Answers to Mini-Exam #6
(1) D
(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.
(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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