# Show that

Question:

Show that $f(x)=e^{2 x}$ is increasing on $R$.

Solution:

Given:- Function $f(x)=e^{2 x}$

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 $(a, b)$

Algorithm:-

(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,

$f(x)=e^{2 x}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{2 \mathrm{x}}\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{e}^{2 \mathrm{x}}$

For $f(x)$ to be increasing, we must have

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

$\Rightarrow 2 \mathrm{e}^{2 \mathrm{x}}>0$

$\Rightarrow \mathrm{e}^{2 x}>0$

since, the value of e lies between 2 and 3

so, whatever be the power of e (i.e $x$ in domain $R$ ) will be greater than zero.

Thus $f(x)$ is increasing on interval $R$