Show that the function given by f(x) = x x has maximum at x = e. - Vidyadip

Question

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

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Answer

It is given that f(x) = $\frac{\log x}{x}$
Then, f'(x) = $\frac{\mathrm{x}\left(\frac{1}{\mathrm{x}}\right)-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}$
Now, f'(x) = 0
$\Rightarrow$ 1 - log x = 0
$\Rightarrow$ log x = 1
$\Rightarrow$ log x = log e
$\Rightarrow$ x = e
Further, $f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}$
= $\frac{-x-2 x(1-\log x)}{x^{4}}$
= $\frac{-3+2 \log x}{x^{3}}$
Now, $f^\prime{^\prime}$(e) = $\frac{-3+2 \log e}{e^{3}}=\frac{-3+2}{e^{3}}=\frac{-1}{e^{3}}<0$
Therefore, by second derivative test, f is the maximum at x = e.

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