the maximum and minimum value of the function y=x cube-3xsquare+6,are
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dy/dx=3x^2-6x
double differentiation
d^2y/dx^2 =6x-6
to find minima Maxima
dy/dx =0
therefore x=0
3x-6=0
x=2
if u get
d^2y/dx^2 >0 it's minimum
if it is less than 0 then it's maximum
double differentiation
d^2y/dx^2 =6x-6
to find minima Maxima
dy/dx =0
therefore x=0
3x-6=0
x=2
if u get
d^2y/dx^2 >0 it's minimum
if it is less than 0 then it's maximum
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