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Show that the function given by $f(x)=\frac{\log x}{x}$ has maximum at $x=e$.
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$f(x)=\frac{\log x}{x} f^{\prime}(x)=\frac{1-\log x}{x^2} 1$
For maxima and minima, $\mathrm{f}^{\prime}(\mathrm{x})=0$
$\Rightarrow \log \mathrm{x}=0$ or $\log \mathrm{x}=1 \therefore \quad \mathrm{x}=\mathrm{e}^{\prime}=\mathrm{e}$
$\mathrm{f}^{\prime \prime}(\mathrm{k})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^2}\right)=-\frac{3-2 \log \mathrm{x}}{\mathrm{x}^3}$ $\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{3-2 \log _{\mathrm{e}} \mathrm{e}}{\mathrm{e}^3}=-\frac{3-2}{\mathrm{e}^3}=-\frac{1}{\mathrm{e}^3}=-\mathrm{ve}$
$\therefore$ f is maximum at $\mathrm{x}=\mathrm{e}$.
For maxima and minima, $\mathrm{f}^{\prime}(\mathrm{x})=0$
$\Rightarrow \log \mathrm{x}=0$ or $\log \mathrm{x}=1 \therefore \quad \mathrm{x}=\mathrm{e}^{\prime}=\mathrm{e}$
$\mathrm{f}^{\prime \prime}(\mathrm{k})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^2}\right)=-\frac{3-2 \log \mathrm{x}}{\mathrm{x}^3}$ $\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{3-2 \log _{\mathrm{e}} \mathrm{e}}{\mathrm{e}^3}=-\frac{3-2}{\mathrm{e}^3}=-\frac{1}{\mathrm{e}^3}=-\mathrm{ve}$
$\therefore$ f is maximum at $\mathrm{x}=\mathrm{e}$.
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