Let f(x)=(x-a)^2+(x-b)^2+(x-c)^2 then f(x) has a minimum value at x=

The correct answer is Option (4) → \( \displaystyle\frac{a+b+c}{3} \)

At a minimum the first derivative is zero:

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

    \( \displaystyle =2\big(3x-(a+b+c)\big) \)

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

Since \( f''(x)=6\gt 0 \), this point gives the minimum.

Tip: The sum of squared distances from x to given numbers is least at their arithmetic mean; the result holds for any number of terms.