# Implied Domain

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

Thanks heaps
• June 14th 2009, 03:11 AM
mr fantastic
Quote:

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

Thanks heaps

For the first one you require all values of $x$ such that $x + 1 \neq 0$.

For the second one, you require all values of $x$ such that $\frac{x - 1}{x + 2} \geq 0$.
• June 14th 2009, 03:29 AM
Stroodle
Thanks. I get the first one now, but my problem is solving $\frac{x-1}{x+2}\geq0$

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

What's the command for squiggly brackets in latex btw? :)
• June 14th 2009, 03:38 AM
AMI
I think for the first one there are no restrictions, because $\frac{x^2-1}{x+1}=x-1$ so the implied domain should be $\mathbb{R}$.
• June 14th 2009, 03:54 AM
Stroodle
For the first one I'm getting $\frac{x^2-1}{x+1}=2+\frac{-2}{x+1}$
• June 14th 2009, 04:00 AM
AMI
Quote:

Originally Posted by Stroodle
For the first one I'm getting $\frac{x^2-1}{x+1}=2+\frac{-2}{x+1}$

That's not correct.
$2+\frac{-2}{x+1}=\frac{2x}{x+1}$
• June 14th 2009, 04:09 AM
Raoh
hi
for the first one : $\mathbb{R}$
for the second : $]-\infty,-2[ \cup [1,+\infty[$
• June 14th 2009, 04:10 AM
Stroodle
Hmm. I'm definitely doing something wrong here. I'm getting $\frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1}=1+\frac{x-1}{x+1}=1+\frac{(x+1)-2}{x+1}=2+\frac{-2}{x+1}$
Which must be wrong. But the answer is definitely R/{-1}
• June 14th 2009, 04:20 AM
Stroodle
Quote:

Originally Posted by Raoh
hi
for the first one : $\mathbb{R}$
for the second : $]-\infty,-2[ \cup [1,+\infty[$

Thanks Raoh.
How'd you get the second answer?
• June 14th 2009, 04:23 AM
mr fantastic
Quote:

Originally Posted by AMI
I think for the first one there are no restrictions, because $\frac{x^2-1}{x+1}=x-1$ so the implied domain should be $\mathbb{R}$.

Incorrect. If $x = -1$ then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT $x = -1$. The graph of $y = \frac{x^2 - 1}{x + 1}$ has a 'hole' at $x = -1$.

Quote:

Originally Posted by Raoh
hi
for the first one : $\mathbb{R}$
[snip]

Incorrect. See above.

Quote:

Originally Posted by Raoh
[snip]
for the second : $]-\infty,-2[ \cup [1,+\infty[$

Your bracket notation needs some minor corrections:

${\color{red} (}-\infty,-2{\color{red})} \cup [1,+\infty{\color{red})}$
• June 14th 2009, 04:27 AM
mr fantastic
Quote:

Originally Posted by Stroodle
Hmm. I'm definitely doing something wrong here. I'm getting $\frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1}$ Mr F says: Correct. Now cancel the common factor of (x + 1) to get x - 1. You can do this provided ${\color{red}x \neq -1}$ (otherwise you're dividing by zero).

$=1+\frac{x-1}{x+1}=1+\frac{(x+1)-2}{x+1}=2+\frac{-2}{x+1}$ Mr F says: Wrong.
Which must be wrong. But the answer is definitely R/{-1}

Quote:

Originally Posted by Stroodle
Quote:

Originally Posted by Raoh
hi

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

for the second : $]-\infty,-2[ \cup [1,+\infty[$

Thanks Raoh.

How'd you get the second answer?

Quote:

Originally Posted by mr fantastic
[snip]
For the second one, you require all values of $x$ such that $\frac{x - 1}{x + 2} \geq 0$.

Case 1: $x - 1 \geq 0$ AND $x + 2 > 0 \Rightarrow x \geq 1$.

Case 2: $x - 1 \leq 0$ AND $x + 2 < 0 \Rightarrow x < -2$.
• June 14th 2009, 04:36 AM
Raoh
$\mathbb{R}/\left \{ -1 \right \}$ (sorry about later)
for the second one : study the sign of $x-1$ and $x+2$ then study $\frac{x-1}{x+2}$
and see where $\frac{x-1}{x+2}$ is positive.
• June 14th 2009, 04:43 AM
Stroodle
Awesome. Thanks. I made some pretty silly errors there :)
• June 14th 2009, 05:35 AM
AMI
Quote:

Originally Posted by mr fantastic
Incorrect. If $x = -1$ then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT $x = -1$. The graph of $y = \frac{x^2 - 1}{x + 1}$ has a 'hole' at $x = -1$.

My judgement was that we don't actually divide by $x+1$ since the final expression is $x-1$, but I understand what you mean.
So, for example, $\frac{x^2}{x}$ will not be defined in $0$ even though it's equal to $x$??
• June 14th 2009, 06:14 AM
AMI
Oh, nevermind, I think I got it.
I think I was saying something like "we don't need any restrictions for the function $x\mapsto (\sqrt{x})^2$ because it's equal to $x$", which is obviously false :mad: