Question: Answer Key attached. Please Show all work for all problems. Write neatly and show work simply as possible. Please complete both the MC and FRQ

 Answer Key attached. Please Show all work for all problems. Writeneatly and show work simply as possible. Please complete both the MCand FRQ portion. I am not turning this assignment in for agrade, I'm only using it for studying purpose so please do notreport for academic dishonesty. Thanks for your support! There is no informationmissing all the information is on the pictures. Multiple Choice 16 -

Answer Key attached. Please Show all work for all problems. Write neatly and show work simply as possible. Please complete both the MC and FRQ portion. I am not turning this assignment in for a grade, I'm only using it for studying purpose so please do not report for academic dishonesty. Thanks for your support! There is no information missing all the information is on the pictures.

17.9 95% Calculator 15 - 15.9 90% 1. A 13 - 14.985% 2. D 3. C 11 - 12.9 80% 4. D 5.B 9 - 10.9 75% 6. E 7 - 8.9 70% 7.C 5 -6.9 65% 0 - 4.9* 60% (With serious artempt) CalculatorNOT Permitted Free Response Part A - 3 points total 1 8(3)= [f(1)di =4x(2)2 -2 (1)(1) =1 -2 1 8'(3) = A(3) =-1

Multiple Choice 16 - 17.9 95% Calculator 15 - 15.9 90% 1. A 13 - 14.9 85% 2. D 3. C 11 - 12.9 80% 4. D 5. B 9 - 10.9 75% 6. E 7 - 8.9 70% 7. C 5 -6.9 65% 0 - 4.9* 60% (With serious artempt) Calculator NOT Permitted Free Response Part A - 3 points total 1 8(3) = [f(1)di =4x(2)2 -2 (1)(1) =1 -2 1 8'(3) = A(3) =-1 1 g"(3) = undefined because g'(x) = f(x) is not differentiable at x = 3. Calculator NOT Permitted Free Response Part B - 3 points total 1 g(x) has a point of inflection when g"(x) changes signs. 1 g"(x) changes signs when the graph of g'(x) = f(x) has a relative maximum or minimum 1 Thus, g has a point of inflection when x = 0 and x = 3. Calculator NOT Permitted Free Response Part C -3 points total 1 Correctly finds the values of g(-2) and g(5), as x = -2 and x = 5 are the endpoints of the interval. 8 (-2 ) - fund - - ff()di - -47(2)2 --1 8(5) = [f()at = 4x(2)2 -2 (1)(1) =1-2 1 Correctly finds the value of g(4), as x = 4 is the only relative minimum of g on the interval 8 (4 ) = [f(1)dt = 4 x (2) 2 - 2 (2)(1) = 1-1 1 According to the Extreme Value Theorem, the absolute minimum value of g on -2

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