Mathematics
OpenStudy (anonymous):

How do you find the domain of f(r,s)=sqrt(1-r)-e^(r/s)?

OpenStudy (anonymous):

$f(r,s)=\sqrt{1-r}-e^{r/s}$ correct?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Okay, what's the domain of the square root function? Given $$f(x)=\sqrt x$$, clearly, the domain is $$x\ge0$$, right? How would this translate to your function?

OpenStudy (anonymous):

For the first term, that is.

OpenStudy (anonymous):

r cannot be greater than one?

OpenStudy (anonymous):

Right! Now onto the next term. What do you know about the exponential function, $$f(x)=e^x$$ ? What's the domain here, in terms of $$x$$?

OpenStudy (anonymous):

all real numbers?

OpenStudy (anonymous):

Right again! However, the exponent isn't just $$x$$, or $$r$$ or $$s$$. Since it's $$\dfrac{r}{s}$$, you have to correct for this. What values of $$r$$ or $$s$$ can't be used here?

OpenStudy (anonymous):

well, r can be anything...even zero. but s can't be zero?

OpenStudy (anonymous):

Yes. So what would the overall domain of $$f(r,s)$$ be?

OpenStudy (anonymous):

$s \neq 0$and$r \ge 1$but is there a certain way that it needs to be formatted to be "correct?"

OpenStudy (anonymous):

Correct! There are a few ways of writing the domain set, but here's one of the more common forms for 2-variable functions like this one: $f(x)=\left\{(r,s)\in\mathbb{R}^2~|~r\ge1,~s\not=0\right\}$ Which would translate, roughly, to "the set of all coordinate points in the x-y (r-s, or real) plane such that r is at least 1 and s is non-zero."

OpenStudy (anonymous):

okay, thank you so much!

OpenStudy (anonymous):

You're welcome!