Calculus multiple-choice question antiderivative

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Calculus multiple-choice question

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Evaluate the limit lim (2-2)(11+52) 3-0 (3-3x)(11+62) 3 5 18 9 2. If f(x) = sin*x, find f|(1). 2 sin ? (1) cos (1) sin' (1) cos (1) 4 sinº (1) cos (1) 4 sin (1) 3. Compute the following limit using l’Hôpital's rule if appropriate. lim (1 - 0)*. 200 e 1 Does not exist 12. Let f(x) = -3e-5. Find f(9)(-1). e 3 39 --- e 3 3 13. Let læ\ < and y = (cos-+~)*. Then find the derivative Dry. (cos-13) V1-22 -8(cos-12) V1-3? -8(cos-12) 1-2 8(COS-12) V1-27 14. 210x+9 Evaluate the limit lim 2-1 2+1 Does not exist 8 O 4. Let f(x) = 3-7. Find f(x). 3-2 -(In 3)3– (In 3)3–2 -3- 5. Evaluate the limit using l’Hôpital's rule. lim 2-0e7 0 7 14 Does not exist 6. Let f(x) = 4 cos 6 In (2)). Find f (1). OO 6 -24 O 7. Evaluate lim - 4.6 + 703 – 5. 100 -2 -5 8. Evaluate the indefinite integral Sz’e-53" dx. -52-3 +C -5. e +C 15 Israe -5° +C 'e e 53 15 +C 9. Let f(a) sin and g(x) = 1 Find f(x) and g(x). sin cos f(x) = and g(x) COS sin' f(x) = cos 22 and g(x) = COS 2 sin cos! f(x) = and g(x) = COS 3 2 sin cos fi() and g(x) ? 1 sin' 10. Consider the function f(x) = 2x2 – 2x3 + 4x. Find the average slope m of this function on the interval (-2,4). By the Mean Value Theorem, we know there exists a c in the open interval (-2, 4) such that f(c) is equal to this mean slope. Find the two values of c in the interval which works. m = -16, c = 2.54 or - 1.72 m= -16,c= 2.19 or - 1.52 0 m = 40 3' c=2.19 or 1.52 m = 40 c= 2.54 or – 1.72 3 11. If f(x) = = 22x, find f (c). V22 V22 2 1 2225 V11 va
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