SOLUTION: f(x)= 2/3 x^3 - x^2 - 4x+2 Find the local max and local min of f(×) if exist

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Question 1082587: f(x)= 2/3 x^3 - x^2 - 4x+2
Find the local max and local min of f(×) if exist

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
2/3 x^3 - x^2 - 4x+2
derivative is 2x^2-2x-4=0
x^2-x-2=0
(x-2)(x+1)=0
x=2, -1; local points are (2, -14/3) which is a minimum and (-1, 13/3)
Second derivative is 4x-2; when x=2, f''(2)=6, so x=2 is a minimum.
when x=-1, f''(-1)=-6 maximum


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