Function $\mathrm{f}(\mathrm{x})=\mathrm{e}^{-1 / \mathrm{x}}$ is strictly increasing for all $\mathrm{x}$…
Function $\mathrm{f}(\mathrm{x})=\mathrm{e}^{-1 / \mathrm{x}}$ is strictly increasing for all $\mathrm{x}$ where
$\mathrm{x}$ is only positive real number
$\mathrm{x}$ is only negative real number
$\mathrm{x}$ is a real number
$\mathrm{x}$ is a non - zero real number
Solution
$\begin{aligned}
& f(x)=e^{-\frac{1}{x}} \\
& f^{\prime}(x)=e^{-\frac{1}{x}}(-1)\left(\frac{-1}{x^2}\right)=\frac{1}{x^2 e^{\frac{1}{x}}}
\end{aligned}$
When $\mathrm{f}^{\prime}(\mathrm{x})>0, \mathrm{x}^2 \mathrm{e}^{\frac{1}{\mathrm{x}}}>0$ and $\mathrm{x} \neq 0$
Now $\mathrm{x}^2>0$ and $\mathrm{e}^{\frac{1}{\mathrm{x}}}>0$ for $\forall \mathrm{x} \in \mathrm{R}$
Hence $f(x)$ is an increasing function for $\forall x$, except $x=0$