The function
Question:

The function $f(x)=\sum_{r=1}^{5}(x-r)^{2}$ assumes minimum value at $x=$

(a) 5

(b) $\frac{5}{2}$

(c) 3

(d) 2

Solution:

(c) 3

Given : $f(x)=\sum_{r=1}^{5}(x-r)^{2}$

$\Rightarrow f(x)=(x-1)^{2}+(x-2)^{2}+(x-3)^{2}+(x-4)^{2}+(x-5)^{2}$

$\Rightarrow f^{\prime}(x)=2(x-1+x-2+x-3+x-4+x-5)$

$\Rightarrow f^{\prime}(x)=2(5 x-15)$

For a local maxima and a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 2(5 x-15)=0$

$\Rightarrow 5 x-15=0$

$\Rightarrow 5 x=15$

$\Rightarrow x=3$

Now,

$f^{\prime \prime}(x)=10$

$f^{\prime \prime}(x)=10>0$

So, $x=3$ is a local minima.