Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 5 maths textbook solution

f(x) is decreasing function for all x≠0

Given:

$f(x)=e^{\frac{1}{x}} ; x \neq 0$

To show:

We have to show that f(x) is decreasing function for all x≠0.

Hint:

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

Solution:

We have

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

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(e^{\frac{1}{x}}\right) \\ &=e^{\frac{1}{x}}\left(-\frac{1}{x^{2}}\right) \\ &\Rightarrow f^{\prime}(x)=-\frac{e^{\frac{1}{x}}}{x^{2}} \\ &\text { As given } x \in R, x \neq 0 \\ &\Rightarrow \frac{1}{x^{2}}>0 \text { and } e^{\frac{1}{x}}>0 \end{aligned}

Their ratio is also greater than 0.

\begin{aligned} &\Rightarrow \frac{e^{\frac{1}{x}}}{x^{2}}>0 \\ &\Rightarrow-\frac{e^{\frac{1}{x}}}{x^{2}}<0 \end{aligned}

As by applying negative sign change in comparison sign.

\begin{aligned} &\Rightarrow f'(x)< 0 \end{aligned}

Hence f(x) is decreasing for all x≠0.