Finding Range and Domain of the function $g(x,y) = \sqrt{x - y} + \sqrt{x^{2} + y^{2} - 6}$

Question

I am trying to find domain and range of a function with equation

$$g(x, y) = \sqrt{x − y} + \sqrt {x^2 + y^2 − 6}$$

I have reduced this statement, when $g(x, y) = c = 0 $

$$ g(x, y) = (x+0.5)^2+(y-0.5)^2 = \sqrt{\frac{26}4}$$

Now this is a equation of a circle with centre at $(-0.5,0.5)$ and radius $\sqrt{\frac{26}4}$

But this is when $c=0$ for different values of $c$ we get different radius If I solve the equation by using arbitrary $c$,

then radius is $\sqrt{\frac{26+4c^2}4} $ now $\sqrt{\frac{26+4c^2}4}\ge0$ has no real roots In this case what would be the domain and range for this function

Answer 1

The domain of such function is given by:

\begin{align*} D_{g} & = \{(x,y)\in\mathbb{R}^{2} \mid x - y \geq 0\}\cap\{(x,y)\in\mathbb{R}^{2} \mid x^{2} + y^{2} - 6 \geq 0\}\\\\ & = \{(x,y)\in\mathbb{R}^{2} \mid x \geq y\}\cap\{(x,y)\in\mathbb{R}^{2} \mid x^{2} + y^{2} \geq 6\} \end{align*}

Here it is a picture which may help you visualize it better:

Answer 2

Regarding the range.

If you choose $x=y= \sqrt 3$ or $x=y= −\sqrt 3$ both terms in $g$ are zero and hence $g(x,y)=0$. Now obviously both square root terms must be larger or equal to zero, so it follows immediately that the minimum value of the range is equal to $0$.

For the upper value of $g(x,y)$, just consider the case of $y=0$ and $x$ going to $+\infty$. This shows that the maximum of the range is $+\infty$.