The function `f(x)=(x^2-4)^n(x^2-x+1),n in N ,` assumes a local minimum value at `x=2.` Then find the possible values of `n`

Explanation

f(x) =`(x^(2)-4)^(n)(x^(2)-x+1)`
f(2)=0
Now `x^(2)-x+1gt0forall x`
`f(2^(+))=lim(x^(2)-4)^(n)(x^(2)-x+1)`
`=underset(hrarr0)3lim(h+2)^(2)-4^(n)`
`=underset(hrarr0)3lim (4h+h^(2^(n)))`
`gt0`
`f(2^(-))=underset(hrarr0)3lim (h-2)^(2)-4^(n)`
`=underset(hrarr0)3lim (h^(2)-4h^(n))`
`=3xx("very small negative value")^(n)`
For x = 0 to be a point of minima we must have `f(2^(-))gt0` for which n must be an even interger.