8) Find the absolute maximum and absolute minimum values of f on the given interval.?

Question

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = 2x3 - 3x2 - 36x + 8
[-3, 4]
1 (min)
2 (max)

Answer 1

Best Answer - Chosen by Voters

Differentiating gives f'(x)=6x^2 - 6x - 36 = 6(x^2-x-6) = 0 for x=-2 and x=3.
Now we have:
f(-3) = 2*(-3)^3 - 3(-3)^2 -36*(-3) + 8 = -54 - 27 + 108 + 8 = 35
f(-2) = 2*(-2)^3 - 3(-2)^2 -36(-2) + 8 = -16 -12 + 72 +8 = 52
f(3) = 2*3^3 - 3*3^2 -36*3 + 8 = 54 - 27 - 108 + 8 = -73
f(4) = 2*4^3 -3*4^2 - 36*4 + 8 = 128 - 48 - 144 + 8 = -56
So the absolute minimum is -73 for x=3
and the absolute maximum is 52 for x=2