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Mathematics 9 Online
OpenStudy (anonymous):

Pls. help me answer this.. Maxima & Minima first derivative test y=x^3+3x^2+3x y=2x^3-9x^2+12x-2 y=(x-6)^2(9-x)

OpenStudy (anonymous):

1.) x^3+3x^2+3x =\[3x ^{2}+6x+3\] =3(x^2 +2x+1) =3(x-2x)(x+4) Im not sure of this and i'm confused

OpenStudy (anonymous):

To find a max and min, you take the first and second derivatives. Where the first derivative is 0, the slope is flat, so those points are either max or min. The second derivative is the curvature... if the second derivative is positive the curve is a bowl, so that means a min. If the second derivative is negative, that means a max. So, take first derivative and solve for all points where its 0. Then, at those points calc the second derivative.

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