another domain and range problem

renegade05

renegade05

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Question: Using proper set notation, describe the domain of \(\displaystyle f(x,y)=cose^{-1}(2x^2+y^2-3)\). What is the domain, what is the range of \(\displaystyle f\)?

So, I started off with realizing that \(\displaystyle 2x^2+y^2-3\) has to be between -1 and 1. Or equal to -1 or 1.

So, how do I go about doing this one? Solve the inequality? \(\displaystyle -1<=2x^2+y^2-3<=1\) ? And then what?

And isn't the range just between 0 and \(\displaystyle \pi\) ?

Little help..
 
Perhaps you should have a good close look at 2x^2 + y^2 = 3 before you think you are done.
 
Isnt that just an elliptic cylinder in R^3 ?

What are you suggesting i do with that?
 
Split it up so I get:

\(\displaystyle y^2+2x^2\leq4\)
and
\(\displaystyle 2\leq2x^2+y^2\)

So it appears that the domain is a ring in the xy plane.

How is my logic? And how do i display my answer in set notation?
 
I'd go with "Elliptical Ring", but that's pretty good. It was the hole in the middle that I wanted to make sure you found.

How about \(\displaystyle \{(x,y)|x\in\mathbb{R}, y\in\mathbb{R},2\le 2x^{2}+y^{2}\le 4\}\)?
 
renegade05 said:
Question: Using proper set notation, describe the domain of \(\displaystyle f(x,y)=cose^{-1}(2x^2+y^2-3)\). What is the domain, what is the range of \(\displaystyle f\)?

So, I started off with realizing that \(\displaystyle 2x^2+y^2-3\) has to be between -1 and 1. Or equal to -1 or 1.

So, how do I go about doing this one? Solve the inequality? \(\displaystyle -1<=2x^2+y^2-3<=1\) ? And then what?

And isn't the range just between 0 and \(\displaystyle \pi\) ?

Little help..
Click to expand...

What does cose-1 mean - I have not seen that expression (cose) before. Is it "cosh" by any chance?
 
What does cose-1 mean - I have not seen that expression (cose) before. Is it "cosh" by any chance?
Click to expand...

Nope, just a consequence of posting at 2 am.

It is arc-cosine.
 
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