f(x,y) = ((ln(2x-y))sqrt(y))/(sqrt(4-x))

How do you find the domain and range and properly state it?

My work:

ln (2x-y)

2x-y > 0

2x > y.

sqrt(4-x)

x > 4.

Domain: 2x > y, but x > 4. Don't think this is right but probably on the right track..

Printable View

- March 27th 2010, 01:36 AMAlphaRockHow To Find The Domain And Range Of A Two-Variable Function?
f(x,y) = ((ln(2x-y))sqrt(y))/(sqrt(4-x))

How do you find the domain and range and properly state it?

My work:

ln (2x-y)

2x-y > 0

2x > y.

sqrt(4-x)

x > 4.

Domain: 2x > y, but x > 4. Don't think this is right but probably on the right track.. - March 27th 2010, 01:51 AMProve It
- March 27th 2010, 02:12 AMAlphaRock
Yes. You got it. Not sure if it makes a difference (it's probably the way I wrote it on here), but in the original function, there was no square brackets [].

- March 27th 2010, 02:17 AMOpalg
- March 27th 2010, 02:20 AMProve It
I added them for readability.

Anyway, you're on the right track.

First, you can't have the denominator = 0.

So

.

Next, you can't have the square root of a negative number (I assume you are dealing only with real numbers).

So

.

But since we already said

that means .

Now, you should know that you can only take the logarithm of a positive number.

So

.

And remembering that , this means .

Finally, you know that you can't have the square root of a negative number. So .

Putting it all together, you have:

Domain:

Range: . - March 27th 2010, 08:51 AMOpalg
As Prove It says, the inequalities are . But this is a function from two-dimensional space to the real numbers. So the domain must be a subset of the plane. In fact it is a triangular region bounded by the lines y = 0 (the x-axis), y = 2x, and x = 4. The range of the function is the set of values that it takes, which can go from (as you approach the line y = 2x) to (as you approach the line x = 4). The function is continuous, so it takes all intermediate values. In other words, the range is the whole real line.

- March 27th 2010, 01:41 PMAlphaRock
Thanks guys.

Out of curiosity, How would we graph the function's domain? Can somebody show me what it would look like? - March 27th 2010, 01:52 PMOpalg
See my previous comment. Draw the three lines y = 0 (the x-axis), y = 2x, and x = 4. They form the sides of a triangle, with vertices at (0,0), (4,0) and (4,8). Inside that triangle the inequalities , and are all satisfied. The domain of your function consists of the interior of the triangle, together with one of its sides (the x-axis).