The function f(x) = frac{log x}{x} is increas | vedclass

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(B) Given the function $f(x) = \frac{\log x}{x}$.To find the interval where the function is increasing,we first find the derivative $f'(x)$ using the

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$(0, e)$

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(B) Given the function $f(x) = \frac{\log x}{x}$.To find the interval where the function is increasing,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}$$f'(x) = \frac{x \cdot (\frac{1}{x}) - \log x \cdot (1)}{x^2} = \frac{1 - \log x}{x^2}$.For the function to be increasing,we must have $f'(x) > 0$.Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.$1 > \log x \implies \log_e x Since the domain of $\log x$ is $x > 0$,the function is increasing in the interval $(0, e)$.

The function $f(x) = \frac{\log x}{x}$ is increasing in the interval

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