The set of real values of x for which f(x) = | vedclass

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(A) Given the function $f(x) = \frac{x}{\log x}$.First,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(\log x)(1) - (x)(\frac{1

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$\{x: x \geq e\}$

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(A) Given the function $f(x) = \frac{x}{\log x}$.First,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.For the function to be increasing,we must have $f'(x) > 0$.Since $(\log x)^2$ is always positive for $x > 0$ and $x 
eq 1$,the condition $f'(x) > 0$ implies $\log x - 1 > 0$.$\log x > 1$.Since the base of the logarithm is $e$,we have $\log_e x > \log_e e$.Therefore,$x > e$.Thus,the set of values is $\{x: x > e\}$.

The set of real values of $x$ for which $f(x) = \frac{x}{\log x}$ is increasing,is

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