Question

Show that f(x) = e^1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Answer 1

$$\begin{gathered} f\left(x\right)=e^{\frac{1}{x}} \\ f\text{'}\left(x\right)=e^{\frac{1}{x}}\frac{d}{dx}\left(\frac{1}{x}\right) \\ =e^{\frac{1}{x}}\left(\frac{-1}{x^2}\right) \\ =-\frac{e^{{\displaystyle \frac{1}{x}}}}{x^2} \\ \text{Here,}{e}^{\frac{1}{x}}>0\text{and}{x}^2>0,\text{for any real value of}\mathit{\text{x ≠ 0.}} \\ \text{∴}f\mathit{\text{'}}\left(x\right)\mathit{=}\mathit{-}\frac{e^{\frac{\mathit{1}}{x}}}{x^{\mathit{2}}}\mathit{<}\mathit{0}\mathit{,}\forall x\mathit{\in }R,x\mathit{\ne }\mathit{0} \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) is a decreasing function.} \end{gathered}$$