MCQ
The function $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$ assume minimum value at $x =$
- A$5$
- B$\frac{5}{2}$
- ✓$3$
- D$2$
✓
Answer
Correct option: C.
$3$
Given, $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$lmplies that $f(x) = (x - 1)^2 + (x - 2)^2+ (x - 3)^2 + (x - 4)^2 + x - 5^2$
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.
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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