we have,f(x) = ax + b, a < 0let x1,x2 R and x1 > x2⇒ ax1 < ax2 for some a > 0⇒ ax1 + b< ax2 + b for some b⇒ f(x1) < f(x2)Hence, x1 > x2⇒ f(x1) < f(x2)So, f(x) is decreasing function of R

Q. 45.0( 2 Votes )

<span lang="EN-US

Class 12th RD Sharma - Volume 1 17. Increasing and Decreasing Functions

Answer :

we have,

f(x) = ax + b, a < 0

let x₁,x₂ R and x₁ > x₂

⇒ ax₁ < ax₂ for some a > 0

⇒ ax₁ + b< ax₂ + b for some b

⇒ f(x₁) < f(x₂)

Hence, x₁ > x₂⇒ f(x₁) < f(x₂)

So, f(x) is decreasing function of R

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

PREVIOUS<span lang="EN-US NEXT<span lang="EN-US

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation

view all courses