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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} $


$\ 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...

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You can check the continuity properties of the function to see where it can go wrong for the range. – user13838 Nov 6 '11 at 2:27
Yeah, but which point you suggest. – nEAnnam Nov 6 '11 at 2:39
Shaun's answer explains it, I am just curious why did you feel you need to rationalize.... – N. S. Nov 6 '11 at 2:43
Is this really a recommended way to find ranges? In my day we'd have identified critical points and then taken the union of the range of every monotonic interval. – Henning Makholm Nov 6 '11 at 2:45
@user9176 yeah, i was doing wrong, just needed to isolate the x. thank you man. – nEAnnam Nov 6 '11 at 2:50
up vote 1 down vote accepted

Check what happens when you plug in $y=-3$ to $\sqrt{16 + y^2} + 2 - y$. (Spoiler: you don't get 0). What happened is that your step 3 in the finding of the roots is a non-invertible algebraic step. While the two sides of the equation remain equal after squaring, there is the unfortunate side-effect of introducing extraneous solutions.

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ooo yeah, i remember, lol. So do you know how can i conclude that the range is all Real Numbers. Thank you – nEAnnam Nov 6 '11 at 2:44
By the way, the rationalizing step is unnecessary (and could have led to even more extraneous solutions). Since the $y$-domain of $\frac{\sqrt{16+y^2} - 2 + y}{2}$ is $\mathbb{R}$, the range of the original function is $\mathbb{R}$. In essence, to obtain any given $y$-value, I simple let $x$ be the corresponding $x = \frac{\sqrt{16+y^2} - 2 + y}{2}$, – Shaun Ault Nov 6 '11 at 2:45
Absolutely true, thanks man, really. – nEAnnam Nov 6 '11 at 2:47

If you know calculus, you could argue this way: the function has a vertical asymptote at $x=-1$, and goes to negative infinity there when you approach from the right.

Now look at the derivative for $x > -1$: show that it is always positive (the numerator is quadratic, and has no real roots), and even more, is always greater than $1$ -- so that the function must in fact increase and go to infinity as $x$ goes to infinity. And since it's continuous for those $x$ values, its range is all of $\mathbb{R}$.

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