The minimum value of `(x-p)^(2) + (x-q)^(2) + (x-r)^(2)` will be at x equals to

Options

A. pqr
B. `root3(pqr)`
C. `(p + q + r)/(3)`
D. `p^(2) + q^(2) + r^(2)`

Answer

To find the minimum value of the function \( f(x) = (x - p)^2 + (x - q)^2 + (x - r)^2 \), we will follow these steps:

### Step 1: Write the function
We start with the function:
\[ f(x) = (x - p)^2 + (x - q)^2 + (x - r)^2 \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 2(x - p) + 2(x - q) + 2(x - r) \]
This simplifies to:
\[ f'(x) = 2[(x - p) + (x - q) + (x - r)] = 2[3x - (p + q + r)] \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 2[3x - (p + q + r)] = 0 \]
Dividing both sides by 2 gives:
\[ 3x - (p + q + r) = 0 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \):
\[ 3x = p + q + r \]
\[ x = \frac{p + q + r}{3} \]

### Conclusion
The minimum value of the function \( f(x) \) occurs at:
\[ x = \frac{p + q + r}{3} \]

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