The minimum value of frac{log x}{x} in the in | vedclass

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(D) Let $f(x) = \frac{\log x}{x}$.Find the derivative: $f'(x) = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.Set $f'(x) = 0$ to fi

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Does not exist

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(D) Let $f(x) = \frac{\log x}{x}$.Find the derivative: $f'(x) = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.Set $f'(x) = 0$ to find critical points: $1 - \log x = 0 \implies \log x = 1 \implies x = e$.Since $e \approx 2.718$, $x = e$ lies in the interval $(2, \infty)$.Check the sign of $f'(x)$ around $x = e$: for $x  0$ (increasing); for $x > e$, $f'(x) Thus, $x = e$ is a point of local maximum.As $x 	o 2^+$, $f(x) 	o \frac{\log 2}{2} \approx 0.346$.As $x 	o \infty$, $f(x) 	o 0$.Since the function increases until $x = e$ and then decreases towards $0$ as $x 	o \infty$, the infimum of the function on $(2, \infty)$ is $0$, but it is never attained. Therefore, the minimum value does not exist.

The minimum value of $\frac{\log x}{x}$ in the interval $(2, \infty)$ is

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