Find the absolute maximum and absolute minimum values of the functionf(x)= x3 + 6x2 −63x +8over each of the indicated intervals.(a) interval = [−8,0].1. absolute maximum = 2. absolute minimum = (b) interval = [−5,4].1. absolute maximum = 2. absolute minimum = (c) interval = [−8,4].1. absolute maximum = 2. absolute minimum =

  (a) maximum = 400; minimum = 8

  (b) maximum = 348; minimum = -100

  (c) maximum = 400; minimum = -100

Step-by-step explanation:

The absolute extrema will be either at the ends of the interval or at a turning point in the interval. Here, the turning points can be found from the derivative:

  f'(x) = 3x^2 +12x -63 = 3(x^2 +4x -21)

  f'(x) = 3(x +7)(x -3)

The derivative is zero at x=-7 and at x=3. Since the cubic has a positive leading coefficient, the extreme at x=-7 is a maximum; that at x=3 is a minimum.

So, the values we are concerned with are ...

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(a)

  1. The left turning point is in the interval, so the absolute maximum is f(-7) = 400.

  2. The absolute minimum is at the right end of the interval, at x=0. Its value is f(0) = 8.

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(b)

  1. The absolute maximum is at the left end of the interval. Its value is f(-5) = 348.

  2. The absolute minimum is at the right turning point: f(3) = -100.

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(c)

 1. The absolute maximum is at the left turning point: f(-7) = 400.

 2. The absolute minimum is at the right turning point: f(3) = -100.

polyhedron???

Step-by-step explanation:

A polyhedron with two congruent faces, called bases, that lie in parallel planes. The other faces, called lateral faces, are parallelograms formed by connecting the corresponding vertices of the bases. The segments connecting the vertices are lateral edges.

I mean im not 100% sure but might be the correct answer? Hope this helps and hope you have a great day.