Show that the function given by $f(x) = e^{2x}$ is strictly increasing on $R$.
(N/A) Let $x_{1}$ and $x_{2}$ be any two real numbers in $R$ such that $x_{1} < x_{2}$.
Multiplying both sides by $2$,we get $2x_{1} < 2x_{2}$.
Since the exponential function $f(t) = e^{t}$ is a strictly increasing function,we have $e^{2x_{1}} < e^{2x_{2}}$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = e^{2x}$ is strictly increasing on $R$.
Multiplying both sides by $2$,we get $2x_{1} < 2x_{2}$.
Since the exponential function $f(t) = e^{t}$ is a strictly increasing function,we have $e^{2x_{1}} < e^{2x_{2}}$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = e^{2x}$ is strictly increasing on $R$.
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