The function $f(x)=\sum\limits_{k=1}^5(x-k)^2$ assumes minimum value for x given by

5 3 $\frac{5}{2}$ 2

3

$f(x)=(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$ $\Rightarrow f'(x)=2[x-1+x-2+x-3+x-4+x-5] \Rightarrow f'(x)$ $=2[x-1+x-2+x-3+x-4+x-5]$ $=2(5 x-15) \Rightarrow f''(x)=10>0$  for each x ∴ f'(x) = 0  ⇒  x = 3 ∴ f (x) is min. for x = 3.