So I know domain implies input, but how would one go about finding the domain of a problem such as this:

$\displaystyle y=\frac{\sqrt{1-x}}{x}$

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- Sep 30th 2009, 12:59 PMSHiFTDomain of a Function
So I know domain implies input, but how would one go about finding the domain of a problem such as this:

$\displaystyle y=\frac{\sqrt{1-x}}{x}$ - Sep 30th 2009, 01:04 PMe^(i*pi)
- Sep 30th 2009, 01:05 PMpflo
Two things about functions such as this one: 1) it won't exist when there is a negative inside the square root and 2) it won't exist when you're dividing by zero.

Based on the first thing:

$\displaystyle 1-x\ge0$

So $\displaystyle x\le1$

Based on the second thing, the denominator cannot be zero. Since the denominator is x, $\displaystyle x\ne0$

The domain is all real numbers less than or equal to 1 except 0. - Sep 30th 2009, 01:14 PMSHiFT
Thanks a lot guys, It's hard for me to understand a problem when I'm looking at it, but once I see the solution it makes so much sense.

- Sep 30th 2009, 10:51 PMA Beautiful Mind
If you get something like this, it's just $\displaystyle x \neq$ anything in the denominator that's going to make it zero.

Since it was just plain old $\displaystyle x$, you'd think about making it zero since any number you could possibly plug in would not make the denominator zero unless it was 0 itself.

This goes for anything like that pretty much.

Like take for instance:

$\displaystyle f(x) = \frac{x+3}{x^2-9}. $

What's gonna make it zero?

Well a negative and a positive squared is always going to end up positive and to make it zero what would you need to get make it that way? Well, you'd need some number $\displaystyle x $squared to make 9. What number $\displaystyle x^2 = 9?$ $\displaystyle 3. -3$ also though too because $\displaystyle (-3)^2 = (-3)(-3) = 9. $

$\displaystyle f(x) = \frac{x+3}{9-9}. $

Can't divide by zero.

$\displaystyle x \neq 3$ $\displaystyle or -3.$