Show that the function given by $f(x) = 3x + 17$ is strictly increasing on $R$.

(A) Let $x_{1}$ and $x_{2}$ be any two real numbers such that $x_{1} < x_{2}$.
Multiplying both sides by $3$,we get $3x_{1} < 3x_{2}$.
Adding $17$ to both sides,we get $3x_{1} + 17 < 3x_{2} + 17$.
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 $f(x) = 3x + 17$ is strictly increasing on $R$.
Alternative Method:
Find the derivative of the function: $f'(x) = \frac{d}{dx}(3x + 17) = 3$.
Since $f'(x) = 3 > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing on $R$.