Show that


Show that $f(x)=e^{\frac{1}{x}}, x \neq 0$ is a decreasing function for all $x \neq 0$.


Given:- Function $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\frac{1}{x}}$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,


$\Rightarrow f(x)=\frac{d}{d x}\left(e^{\frac{1}{x}}\right)$

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

$\Rightarrow f(x)=-\frac{e^{\frac{1}{x}}}{x^{2}}$

As given $x \in R, x \neq 0$

$\Rightarrow \frac{1}{x^{2}}>0$ and $e^{\frac{1}{x}}>0$

Their ratio is also greater than 0

$\Rightarrow \frac{e^{\frac{1}{x}}}{x^{2}}>0$

$\Rightarrow-\frac{e^{\frac{1}{x}}}{x^{2}}<0$; as by applying -ve sign change in comparision sign

$\Rightarrow f^{\prime}(x)<0$

Hence, condition for $f(x)$ to be decreasing

Thus $f(x)$ is decreasing for all $x \neq 0$

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