### Answers to Mini-Exam #6

(1) D

(2) D

(3) A

(4) B

(5) A

(a) Because g is the derivative of the function ƒ, ƒ will attain a relative minimum at ta point where g=0 and where g is negative to the left of that point and positive to the right of it. This occurs at x=6.

(b) Bcause g is the derivative of the function ƒ, ƒ will attain a relative amximum at a point where g=0 and where g is positive to the left of that point and negative to the right of it. This occurs at x=3.

(c) We are trying to find the area between the graph and the x-axis from x=-3 to x=6. From x=-3 to x=3, the region is a semicircle of radius 3, so the area is 9π/2.

From x=3 to x=6, the region is a semicircle of radius 3/2, so the area is 9π/8.

We substract the latter region from the former to obtain: (9π/2) - (9π/8) = (27π/8)

(d) Because ƒ''(x) = g'(x), we are looking for points where the derivative of g is zero. This occurs at the horizontal tangent lines at x=0, x=4.5, and x=7.5.

(2) D

(3) A

(4) B

(5) A

*free response question*(a) Because g is the derivative of the function ƒ, ƒ will attain a relative minimum at ta point where g=0 and where g is negative to the left of that point and positive to the right of it. This occurs at x=6.

(b) Bcause g is the derivative of the function ƒ, ƒ will attain a relative amximum at a point where g=0 and where g is positive to the left of that point and negative to the right of it. This occurs at x=3.

(c) We are trying to find the area between the graph and the x-axis from x=-3 to x=6. From x=-3 to x=3, the region is a semicircle of radius 3, so the area is 9π/2.

From x=3 to x=6, the region is a semicircle of radius 3/2, so the area is 9π/8.

We substract the latter region from the former to obtain: (9π/2) - (9π/8) = (27π/8)

(d) Because ƒ''(x) = g'(x), we are looking for points where the derivative of g is zero. This occurs at the horizontal tangent lines at x=0, x=4.5, and x=7.5.

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