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


Let x1 and x2 be any two numbers in R.

Then, we have,


x1 < x2


3x1 < 3x2


3x1 +17 < 3x2 +17


f(x1) < f(x2)


Therefore, f is strictly increasing on R.


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