Show that logx x has a minimum value at x = e. - Vidyadip

Question

Show that $\frac{\text{logx}}{\text{x}}$ has a minimum value at x = e.

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Answer

Here, $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1-\log\text{x}}{\text{x}^{2}}$
For the local maxima or minima, We have f'(x) = 0
$\Rightarrow\frac{1-\log\text{x}}{\text{x}^{2}}=0$
$\Rightarrow1=\log\text{x}$
$\Rightarrow\log\text{e}=\log\text{x}$
$\Rightarrow\text{x}=\text{e}$
Now, $\text{f}''(\text{x})=\frac{\text{x}^{2}(\frac{-1}{\text{x}})-2\text{x}(1-\log\text{x})}{\text{x}^{4}}=\frac{-3+2\log\text{x}}{\text{x}^{3}}$
$\Rightarrow\text{f}''(\text{e})=\frac{-3+2\log\text{x}}{\text{x}^{3}}=\frac{-1}{\text{e}^{3}}<0$
So, x = e is the point of local maximum.

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