Let $f(x) = (x-2)^2 x^n, n \in N$. Then, f(x) has a minimum at

(a) x = 2 for all n ∈ N
(b) x = 2, if n is even
(c) x = 0, if n is even
(d) x = 0, if n is odd

(a), (c)

We have,

$f(x)=(x-2)^2 x^n$

$\Rightarrow f'(x)=2(x-2) x^n+n(x-2)^2 x^{n-1}$

$\Rightarrow f'(x)=(x-2) x^{n-1}\{2 x+n(x-2)\}$

$\Rightarrow f'(x)=\{(n+2) x-2 n\}(x-2) x^{n-1}$

∴  $f'(x)=0 \Rightarrow x=0, x=2, x=\frac{2 n}{n+2}$

Now,

$f'(0-h)=\{(n+2) h+2 n\}(2+h)(-1)^{n-1} h^{n-1}$

$f'(0+h)=\{(n+2) h-2 n\}(h-2) h^{n-1}$

$\Rightarrow f'(0+h)=\{2 n-(n+2) h\}(2-h) h^{n-1}$

Clearly, f'(x) changes its sign from negative to positive if n is even. Therefore, f(x) attains minimum at x = 0, if n is even. If n is odd, then there is no change in the sign of f'(x) in the neighbourhood of x = 0. So, f(x) does not attain a maximum or minimum.

Similarly, we have

$f'(2-h)=(-h)(2-h)^{n-1}(4-2 h-n h)<0$ for all $n \in N$

and, $f'(2+h)=h(2+h)^{n-1}(4+2 h+n h)>0$ for all $n \in N$

Therefore, f'(x) changes its sign from negative to positive in the neighbourhood of x = 2.

So, x = 2 is point of local minimum for all $n \in N$.