Browse Questions

# Show that the function given by $f(x) = e^{2x}$ is strictly increasing on $R$.

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Given : $f(x)=e^{2x}$
Differentiating on both sides we get,
$f'(x)=2e^{2x}$
Let us consider the three cases,
Step 2:
Case (i) When $x > 0$,then
$f'(x)=2e^{2x}$
$\quad\;\;\;\;=2[1+(2x)+\large\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+......]$
Therefore $f'(x)>0$ for all $x>0$
Step 3:
Case (ii) When $x = 0$,then
$f'(x)=2e^0$
But $e^0=1$
Then $f'(x)=2$
$f'(x)>0$,when $x=0$
Step 4:
Case (iii) When $x < 0$,then
$f'(x)=2e^{-2x}$
$\quad\;\;\;\;=\large\frac{2}{e^{2x}}$
Again this is a positive quantity.
Therefore $f'(x) > 0$ when $x < 0$