If f(x)=(x-a)^2+(x-b)^2+(x-c)^2 , then f(x) has minimum at x equal to:

The correct answer is Option (3) → \( \displaystyle\frac{a+b+c}{3} \)

At a minimum, f′(x) = 0.

    \( \displaystyle f'(x)=2(x-a)+2(x-b)+2(x-c) \)

    \( \displaystyle =2\left[3x-(a+b+c)\right] \)

    \( \displaystyle f'(x)=0\ \Rightarrow\ x=\frac{a+b+c}{3} \)

    \( \displaystyle f''(x)=6\gt 0 \)

So f has its minimum at \( x=\frac{a+b+c}{3} \), the arithmetic mean of a, b and c.

Tip: The sum of squared deviations \( \sum(x-a_i)^2 \) is always least at the arithmetic mean of the \( a_i \); this is a standard result in statistics too.