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f(x)=2x^3-3x^2-12x+1 [-2,3]

I need to use the extreme value theorm to find the absolute extreme values of this function.

So far, I've got:

f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x-2)(x+1)

x= 2,-1

Now what do I do? Please explain in a manner a non-math student can comprehend please.

Thank you!

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# Extreme value theorm

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