The maximum and minimum values of the function y=x^3-3x^2+6 are
=> answer is 6,2
please give a valid explanation..
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d/dx =0
d/dx x^3-3x^2+6=0 ......(1 eq.)
3x^2-6x=0
x=2 & x=0
and putting values of x in (1 eq)
And ur answer will be 6 & 2
d/dx x^3-3x^2+6=0 ......(1 eq.)
3x^2-6x=0
x=2 & x=0
and putting values of x in (1 eq)
And ur answer will be 6 & 2
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