Question

The function $f\left(x\right)={\sum }_{K=1}^5(x-K{)}^2$ assumes the minimum value of $x$ given by-

• A. $5$
• B. $\frac{5}{2}$
• C. $3$
• D. $2$

Answer 1

Answer: (C)

$3$

The given equation is:

$f\left(x\right)={\sum }_{K=1}^5(x-K{)}^2$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow f^{\prime }\left(x\right)=2{\sum }_{K=1}^5(x-K)$

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

$\Rightarrow f^{\prime }\left(x\right)=10x-30$

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

$\Rightarrow 10x-30=0$

$\Rightarrow x=3$

Now to find whether at the critical points we find a maxima or minima we use the second derivative test

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

$\Rightarrow f^{\prime \prime }\left(3\right)=10$

$\Rightarrow f^{\prime \prime }\left(3\right)>0$

From the second derivative test it is clear that at $x=3$ we get a minimum value of f(x). ...Answer