Question

Minimum value of the function $f\left(x\right)={\sum }_{k=1}^5{(x-k)}^2$ is at-

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

Answer 1

The correct option is C $x=3$

$f\left(x\right)=(x-1{)}^2+(x-2{)}^2+(x-3{)}^2+(x-4{)}^2+(x-5{)}^2$
Now
$f^{\prime }\left(x\right)=2[5x-\frac{5(5+1)}{2}]=0$
Or
$2[5x-15]=0$
Or
$x=3$
Hence extremum occurs at $x=3$.