September 2nd, 2014, 09:45 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  range of y
Find range of y=(1x^2)/(1+x^2) Last edited by skipjack; September 2nd, 2014 at 11:47 AM. 
September 2nd, 2014, 10:16 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
$$\lim_{x \to \infty} y = \lim_{x \to +\infty} y = 1 \\ \frac{\mathbb{d}y}{\mathbb{d}x} = 0 \implies x=0 \\ y(0) = 1$$ and $y$ is bounded everywhere. So the range of $y$ is $(1, 1]$. 
September 2nd, 2014, 11:59 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
As y = 1 + 2/(1 + x²) = 1  2x²/(1 + x²), y ∈ (1,1].
