The function $f(x)=\frac{x}{\log x}$ increases on the interval
Answer & explanation
Correct answer: option 3
Clearly, $f(x)$ is defined for $x > 0$
Now, $f(x)=\frac{x}{\log x}⇒f'(x)=\frac{\log x-1}{(\log x)^2}$
$∴f '(x) > 0⇒\log x −1> 0⇒\log x > 1⇒x > e⇒x∈(e, ∞)$