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

Question

Philoid · Accepted Answer

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.