What is the local maximum value of frac{log x | vedclass

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

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$1/e$

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(C) Let $f(x) = \frac{\log x}{x}$.To find the local maximum,we first find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.For local extrema,set $f'(x) = 0$:$\frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e$.Now,check the second derivative $f''(x)$ at $x = e$:$f''(x) = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4}$.At $x = e$,$f''(e) = \frac{-3e + 2e(1)}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} Since $f''(e) The local maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

What is the local maximum value of $\frac{\log x}{x}$?

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