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

Question

Show that $\text{f}(\text{x})=\text{e}^{\frac{1}{\text{x}}},\text{x}\neq0$ is a decreasing function for all $\text{x}\neq0.$

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Answer

We have,
$\text{f}(\text{x})=\text{e}^{\frac{1}{\text{x}}},\text{x}\neq0$
$\text{f}'(\text{x})=\text{e}^{\frac{1}{\text{x}}}\times\Big(\frac{-1}{\text{x}^2}\Big)$
$\therefore\ \text{f}'(\text{x})=-\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}$
Now,
$\text{x}\in\text{R},\text{x}\neq0$
$\Rightarrow\frac{1}{\text{x}^2}>0\text{ and }\text{e}^{\frac{1}{\text{x}}}>0$
$\Rightarrow\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}>0$
$\Rightarrow-\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}<0$
$\Rightarrow\text{f}'(\text{x})<0$
Hence, f(x) is decreasing function for all $\text{x}\neq0.$

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