Question
Show that the function given by f(x) $\frac{\log\text{x}}{\text{x}}$ has maximum at $\text{x}=\theta.$

Answer

The given function is $\text{f}\text{(x)}=\frac{\log\text{x}}{\text{x}}.$
$\text{f}\text{(x)}=\frac{\text{x}\Big(\frac{1}{\text{x}}\Big)-\log\text{x}}{\text{x}^2}=\frac{1-\log\text{x}}{\text{x}^2}$
Now, f(x) = 0
$\Rightarrow\ 1-\log \text{x}=0$
$\Rightarrow\ \log \text{x}=1$
$\Rightarrow\ \log\ \text{x}=\log\text{e}$
$\Rightarrow\ \text{x}=\text{e}$
Now, $\text{f}''\text{x}=\frac{\text{x}^2\Big(-\frac{1}{\text{x}}\Big)-(1-\log\text{x})(2\text{x})}{\text{x}^4}$
$=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^4}$
$=\frac{-32\log\text{x}}{\text{x}^3}$
Now, $\text{f}''\text{(e)}=\frac{-3+2\log\text{e}}{\text{e}^3}=\frac{-3+2}{\text{e}^3}=\frac{-1}{\text{e}^3}<0$
Therefore, by second derivative test, f is the maximum at x = e.

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