Where does the function f(x)=\sum_{j=1}^{7}(x-j)^{2} attain its <strong>MINIMUM</strong> value ?
- A. x=3.5
- B. x=4 ✓
- C. x=4.5
- D. x=5
Correct Answer: B. x=4
Explanation
The sum of squared deviations \sum (x-x_i)^2 is minimized at the mean of the data points. The mean of the first 7 natural numbers is \frac{1+2+3+4+5+6+7}{7} = \frac{28}{7} = 4.
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