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..

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- Mar 27th 2010, 12:36 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.. - Mar 27th 2010, 12:51 AMProve It
If you are using nested brackets, they're written like this:

$\displaystyle \left\{\left[\left(\right)\right]\right\}$

In other words, round brackets, then square brackets around them, and then curly brackets around them... Please use this system to make things easier to read...

Anyway, is your function

$\displaystyle f(x, y) = \frac{[\ln{(2x - y)}]\sqrt{y}}{\sqrt{4 - x}}$? - Mar 27th 2010, 01:12 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 [].

- Mar 27th 2010, 01:17 AMOpalg
- Mar 27th 2010, 01:20 AMProve It
I added them for readability.

Anyway, you're on the right track.

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

So $\displaystyle \sqrt{4 - x} \neq 0$

$\displaystyle 4 - x \neq 0$

$\displaystyle x \neq 4$.

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

So $\displaystyle 4 - x \geq 0$

$\displaystyle x \leq 4$.

But since we already said $\displaystyle x \neq 4$

that means $\displaystyle x < 4$.

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

So $\displaystyle 2x - y > 0$

$\displaystyle y < 2x$.

And remembering that $\displaystyle x < 4$, this means $\displaystyle y < 8$.

Finally, you know that you can't have the square root of a negative number. So $\displaystyle y \geq 0$.

Putting it all together, you have:

Domain: $\displaystyle x < 4$

Range: $\displaystyle 0 \leq y < 8$. - Mar 27th 2010, 07:51 AMOpalg
As Prove It says, the inequalities are $\displaystyle 0\leqslant y<2x<8$. 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 $\displaystyle -\infty$ (as you approach the line y = 2x) to $\displaystyle +\infty$ (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.

- Mar 27th 2010, 12:41 PMAlphaRock
Thanks guys.

Out of curiosity, How would we graph the function's domain? Can somebody show me what it would look like? - Mar 27th 2010, 12:52 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 $\displaystyle 0\leqslant y$, $\displaystyle y<2x $ and $\displaystyle 2x<8$ are all satisfied. The domain of your function consists of the interior of the triangle, together with one of its sides (the x-axis).