Question

find the local maxima of y(x)=x^3-3x^2+6

Answer 1

First differentiating it
So y'=3x^2-6x
The equateto 0
Therefore
3x^2-6x=0
X=2
Which is maxima
And max. Value is 2^3-3×2^2+6=8-12+6=10