Question

Show that $f\left(x\right)=e^{1/x}$, $x\ne 0$ is a decreasing function for all $x\ne 0$.

Answer 1

$f\left(x\right)=e^{1/x}x\ne 0$

$f^{\prime }\left(x\right)=e^{1/x}-\frac{1}{x^2}=\frac{-e^{1/x}}{x^2}$

for the function be decreasing or increasing $f^{\prime }\left(x\right)$ should be less than or more than equal to $0$

since the function is decreasing

$e^{1/x}>0\forall x\ne 0$ and $-\frac{1}{x^2}<0\forall \ne 0$

$\therefore f^{\prime }\left(x\right)=-\left(\frac{1}{x^2}\right)e^{1/x}<0\forall x\ne 0$