Question
Write the maximum value of $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$, if it exists.

Answer

Given, $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$
limplies that $\text{f}'(\text{x})=\frac{1-\log\text{x}}{\text{x}^{2}}$
For a local maxima or a local minima, we must have f'(x) = 0
lmplies that $\frac{1-\log\text{x}}{\text{x}^{2}}=0$
lmplies that $1-\log\text{x} = 0$
lmplies that $\log\text{x}=1$
lmplies that $\log\text{x} =\log\text{e}$
limplies that x = e
Now, $\text{f}''(\text{x})=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^{4}}$
$=\frac{-3\text{x}-2\text{x}\ \log\text{x}}{\text{x}^{4}}$
At x = e
$\text{f}''(\text{e})=\frac{-3\text{x}-2\text{x}\ \log\text{e}}{\text{e}^{4}}\text{e}$
$=\frac{-5}{\text{e}^{3}}<0$
Therefore, x = e is a point of local maximum.
Thus, the local maximum value is given by
$\text{f}(\text{e})=\frac{\log\text{e}}{\text{e}}=\frac{1}{\text{e}}$.

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