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: Find the critical point.
Differentiate \(f\).
\[ f'(x) = 2(x-a)+2(x-b)+2(x-c) = 2(3x-a-b-c) \] Set \(f'(x)=0\). This gives \(x=\frac{a+b+c}{3}\).

Step 2: Test for a minimum.
\(f''(x) = 6 > 0\) for every \(x\). So the critical point is a minimum.

Step 3: Check the options.
Option 1 (\(3\sqrt{abc}\)) and option 2 (\(3abc\)) involve products, but \(f'\) only gives sums of \(a,b,c\). Option 4 is the harmonic mean, which does not solve \(f'(x)=0\). Option 3 is the arithmetic mean, and it does.

Final Answer:
The minimum is at \(x=\frac{a+b+c}{3}\), option 3. \[ \boxed{\frac{a+b+c}{3}} \]