How to find the Range of function $$f(x)= \frac{x^2+2x-3}{x^2 - 3x +2}$$
I made a quadratic equation in terms of y which came to be:
$$ y(x^2 - 3x +2)= x^2+2x-3 $$
$$\implies x^2(y-1)-x(3y+2)+2y+3=0$$
Now I made two cases :
When y=1
This means the quadratc expression reduces to a linear expression and gives x=1. But x=1 is not possible because no value exists there.So y=1 is not possible
When $y\neq1$
I set $D \geq0$
Which gives me $$(y+4)^2 \geq0$$
So this is true for all values of y .
Combining the results of two cases gives me range to be $y\in (R-1)$ but y is also not equal to {-4}.Can you tell me why is this true
Defining $f$ as
$$ f(x)= \frac{x^2+2x-3}{x^2 - 3x +2} $$
is equivalent to defining $f$ as
$$ f(x)=\frac{x+3}{x-2}\text{ for }x\ne1 $$
The range of $f$ equals the domain of $f^{-1}$ and the equation of $f^{-1}$ can be written as
$$ x=\frac{y+3}{y-2}\text{ for }y\ne1 $$
Solving for $y$ gives the equation
$$ y=\frac{2x+3}{x-1}\text{ for }y\ne1 $$
But $y\ne1$ is equivalent to $x\ne-4$. So the equation of $f^{-1}$ is
$$ f^{-1}(x)=\frac{2x+3}{x-1}\text{ for }x\ne-4 $$
The domain of $f^{-1}$ is $(-\infty,-4)\cup(-4,1)\cup(1,\infty)$ which is the range of $f$.