Question:
Show that $\frac{\log x}{x}$ has a maximum value at $x=e$.
Solution:
Here,
$f(x)=\frac{\log x}{x}$
$\Rightarrow f^{\prime}(x)=\frac{1-\log x}{x^{2}}$
For the local maxima or minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow \frac{1-\log x}{x^{2}}=0$
$\Rightarrow 1=\log x$
$\Rightarrow \log e=\log x$
$\Rightarrow x=e$
Now,
$f^{\prime \prime}(x)=\frac{x^{2}\left(\frac{-1}{x}\right)-2 x(1-\log x)}{x^{4}}=\frac{-3+2 \log x}{x^{3}}$
$\Rightarrow f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^{3}}=\frac{-1}{e^{3}}<0$
So, $x=e$ is the point of local maximum.
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