# Range of rational function

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

• Note: $f(x)=\dfrac{(x+3)(x-1)}{(x-1)(x-2)}$ – J. W. Tanner Apr 23 at 3:45

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 $$f$$.

• Ah! Never thought we can analyse it this way.Thanks for making it clear – Scáthach Apr 23 at 6:25