Im confused with this function, i need to find the Range...
Original Function
$\ f(x)= \frac{x^2+2x-3}{x+1} $
In terms of y:
$\ y= \frac{x^2+2x-3}{x+1} $
Then x isolated: $\ x= \frac{\sqrt{16+y^2}-2+y}{2} $
Rationalizing:
$\ x= \frac{2y+6}{\sqrt{16+y^2}+2-y} $
Denominator should not be 0
, so we search wich value do that.
$\ \sqrt{16+y^2}+2-y=0 $
$\ \sqrt{16+y^2}=-2+y $
$\ (\sqrt{16+y^2})^2=(-2+y)^2 $
$\ 16+y^2=4-4y+y^2 $
$\ 12=-4y $
$\ y=-3 $
It's supposed that the Range is:
$\ \mathbb{R}-\{-3\}$
But i have 2 problems:
1.- The graphic shows that the Range is all $\ \mathbb{R}$
2.- wolframalpha says that $\ \sqrt{16+y^2}+2-y=0 $ have no solutions.
I want to know if i'm wrong going this way...