Question

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

Answer 1

Let $x_1$ and $x_2$ be any two numbers in $R$.
Then, we have:
$x_1<x_2=3x_1<3x_2=3x_1+17<3x_2+17=f(x_1)<f(x_2)$
Hence, $f$ is strictly increasing on $R$.
Alternate method: $f'(x) = 3 > 0$, in every interval of $R$. Thus, the function is strictly increasing on $R$.