The maximum value of frac{log _{e} x}{x},if x | vedclass

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

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$\frac{1}{e}$

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(C) Let $f(x) = \frac{\log _{e} x}{x}$.To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log _{e} x) - \log _{e} x \cdot \frac{d}{dx}(x)}{x^{2}} = \frac{x \cdot \frac{1}{x} - \log _{e} x \cdot 1}{x^{2}} = \frac{1 - \log _{e} x}{x^{2}}$.For critical points,set $f'(x) = 0$:$1 - \log _{e} x = 0 \Rightarrow \log _{e} x = 1 \Rightarrow x = e$.Now,we check the second derivative $f''(x)$ to confirm the maximum:$f''(x) = \frac{x^{2}(-\frac{1}{x}) - (1 - \log _{e} x)(2x)}{x^{4}} = \frac{-x - 2x(1 - \log _{e} x)}{x^{4}}$.At $x = e$,$f''(e) = \frac{-e - 2e(1 - 1)}{e^{4}} = \frac{-e}{e^{4}} = -\frac{1}{e^{3}} Since $f''(e) The maximum value is $f(e) = \frac{\log _{e} e}{e} = \frac{1}{e}$.

The maximum value of $\frac{\log _{e} x}{x}$,if $x>0$ is

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