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: $f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.Set $f'(x)

Vedclass · Accepted Answer

Does not exist

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(D) Let $f(x) = \frac{\log x}{x}$.Find the derivative: $f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.Set $f'(x) = 0$ to find critical points: $1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.Since $e \approx 2.718$,$x = e$ lies in the interval $[2, \infty)$.Evaluate the function at the critical point and the boundary:$f(e) = \frac{\log e}{e} = \frac{1}{e} \approx 0.367$.$f(2) = \frac{\log 2}{2} \approx \frac{0.693}{2} = 0.346$.As $x 	o \infty$,$\frac{\log x}{x} 	o 0$ (by $L$'Hopital's rule).Since $f(2) \approx 0.346$ and the limit as $x 	o \infty$ is $0$,and the function decreases for $x > e$,the function values approach $0$ but never reach it in the interval $[2, \infty)$.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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