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Unformatted text preview: Horstman (mdh995) – HW03 – Radin – (58505) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A function h has graph 2 2 2 2 on ( 4 , 4). If f is defined on ( 4 , 4) by f ( x ) =  1 , 4 < x < 3 , integraldisplay x − 3 h ( t ) dt, 3 ≤ x < 4 , which of the following is the graph of f ? 1. 2 2 2 2 2. 2 2 2 4 2 3. 2 2 2 2 4 4. 2 2 2 2 correct 5. 2 2 2 2 Explanation: Horstman (mdh995) – HW03 – Radin – (58505) 2 Since f ( 3) = integraldisplay − 3 − 3 h ( t ) dt = 0 , two of the five graphs can be eliminated im mediately. On the other hand, by the Fun damental Theorem of Calculus, f ′ ( x ) = h ( x ) on ( 3 , 4); in particular, the critical points of f occur at the xintercepts of the graph of h . As these xintercepts occur at 1 , , 2, this eliminates a third graph. Thus the remain ing two possible graphs for f both have the same critical points and to decide which one is the graph of f we can use the first derivative test because the graph of f will have a local maximum at an xintercept of the graph of h where it changes from positive to negative values, and a local minimum at an xintercept where h changes from negative to positive val ues. Consequently, the graph of f must be 2 2 2 2 keywords: 002 10.0 points If the function F is defined by F ( x ) = d dx parenleftBig integraldisplay x 3 1 6 t 5 dt parenrightBig , determine the value of F (1). Correct answer: 18. Explanation: By the Fundamental Theorem of Calculus, integraldisplay x 3 1 6 t 5 dt = bracketleftBig t 6 bracketrightBig x 3 1 = ( x 18 1) . Thus F ( x ) = d dx parenleftBig ( x 18 1) parenrightBig = 18 x 17 Consequently, F (1) = 18. 003 10.0 points If f is a continuous function such that integraldisplay x f ( t ) dt = 9 x 7 x 2 + 1 , find the value of f (1). 1. f (1) = 27 32 correct 2. f (1) = 25 32 3. f (1) = 51 64 4. f (1) = 13 16 5. f (1) = 53 64 Explanation: By the Fundamental Theorem of Calculus, d dx parenleftBig integraldisplay x f ( t ) dt parenrightBig = f ( x ) . So by the Quotient Rule, f ( x ) = d dx parenleftBig 9 x 7 x 2 + 1 parenrightBig = 9 63 x 2 (7 x 2 + 1) 2 . In this case, f (1) = 27 32 . Horstman (mdh995) – HW03 – Radin – (58505) 3 keywords: indefinite integral, Fundamental Theorem Calculus, FTC, function value, Quo tient Rule, rational function, 004 10.0 points Determine F ′ ( x ) when F ( x ) = integraldisplay √ x 2 2 cos t t dt . 1. F ′ ( x ) = 2 cos x √ x 2. F ′ ( x ) = cos( √ x ) √ x 3. F ′ ( x ) = sin( √ x ) x 4. F ′ ( x ) = 2 sin x x 5. F ′ ( x ) = 2 sin x √ x 6. F ′ ( x ) = cos x x 7. F ′ ( x ) = 2 sin( √ x ) √ x 8. F ′ ( x ) = cos( √ x ) x correct Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx parenleftBig integraldisplay g ( x ) a f ( t ) dt parenrightBig = f ( g ( x )) g ′ ( x ) ....
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This note was uploaded on 11/30/2008 for the course M 58510 taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

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