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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!

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