Trying to find the right answer to this..

Let x 1 , x 2 x ∊ R and let x 1 < x 2 Now x 1 < x 2 ⇒ a x 1 < a x 2: [ ∵ a > 0] ⇒ a x 1 + b < a x 2 + b ⇒ (x 1 ) < f (x 2 ) ∴ x 1 < x 2 ⇒ f (x 1 ) < f (x 2 ) ⇒ f is a strictly increasing function in R.

Application of Derivatives

Question

Wired Faculty App

Prove that f (x) = ax + b, where a and b are constants and a > 0 is an strictly increasing function for all real values of x. without using the derivative.

Answer

Let x₁ , x₂ x ∊ R and let x₁ < x₂Now x₁ < x₂⇒ a x₁ < a x_2: [ ∵ a > 0]
⇒ a x₁+ b < a x₂+ b ⇒ (x₁) < f (x₂)
∴ x₁ < x₂ ⇒ f (x₁) < f (x₂)
⇒ f is a strictly increasing function in R.