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..
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..
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: .
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.
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).