Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
    \frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
I already know that:
\begin{equation}
    dom f = \{ f(x,y) ∈ \mathbb{R}^2\backslash(x^2y^2-4xy^2+3y^2=0) : \sqrt{4x-x^2-y^2} \geq 0 \}
\end{equation}
But of course this needs to be written in a more simpler form.
During class we solve simpler functions  like without fractions and roots, so I don't have anything that can help me.
After this I also need to find:


*

*b) zeros of the function  

*c) Calculating Algebraically, the range of the function:
\begin{equation}
T(x,y) = \sqrt{4x-x^2-y^2}
\end{equation}

*d) Extreme values of the function T


I'm of course not expecting the complete solution but something like a kick start.
 A: To find the zeros of the denominator, note that you can factor out $y^2$ leaving an expression in $x$.  The denominator is zero whenever $y^2=0$, which is $y=0$ or when that expression in $x$ is.  In your expression of $dom f$, you should not have the square root sign.  You need the stuff under the square root to be greater than or equal to zero.  What kind of curve is $4x-x^2-y^2$?  You should complete the square in $x$
A: You factored your first equation correctly. Now determine which values of $y$, $x$, yield a zero denominator:  $$y^2(x-3)(x-1) = 0 \implies y = 0, \text{ or } x = 3, \text{ or } x = 1$$
So all ordered pairs of the form $(x, 0), (3, y), (1, y), \; x, y \in \mathbb R$  must be omitted from the domain. One way to see this is to note that the horizontal line $y = 0$ (i.e., the $x$-axis), and the vertical lines $x = 1, x = 3$ are thus excluded from $\mathbb R^2$.
Also, the second restriction does not require  $\sqrt{4x-x^2-y^2} \geq 0$ (though it certainly will be true, given the proper restriction on the argument). Rather, we need the argument of the square root to be greater than or equal to zero, in order for the square root of the argument to be defined. $$\begin{align} 4x - x^2 - y^2 \geq 0 & \iff -4x + x^2 + y^2\leq 0\\ &\iff  x^2 - 4x + 4 + y^2 \leq 4 \\ &\iff (x - 2)^2 + y^2\leq (2)^2\end{align}$$
Note that $$ (x - 2)^2 + y^2= (2)^2$$ is the equation of a circle of radius 2, centered at the point $(2, 0)$. So $$(x - 2)^2 + y^2\leq (2)^2$$ is the subset of $\mathbb R^2$ consisting of all points on and inside the circle. 
So the domain will be equal to the portion of the plane on and inside circle (all points $(x, y)$ such that $(x - 2)^2 + y^2\leq (2)^2)$, and excluding the lines $y = 0, x=1, x=3$.
A: 
Can anyone give me a hint on how to find the domain in f?

You need to find all the values for which $x^2 y^2-4xy^2+3y^2$ is not equal to zero and $4x-x^2-y^2$ is positive or zero.

zeros of the function

Well you just need to find where $\sqrt{4x-x^2-y^2}=0$, which is just where $4x-x^2-y^2=0$.
A: Hints-kick start:


*

*$4x-x^2-y^2=4-(x-2)^2-y^2$

*$(x-2)^2+y^2=4$ is the equation of a circle

*$x^2y^2-4xy^2+3y^2=y^2((x-2)^2-1)=y^2(x-3)(x-1)$

A: we have for the domian two cases
$4x-x^2-y^2\geq 0$ and $y^2(x^2-4x+3)>0$
or
$4x-x^2-y^2\le 0$ and $y^2(x^2-4x+3)<0$
note that $x^2-4x+3=0$ gives $x_1=1$ or $x_2=3$ and $4x-x^2-y^2=4-(y^2+(x-2)^2)$ thus we get for the first case
$4>y^2+(x-2)^2$ and $x>3$ and $y\neq 0$
or
$4>y^2+(x-2)^2$ and $x<1$ and $y \neq 0$
and for the second case
$(x-2)^2+y^2\geq 4$ and $y\neq 0$ and $1<x<3$
sorry i thougt the problem was  $\sqrt{\frac{4x-x^2-y^2}{x^2y^2-4xy^2+3y^2}}$
b) the zeros of our functions are $(x-2)^2+y^2=4$ and $y^2\neq 0$ and $x\neq 3$ and $x\neq 1$
c) we have to solve
$T_x=\frac{2-x}{\sqrt{4x-x^2-y^2}}=0$
and
$T_y=\frac{y}{\sqrt{4x-x^2-y^2}}=0$
a) for the domain we have $(x-2)^2+y^2\le 4$ and $y\neq 0$ and $x\neq 3$ and $x\neq 1$
