Question

Show that the function given by $f\left(x\right)=3^{2x}$ is strictly increasing on R.

Answer 1

Let $x_1$ and $x_2$ be any two numbers in R, where $x_1<x_2.$
Given $f\left(x\right)=e^{2x}$
Then, we have $x_1<x_2\Rightarrow 2x_1<2x_2\Rightarrow e^{2x_1}<e^{2x_2}\Rightarrow f\left(x_1\right)<f\left(x_2\right)$
Hence, f is strictly increasing on R.
Alternate method
Given $f\left(x\right)=e^{2x}$
On differentiating both sides w.r.t x, we get
$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(e^{2x}\right)\Rightarrow f^{\prime }\left(x\right)=2e^{2x}$
(i) when $x>0$, then $2e^{2x}=2[1+2x+\frac{(2x{)}^2}{2!}+\dots ]>0$
$\therefore f^{\prime }\left(x\right)>0$
(ii) when x=0 then $2e^{2x}=2.1=2>0\therefore f^{\prime }\left(x\right)>0$
(iii) when x<0, then $2e^{2x}=2\times e^{2(-x)}=\frac{2}{e^{2x}}(herex=-z,wherez>0)$
$=\frac{2}{[1+2z+\frac{(2z{)}^2}{2!}+\dots ]}>0\therefore f^{\prime }\left(x\right)>0$
Thus, for all values of x,f'(x)>0. So, f(x) is strictly increasing function on R.