Local maximum value of the function frac{log | vedclass

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

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$\frac{1}{e}$

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(C) Let $f(x) = \frac{\log x}{x}$.To find the critical points,we calculate 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 maxima or minima,set $f'(x) = 0$:$\frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e$.Now,we check the second derivative $f''(x)$ to confirm the nature of the point:$f''(x) = \frac{x^2 \cdot (-\frac{1}{x}) - (1 - \log x) \cdot 2x}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.At $x = e$,$f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3}$.Since $f''(e) The local maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

Local maximum value of the function $\frac{\log x}{x}$ is

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