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.

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)