The minimum value

Question:

The minimum value of $\frac{x}{\log _{e} x}$ is

(a) $\mathrm{e}$

(b) $1 / e$

(c) 1

(d) none of these

Solution:

$(a) e$

Given: $f(x)=\frac{x}{\log _{e} x}$

$\Rightarrow f^{\prime}(x)=\frac{\log _{e} x-1}{\left(\log _{e} x\right)^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{\log _{e} x-1}{\left(\log _{e} x\right)^{2}}=0$

$\Rightarrow \log _{e} x-1=0$

$\Rightarrow \log _{e} x=1$

$\Rightarrow x=e$

Now,

$f^{\prime \prime}(x)=\frac{-1}{x\left(\log _{e} x\right)^{2}}+\frac{2}{x\left(\log _{e} x\right)^{3}}$

$\Rightarrow f^{\prime}(e)=\frac{-1}{e}+\frac{2}{e}=\frac{1}{e}>0$

So, $x=e$ is a local minima.

$\therefore$ Minimum value of $f(x)=\frac{e}{\log _{e} e}=e$