Domain of a function

1. Mar 22, 2016

says

1. The problem statement, all variables and given/known data
Find the domain of the function: f(x,y) = xy*√(x2+y)

2. Relevant equations
The domain is the set of all input values for which the function is defined.

3. The attempt at a solution
D = {(x,y): xy>0}

I plugged the function into Wolfram and it gave the result:

D = {(x,y): x2+y ≥ 0 }

I can see how that solution is correct, but isn't mine as well? I'm saying that the product of x and y is greater than 0, which implies neither x nor y = 0, hence the > symbol.

The wolfram solution seems a little less explicit to me.

2. Mar 22, 2016

BvU

Your D would be the first and the third quadrant. What is f(-1,-3) according to you ?
Sketch the two answers in the xy plane and see that you include areas that should be excludede and vice versa.

3. Mar 22, 2016

says

f(-1,-3) is in the second quadrant of the domain, and according to my domain it maps to 3.

I've stated that D=(x,y): xy>0.that whatever the x or y values take on in the domain, the output will always be greater than 0.

This is incorrect though. The product of the equation states √(x2+y). The point (-3,1) could not be mapped in the reals because √(32+1) = √-8.

√(x2+y) has to be greater than or equal to 0, therefore x2+y ≥ 0

4. Mar 22, 2016

BvU

Correct. For a point x,y to be in the domain, f(x,y) has to exist. It doesn't have to be > 0. (  quote added to indicate what precisely is correct. )

Oops ! I mentioned the point (1, -3) for which x2 + y = -2 < 0 so (-1,3) is not in D.

The point (-3,1), however, happily resides in D: f(-3,1) = - 3 √8 !!

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No. f(-1,-3) does not exist.

You seem to think f(x,y) = x,y but it is not.

Furthermore, you seem to mix up domain and range. Check your notes/textbook.

Last edited: Mar 22, 2016
5. Mar 22, 2016

SammyS

Staff Emeritus

If y=0, then x can be any number.

If x=0, then y≥0 .

The important issue here is that $\ \sqrt{x^2+y\,}\$ gives a real number result so long as $\ x^2+y\ge0\$.

6. Mar 22, 2016

Ray Vickson

Draw the curve $C: \:y = -x^2$ in the $(x,y)$ plane. For a point $p =(x,y)$ to be in the domain of $f$, where must $p$ be located in relation to the curve $C$?

7. Mar 22, 2016

says

I meant to say that the a point (x,y) is in a quadrant on the xy-plane and a function will either map or not map this point. (i.e. it exists or does not exist.)

Last edited by a moderator: Mar 22, 2016
8. Mar 22, 2016

Staff: Mentor

That's pretty vague, to say "a point (x,y) is in a quadrant on the xy-plane". Also, a point in the x-y plane exists whether or not f maps it to some value.

There's a fair amount of confusion in this thread. The domain is not, as has been mentioned, the first and third quadrants. The domain is $\{(x, y) | x^2 + y \ge 0\}$. It would be helpful to sketch a graph of the domain, which is a geometric figure you should already know, together with all the interior points of this figure.

The important thing if for the quantity inside the radical to be >= 0. All you need to say is that $x^2 + y \ge 0$.