The maximum value of frac{log x}{x}, 0

Question

(D) Let $f(x) = \frac{\log x}{x}$.To find the maximum value, we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx

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$e^{-1}$

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(D) Let $f(x) = \frac{\log x}{x}$.To find the maximum value, we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2}$.Set $f'(x) = 0$ to find critical points:$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.Now, we check the second derivative $f''(x)$ to confirm the nature of the point:$f''(x) = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.At $x = e$, $f''(e) = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} Since the second derivative is negative, the function has a local maximum at $x = e$.The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e} = e^{-1}$.

The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is

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