SOLUTION: Please check my work for max/min problem. Find the max/min of: f(x) = 2x^3 - 3x^2 + 6 of f on [-1,1] f'(x) = 6x^2 - 6x f'(x) = 6x(x-1) therefore, x = 0 & x = 1 and includi

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Question 999602: Please check my work for max/min problem.
Find the max/min of:
f(x) = 2x^3 - 3x^2 + 6 of f on [-1,1]
f'(x) = 6x^2 - 6x
f'(x) = 6x(x-1)
therefore, x = 0 & x = 1 and including the boundary x = -1
testing...
f"(x) = 12x - 6
f"(0) = 12(0) - 6 = -6 < 0, MAX value occurs at x = 0
f'(1) = 12(1) - 6 = +6 > 0, MIN value occurs at x = 1
f"(-1) = 12(-1) - 6 = -18 < 0, MAX value occuras at x = -1
This confuses me because now I have one MIN value that occurs at x = 1, but this isn't actually the value of the MIN just where it occurs right?
And then I have two MAX's which one is the real max? because I think I am only dealing with rel max/min so I don't think two would make sense.
Plese explain
Thank you


Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Yes, and are two extrema values.
is the relative maximum.
is the relative minimum.
However since you're in an interval, you need to check the endpoints and there is a definite min and max.


So in the interval, the minimum occurs at and the maximum occurs at .
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