How do I work out the implied domains for $\displaystyle f(x)=\frac{x^2-1}{x+1}$ and also for $\displaystyle f(x)=\sqrt\frac{x-1}{x+2}$

Thanks heaps

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- Jun 14th 2009, 02:02 AMStroodleImplied Domain
How do I work out the implied domains for $\displaystyle f(x)=\frac{x^2-1}{x+1}$ and also for $\displaystyle f(x)=\sqrt\frac{x-1}{x+2}$

Thanks heaps - Jun 14th 2009, 03:11 AMmr fantastic
- Jun 14th 2009, 03:29 AMStroodle
Thanks. I get the first one now, but my problem is solving $\displaystyle \frac{x-1}{x+2}\geq0$

I'm getting $\displaystyle 1-\frac{3}{x+2}$ so is the domain R\{-2} ?

THanks for your help

What's the command for squiggly brackets in latex btw? :) - Jun 14th 2009, 03:38 AMAMI
I think for the first one there are no restrictions, because $\displaystyle \frac{x^2-1}{x+1}=x-1$ so the implied domain should be $\displaystyle \mathbb{R}$.

- Jun 14th 2009, 03:54 AMStroodle
For the first one I'm getting $\displaystyle \frac{x^2-1}{x+1}=2+\frac{-2}{x+1}$

- Jun 14th 2009, 04:00 AMAMI
- Jun 14th 2009, 04:09 AMRaoh
hi

for the first one : $\displaystyle \mathbb{R}$

for the second : $\displaystyle ]-\infty,-2[ \cup [1,+\infty[$ - Jun 14th 2009, 04:10 AMStroodle
Hmm. I'm definitely doing something wrong here. I'm getting $\displaystyle \frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1}=1+\frac{x-1}{x+1}=1+\frac{(x+1)-2}{x+1}=2+\frac{-2}{x+1}$

Which must be wrong. But the answer is definitely R/{-1} - Jun 14th 2009, 04:20 AMStroodle
- Jun 14th 2009, 04:23 AMmr fantastic
Incorrect. If $\displaystyle x = -1$ then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT $\displaystyle x = -1$. The graph of $\displaystyle y = \frac{x^2 - 1}{x + 1}$ has a 'hole' at $\displaystyle x = -1$.

Incorrect. See above.

Your bracket notation needs some minor corrections:

$\displaystyle {\color{red} (}-\infty,-2{\color{red})} \cup [1,+\infty{\color{red})}$ - Jun 14th 2009, 04:27 AMmr fantastic
- Jun 14th 2009, 04:36 AMRaoh
$\displaystyle \mathbb{R}/\left \{ -1 \right \}$ (sorry about later)

for the second one : study the sign of $\displaystyle x-1$ and $\displaystyle x+2 $ then study $\displaystyle \frac{x-1}{x+2}$

and see where $\displaystyle \frac{x-1}{x+2}$ is positive. - Jun 14th 2009, 04:43 AMStroodle
Awesome. Thanks. I made some pretty silly errors there :)

- Jun 14th 2009, 05:35 AMAMI
My judgement was that we don't actually divide by $\displaystyle x+1$ since the final expression is $\displaystyle x-1$, but I understand what you mean.

So, for example, $\displaystyle \frac{x^2}{x}$ will not be defined in $\displaystyle 0$ even though it's equal to $\displaystyle x$?? - Jun 14th 2009, 06:14 AMAMI
Oh, nevermind, I think I got it.

I think I was saying something like*"we don't need any restrictions for the function*$\displaystyle x\mapsto (\sqrt{x})^2$*because it's equal to*$\displaystyle x$", which is obviously false :mad:

Sorry about that..(Speechless)