SOLUTION: Find the critical values of the function and determine the relative max and relative min. y=x^3-3x^2+6x+1 (I dont understand once you get the derivative how to factor it out.

Question 291674: Find the critical values of the function and determine the relative max and relative min.
y=x^3-3x^2+6x+1
(I dont understand once you get the derivative how to factor it out...)
Thanks!
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
the derivative y=x^3-3x^2+6x+1 is 3x^2-6x+6
factor out 3
x^2-2x+2
there are no real roots because it never cross the x axis.(1-i and 1+i)
the vertex is 1,3
so it's low point is 3 when x=1

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