If f(x) = frac{log x}{x} (x 0),then it is i | vedclass

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(A) Given the function $f(x) = \frac{\log x}{x}$.To determine where the function is increasing,we find its derivative $f'(x)$ using the quotient rule:

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$(0, e)$

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(A) Given the function $f(x) = \frac{\log x}{x}$.To determine where the function is increasing,we find its 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 the function to be increasing,we require $f'(x) > 0$.Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.This simplifies to $1 > \log x$,which means $\log x Taking the exponential of both sides,we get $x Given the domain $x > 0$,the function is increasing in the interval $(0, e)$.

If $f(x) = \frac{\log x}{x}$ $(x > 0)$,then it is increasing in

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