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

Algebra ->  Functions -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, x=0 and x=1 are two extrema values.
x=0 is the relative maximum.
x=1 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.
f%281%29=2%281%29%5E3-3%281%29%5E2%2B6=2-3%2B6=5
f%28-1%29=2%28-1%29%5E3-3%28-1%29%5E2%2B6=-2-3%2B6=1
So in the interval, the minimum occurs at x=-1 and the maximum occurs at x=0.
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