Find the minimum and maximum values of the funciton `y=x^3-3x^2+6`. Also find the values of x at which these occur.

Answer

2

Explanation

`(dy)/(dx)=3x^(2)-6x`
For maximum or minimum values of the function, `(dy)/(dx)=0`
`therefore 3x^(2)-6x=0implies3x(x-2)=0`
or, `x=0andx=2`
Now, `(d^(2)y)/(dx^(2))=6x-6`
At `x=0, (d^(2)y)/(dx^(2))=-6lt0` (Maxima)
`therefore y_(max)=(0)^(3)-3(0)^(2)+6=6`
At `x=2,(d^(2)y)/(dx^(2))=12-6=6gt0` (Minima)
`therefore y_(min)=2^(3)-3(2)^(2)+6=2`