If the function f(x) = frac{x}{5} + frac{5}{x | vedclass

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(D) Given the function $f(x) = \frac{x}{5} + \frac{5}{x}$.First, find the derivative $f'(x) = \frac{1}{5} - \frac{5}{x^2}$.Set $f'(x) = 0$ to find cri

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$3$

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(D) Given the function $f(x) = \frac{x}{5} + \frac{5}{x}$.First, find the derivative $f'(x) = \frac{1}{5} - \frac{5}{x^2}$.Set $f'(x) = 0$ to find critical points: $\frac{1}{5} = \frac{5}{x^2} \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$.Find the second derivative $f''(x) = \frac{10}{x^3}$.For $x = 5$, $f''(5) = \frac{10}{125} > 0$, so $x = 5$ is a local minimum.For $x = -5$, $f''(-5) = \frac{10}{-125} Thus, $a = -5$.Now, calculate $\sqrt{a^2 + 2a - 6} = \sqrt{(-5)^2 + 2(-5) - 6} = \sqrt{25 - 10 - 6} = \sqrt{9} = 3$.

If the function $f(x) = \frac{x}{5} + \frac{5}{x}, (x \neq 0)$ attains its relative maximum value at $x = a$, then $\sqrt{a^2 + 2a - 6} = $

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