The function `f(x)=sum_(r=1)^5(x-r)^2` assuming minimum value at `x=` (a)`5` (b) `5/2` (c)` 3` (d)` 2`

Explanation

`sum_{r=1}^{5}(x-r)^{2} `


`f(x)=(x-1)^{2}+(x-2)^{2}+(x-3)^{2}+(x-4)^{2}+(x-5)^{2}`


`f^{prime}(x)=2[5 x-15]`


`f^{prime}(x)=0 ; x=3`



Hence by second derivative test


`f^{prime prime}(x)>0` so it's a point of minimum.


`f^{prime prime}(x)=1>0` so At `x=3` minimum value.