# Thread: How To Find The Domain And Range Of A Two-Variable Function?

1. ## How 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..

2. Originally Posted by AlphaRock
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..
If you are using nested brackets, they're written like this:

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

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...

$f(x, y) = \frac{[\ln{(2x - y)}]\sqrt{y}}{\sqrt{4 - x}}$?

3. 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 [].

4. Originally Posted by AlphaRock
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. That is correct.

sqrt(4-x)

x > 4. Should be x ≤ 4.

Domain: 2x > y, but x > 4. Don't think this is right but probably on the right track..
Don't forget the term $\sqrt y$, which gives you a third inequality. Other than that, you are on the right track.

5. Originally Posted by AlphaRock
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 [].

Anyway, you're on the right track.

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

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

$4 - x \neq 0$

$x \neq 4$.

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

So $4 - x \geq 0$

$x \leq 4$.

But since we already said $x \neq 4$

that means $x < 4$.

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

So $2x - y > 0$

$y < 2x$.

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

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

Putting it all together, you have:

Domain: $x < 4$

Range: $0 \leq y < 8$.

6. As Prove It says, the inequalities are $0\leqslant y<2x<8$. 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 $-\infty$ (as you approach the line y = 2x) to $+\infty$ (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.

7. Thanks guys.

Out of curiosity, How would we graph the function's domain? Can somebody show me what it would look like?

8. Originally Posted by AlphaRock
Thanks guys.

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

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# www.find the domain and range of loge (1/square root x square -9)

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