Question

Show that the maximum value of ${\left(\frac{1}{x}\right)}^x$ is $e^{1/e}$.

Answer 1

Let $y={\left(\frac{1}{x}\right)}^x$

$\Rightarrow Iny=xIn\frac{1}{x}$

Differentiating on both sides gives,

$\Rightarrow \frac{1}{y}.y^{\prime }=In\frac{1}{x}+\frac{x}{\frac{1}{x}}.\left(\frac{-1}{x^2}\right)$

$\Rightarrow \frac{y^{\prime }}{y}=In\frac{1}{x}-1$

$\Rightarrow y^{\prime }={\left(\frac{1}{x}\right)}^x(In\frac{1}{x}-1)$

At maximum value, $y^{\prime }=0,y^{\prime \prime }<0$

$y^{\prime }=0\Rightarrow In\frac{1}{x}=1\Rightarrow \frac{1}{x}=e\Rightarrow x=\frac{1}{e}$

$y^{\prime \prime }={\left(\frac{1}{x}\right)}^x(Inx-1{)}^2+{\left(\frac{1}{x}\right)}^x.\frac{-1}{x}$

$y^{\prime \prime }\left(\frac{-1}{e}\right)<0$.

$\therefore$ At $x=\frac{1}{e},y$ has maximum value equal to $(e{)}^{\frac{1}{e}}$