Question 1204512: Find the absolute max and min values of f(x) = 1 + 96x - 2x^3 on the interval [0,5].
I know I need to find the CP.
f'(x) = 96 - 6x^2
6x^2 = 96
x^2 = 16
x = 4 and x = -4
Then I need to evaluate the functions at X=4, X=-4, X=0 and X=5
f(-4) = 1 + 96(-4) -2(-4)^3 = -255
f(0) = 1 + 96(0) -2(0)^3 = 1
f(4) = 1 + 96(4) - 2(4)^3 = 257
f(5) = 1 + 96(5) - 2(5)^3 = 231
So I know the min is -255 and the max is 257 but then when I put it into the problems under "find the absolute maximum of f on the interval" and "find the absolute minimum of f on the interval" I enter in -255 for min and 257 for max and the program says it is incorrect. Any help would be greatly appreciated :)
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Hey, x= -4 is OUT of the interval of interest [0,5].
So, you should not trouble about f(-4).
You should not calculate it and should not consider it, at all.
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