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

Q: Show that the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.

Explanation: Let $x_1$ and $x_2$ be any two real numbers such that $x_1 < x_2$. Then: \[x_1 < x_2 \Rightarrow 2x_1 < 2x_2\] \[\Rightarrow e^{2x_1} < e^{2x_2}\] \[\Rightarrow f(x_1) < f(x_2)\] Hence, $f(x)$ is strictly increasing on $\mathbb{R}$. \[\] $\bf Alternate Method:$ Differentiating w.r.x. we get \[f'(x) =2 e^{2x}\] It is always positive for any value of x. So $f(x)$ is strictly increasing on $\mathbb{R}$.

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Explanation: Let $x_1$ and $x_2$ be any two real numbers such that $x_1 < x_2$.

Then:
$$x_1 < x_2 \Rightarrow 2x_1 < 2x_2$$
$$\Rightarrow e^{2x_1} < e^{2x_2}$$
$$\Rightarrow f(x_1) < f(x_2)$$

Hence, $f(x)$ is strictly increasing on $\mathbb{R}$.

$\bf Alternate Method:$ Differentiating w.r.x. we get
$$f'(x) =2 e^{2x}$$
It is always positive for any value of x. So $f(x)$ is strictly increasing on $\mathbb{R}$.

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