Question
The function $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$ assume minimum value at x =
- $5$
- $\frac{5}{2}$
- $3$
- $2$
Solution:
Given, $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$
lmplies that f(x) = (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 + x - 52
lmplies that f'(x) = 2(x - 1 + x - 2 + x - 3 + x - 4 + x - 5)
lmplies that f'(x) = 2(5x - 15)
For a local maxima and a local minima, we must have f'(x) = 0
limplies that 2(5x - 15) = 0
limplies that 5x - 15 = 0
limplies that x = 3
Now, f''(x) = 10
f''(x) = 10 > 0
Therefore, x = 3 is a local minima.
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