# How to argue this Range

Function:

$$f(x)= \frac{x-3}{x^2+2x-8}$$

In terms of y:

$$y= \frac{x-3}{x^2+2x-8}$$

Then x isolated: $$x= \frac{\sqrt{36y^2-16y+1}-2y+1}{2y}$$

To find the Range we need to find the Domain of this 'new' Function.

1.- We must look if $\ \sqrt{36y^2-16y+1}$ is a Real number

2.- The Denominator $\ 2y$ should not be 0

1.- $\ \sqrt{36y^2-16y+1}$ will always be positive, so will be Real.

2.- We find the root: $\ 2y=0$ so $\ y=0$

Which left us that this Domain is all Real Numbers except 0, so this is the Range of the original Function.

But look at the graphic:

That wasn't the range, but i've found, i've found the roots of $\ {36y^2-16y+1}$ and those roots are that numbers accordingly to the graphic: .0752 and .03692, but i dont know how to argue this result.

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$$\ x_{1,2}= \frac{\pm\sqrt{36y^2-16y+1}-2y+1}{2y}$$ –  pedja Nov 10 '11 at 5:48
It is clear that $0$ is in the range, either from picture or by setting $x=3$. For solution of the quadratic in $x$, you forgot the $\pm$. It is much easier to use usual calculus max/min techniques (if you know calculus). –  André Nicolas Nov 10 '11 at 5:54
@pedja I dont understand what do you mean :( the another x changes result's? –  nEAnnam Nov 10 '11 at 5:55
you didn't isolate $x$ correctly –  pedja Nov 10 '11 at 5:58
but even with the second solution,it will not change the fact that y=0 and stays as before, sorry if im wrong –  nEAnnam Nov 10 '11 at 6:02

The function inside the square root, $36y^2 - 16y + 1$, is less than zero on the interval $(\frac{1}{18}(4-\sqrt{7}),\frac{1}{18}(4+\sqrt{7})) = (0.075236,0.369208)$ which makes the function undefined on that interval. This appears to correspond to the region outside the range of the graph of your function.
As for why $y$ can be zero when it would make the expression undefined, this is because in order to get the expression for $x$ we had to divide by $y$, which subtly changes the function (since we can't do that if $y=0$) and we have to check separately if $y=0$ will work. It is important to watch out for this sort of slight change when manipulating functions.
When $y=0$, you don't get a quadratic in $x$, you get a linear function in $x$ which corresponds to $x=3$. By solving it as a quadratic you implicitly discard the possibility that $y=0$; you need to do that case separately. –  Arturo Magidin Nov 10 '11 at 6:19