1. ## Largest possible domain

Hey there, I'm just trying to work out how to find the largest possible domain to the following function algebraically.

$\displaystyle f(x)=\sqrt{\frac{x-1}{x+2}}$

The way I've been going about it is:

$\displaystyle x-1$ must be greater than zero, so $\displaystyle x\geq1$

$\displaystyle x+2$ must be greater than zero, so $\displaystyle x\geq-2$

and $\displaystyle \sqrt{x+2}$ can't equal zero, meaning, for this part, $\displaystyle x$ can be any real number besides $\displaystyle -2$

From this I get the domain of $\displaystyle \left [1,\infty\right )$

Which is not the complete domain stated in the answer. I must be missing something...

2. Hi Stroodle
Originally Posted by Stroodle
The way I've been going about it is:

$\displaystyle x-1$ must be greater than zero, so $\displaystyle x\geq1$

$\displaystyle x+2$ must be greater than zero, so $\displaystyle x\geq-2$
Your mistake is here because (x-1) and (x+2) can be less than zero at the same time. What you should do is :

$\displaystyle \frac{x-1}{x+2}\geq 0$

3. Ahh, of course

Thanks heaps for that!

4. To solve $\displaystyle \frac{x-1}{x+2}\geq 0$,

you need to consider three ranges, $\displaystyle x<-2$, $\displaystyle -2<x<1$ and $\displaystyle x>1$.

Choose the range of x s.t. both $\displaystyle (x-1)$ and $\displaystyle (x+2)$ are positive, and that will be $\displaystyle x<-2$ and $\displaystyle x>1$. Also, the fraction is equal to zero when $\displaystyle x=1$.

Hence, the largest possible domain is $\displaystyle (-\infty,-2)\cup[1,\infty)$

5. I'm fine with that part, just stupidly overlooked that both the numerator and denominator could be negative

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