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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

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

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

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