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# hw8so - Create assignment 57321 Homework 8 Mar 29 at 11:11...

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Create assignment, 57321, Homework 8, Mar 29 at 11:11 pm 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC5c03b 51:04, calculus3, multiple choice, > 1 min, wording-variable. 001 A function h has graph -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 on ( - 4 , 4). If f is defined on ( - 4 , 4) by f ( x ) = 1 , - 4 < x < - 3 , Z x - 3 h ( t ) dt, - 3 x < 4 , which of the following is the graph of f ? 1. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 correct 2. -4 -3 -2 -1 0 1 2 3 2 - 2 2 4 - 2 3. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 - 4 4. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 5. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 Explanation:

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Create assignment, 57321, Homework 8, Mar 29 at 11:11 pm 2 Since f ( - 3) = Z - 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 0 ( x ) = h ( x ) on ( - 3 , 4); in particular, the critical points of f occur at the x -intercepts of the graph of h . As these x -intercepts occur at - 1 , 0 , 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 x -intercept of the graph of h where it changes from positive to negative values, and a local minimum at an x -intercept where h changes from negative to positive val- ues. Consequently, the graph of f must be -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 keywords: CalC5c04s 51:04, calculus3, multiple choice, > 1 min, wording-variable. 002 The graph of f is shown in the figure -1 0 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 2 4 6 - 2 If the function g is defined by g ( x ) = Z x 1 f ( t ) dt, for what value of x does g ( x ) have a maxi- mum? 1. x = 1 2. x = 7 3. x = 5 correct 4. x = 6 5. not enough information given 6. x = 2 . 5 Explanation: By the Fundamental theorem of calculus, if g ( x ) = Z x 1 f ( t ) dt, then g 0 ( x ) = f ( x ). Thus the critical points of g occur at the zeros of f , i.e. , at the x - intercepts of the graph of f . To determine which of these gives a local maximum of g we use the sign chart g 0 + - 1 5 7 for g 0 . This shows that the maximum value of g occurs at x = 5
Create assignment, 57321, Homework 8, Mar 29 at 11:11 pm 3 since the sign of g 0 changes from positive to negative at x = 5. keywords: FTC, integral, sign chart, maxi- mum CalC5c08a 51:04, calculus3, multiple choice, < 1 min, wording-variable. 003 If the function F is defined by F ( x ) = d dx Z x 5 0 4 t 3 dt · , determine the value of F (1). 1. F (1) = 40 2. F (1) = 5 3. F (1) = 20 correct 4. F (1) = 10 5. F (1) = 60 Explanation: By the Fundamental Theorem of Calculus, Z x 5 0 4 t 3 dt = h t 4 i x 5 0 = x 20 . In this case, F ( x ) = d dx x 20 · = 20 x 19 Consequently, F (1) = 20 . keywords: indefinite integral, Fundamental Theorem Calculus, FTC, function value CalC5c10a 51:04, calculus3, multiple choice, > 1 min, wording-variable. 004 If f is a continuous function such that Z x 0 f ( t ) dt = 3 x 2 x 2 + 5 , find the value of f (1).

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