Practicing Success
The function $f(x)=\frac{x}{\log x}$ increases on the interval |
$(0, ∞)$ $(0, e)$ $(e, ∞)$ None of these |
$(e, ∞)$ |
Clearly, $f(x)$ is defined for $x > 0$ Now, $f(x)=\frac{x}{\log x}⇒f'(x)=\frac{\log x-1}{(\log x)^2}$ $∴f '(x) > 0⇒\log x −1> 0⇒\log x > 1⇒x > e⇒x∈(e, ∞)$ |