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 \(R\).