Question

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

• A. \(3\sqrt{abc}\)
• B. \(3abc\)
• C. \(\frac{a+b+c}{3}\)
• D. \(\frac{3}{\frac1a+\frac1b+\frac1c}\)

Hint:

Set \(f'(x)=0\). The sum of squared distances is least at the mean.

Answer 1

The Correct Option is C

Step 1: Expand.
$f(x) = 3x^2 - 2(a+b+c)x + (a^2+b^2+c^2)$. This is a parabola opening upward, because the coefficient of $x^2$ is 3.

Step 2: Use the vertex formula.
A parabola $px^2+qx+r$ with $p>0$ has its minimum at $x=-\frac{q}{2p}$.
\[ x = \frac{2(a+b+c)}{2\cdot 3} = \frac{a+b+c}{3} \]

Step 3: Meaning.
The sum of squared distances from a point to $a$, $b$, $c$ is smallest at their mean.

Final Answer:
So the minimum is at $x=\frac{a+b+c}{3}$. \[ \boxed{\frac{a+b+c}{3}} \]