equation of the line passing through point 3,4 and perpendicular to the line8x y 8 0.(A) y 1 8 x 358(B) y 1 8 x 35 8(C) y 8x 35 (D) y 1 8 x 35Solution: (B)Since required line is perpendicular to 8x y 8 0, rewritten y 8x 8, then therequired slope is 1 . The equation of the line is thus8y 4 1 x 3 8 y 1 x 3 8 8 32 8 y 1 x 358 882. Find the slope of the tangent line to the graph of y x 2 at the point76 , 4936.(A) 2 7Solution:Slope dydx(B)x 7 676(D) 2x(C) 2x 7 6 2(D)7673 7 3(E) None of the above.3. Find the derivative of fx x 52 .(A)52 x12 (B)Solution: (B)f x 5 2 x52 1 5 2 x3252 x32 (C)25 x12 (D) 5x 32 (E) None of the above.By the Power Rule4. Compute the limit: limx2x 2 4x 2(A) 2 (B) 0 (C) 4 (D) Does not exist (E) None of the above.Solution:limx2x 2 4x 2 limx2(C)x 2 x 2x 2 lim x 2 2 2 4x21 MAC 2233 - Test 1 Solution

Name: __Solution___5. Differentiate the function fx 5x 4 3x 3 1.(A) f x 20x 3 9x 2 (B) f x 4x 3 3x 2 (C) f x 4x 3 3x 2 1(D) f x 20x 3 9x 2 1 (E) None of the above.Solution: (A)f x 20x 3 9x 2 0 20x 3 9x 26. Differentiate the function fx 5x 3 4x 2 2(A) f x 15x 2 40xx 3 4x 2 2 2(B) f x 15x2 40xx 3 4x 2 2(C) f x 3x 2 8xx 3 4x 2 22(D) f x 53x 2 8x2(E) None of the above.Solution: (A)fx 5x 3 4x 2 2 15 3x 2 8xx 3 4x 2 2 2 f x 5x 3 4x 2 2 2 3x 2 8x15x 2 40xx 3 4x 2 2 27. Find the second derivative of y 1 3x 3(A) 6x 3 (B)Solution:y 1 3 x1 32(C) 3 (D) 13x 3 6x 4(B) dy 1 dx 3 x2 d2 ydx 2(E) None of the above. 2 3 x3 23x 38. Compute d2dt 2 2 t 1t 31(A) 1 (B) 1 (C) (D) (E) None of the above.81682Solution: (B)Let y 2 t 1 2t 1 12 dy 2 t 112dt 2 d2 2 t 1 d2 y 1 t dt 2 dtt 2 2 132 1 3 13223 t 3t 3 1 1 1 1 12 4 32 2 8 1612 MAC 2233 - Test 1 Solution

Name: __Solution___9. The following is the graph of the function fx x 3 1y-axisf(x) = x 3 +110x-axisWhich of the following statements is TRUE?(A) f0 0; (B) f x 0 for x 0;(C) f x 0forx 0; (D) f x 0 for all x;(E) None of the above.Solution: (D)Clearly, f0 1 0, and fx is concave down (opens down) for x 0. So choices (A)and (B) are incorrect.fx is either increasing or turning (at x 0. Sof x 0 for all x.10. The graph of fx 3x 2 6x 4 has a turning point at x 1. Which of the followingstatements is true?(A) fx has a relative minimum at x 1 (B) fx has a relative maximum at x 1(C) fx has an x intercept at x 1 (D) fx has an absolute minimum at x 1(E) None of the above.Solution: (B)f x 6x 6 f x 6, so fx is concave down (opens down) everywhere,and in particular at x 1. So this, combined with the fact that fx turns at x 1 indicatesthat fx has a relative maximum at x 1.3 MAC 2233 - Test 1 Solution

Name: __Solution___11. a. Given that fx x 32i. Compute f16f16 16 32 64SECTION B - Short AnswerDo all questions. Show all work for credit.ii. Compute f 16f x 3 2 x12 f 16 3 2 1612 6b. Find the equation of the tangent line to the curve fx x 32 at x 16.From part i. of a., we have point 16,64 and from part ii. of a., the slope is 6. Theequation is thusy 64 6x 16 y 6x 96 64 y 6x 3212. a. Given the definition of f a, determine the value of a and the function fx.9 h 3f a limh0 hUsing the definition f fa h faa limand comparing terms we get a 9andh0 hfa h 9 h ,so fx x . To check, fa f9 9 3b. Compute the limt limxlimxx 2 2x 32x 2 1 limxx 2 2x 32x 2 1x 2 x 2 2xx 2 3x 22x 2 x 2 1x 2 limx1 2x 3x 22 1x 2 1 213. Suppose gt 1 2t 4 74 .a. Find the slope of the tangent line to gt at t 3.Slope g 3 4 2t 4 73 2 223 7 3 21 3 2t 3b. Compute g 3.g 3 d 22t dt 73 t 3 62t 7 2 2t 3 1223 7 2 124 MAC 2233 - Test 1 Solution

Name: __Solution___14. a. Compute the limit limx5limx5x 2 8x 15x 2 7x 10 limx5x 2 8x 15x 2 7x 10x 5 x 3x 5 x 2 limx5x 3 5 3 2x 2 5 2 3b. Use limits to compute f 5 where fx 3x 1f 5 limh0 limh0 limh0f5 h f5h16 3h 4h3hh 16 3h 4 limh016 3h 416 3h 4 limh035 h 1 35 1h limh0316 3h 416 3h2 42h 16 3h 4316 30 4 limh016 3h 16h 16 3h 415. The graph of f x, the derivative of fx, is given below. Note this is the graph of thederivative of fx NOT the graph of fx. 3 84yy = f ’ (x)321– 1– 2(6,2)•1 2 3 4 5 6 7xa. Identify an x value at which fx attains a relative maximum. Justify your answer.At a relative maximum f x 0andf x 0 (concave down - opens down); thatis the graph of f x is decreasing and crosses the x axis. This occurs at x 3.b. Identify the x value of an inflection point of fx. Justify your answer.An inflection point of fx is a turning point for f x, because the slopes stopincreasing and start decreasing or vice versa. This occurs at x 1andx 4.c. If f0 3, what is the equation of the tangent line to the graph of fx at x 0?The point is 0,f0 0,3. The slope f 0 1 from the graph. The requiredequation is: y 3 1x 0 y x 35 MAC 2233 - Test 1 Solution