Finding the Least Value of Sum of Squares

We are given the expression $f(x) = (x - a_1)^2 + (x - a_2)^2 + \dots + (x - a_n)^2$. We need to find the value of $x$ where this expression is minimized.

Minimization using Calculus

This expression represents a sum of squared differences. To find the minimum value, we can use calculus by finding the derivative of $f(x)$ with respect to $x$ and setting it to zero.

1. Define the function: Let $f(x) = \sum_{i=1}^{n} (x - a_i)^2$.

2. Calculate the first derivative $f'(x)$: $f'(x) = \frac{d}{dx} \left( \sum_{i=1}^{n} (x - a_i)^2 \right)$ $f'(x) = \sum_{i=1}^{n} \frac{d}{dx} (x - a_i)^2$ $f'(x) = \sum_{i=1}^{n} 2(x - a_i)$ $f'(x) = 2 \sum_{i=1}^{n} (x - a_i)$ $f'(x) = 2 \left( \sum_{i=1}^{n} x - \sum_{i=1}^{n} a_i \right)$ $f'(x) = 2 \left( nx - (a_1 + a_2 + \dots + a_n) \right)$

3. Set the derivative to zero to find critical points: $f'(x) = 0$ $2 \left( nx - (a_1 + a_2 + \dots + a_n) \right) = 0$ $nx - (a_1 + a_2 + \dots + a_n) = 0$ $nx = a_1 + a_2 + \dots + a_n$

4. Solve for $x$: $x = \frac{a_1 + a_2 + \dots + a_n}{n}$

5. Verify it's a minimum: The second derivative is $f''(x) = \frac{d}{dx} \left( 2 \left( nx - \sum_{i=1}^{n} a_i \right) \right) = 2n$. Since $n \ge 1$, $f''(x) > 0$, confirming that this value of $x$ corresponds to a minimum.

Therefore, the expression assumes its least value when $x$ is the average (mean) of $a_1, a_2, \dots, a_n$. This matches option 1.