Answer:
Closed and Bounded.
Step-by-step explanation:
Hi there!
1) Let's start by finding the values in which the function is defined. Remember that this can be rewritten:
[tex]f(x,y) = 1 + (4 -y^2)^{\frac{1}{2}} \:id\:est\:f(x,y)=1+\sqrt{(4 -y^2)}[/tex]
Since every quadratic root are defined for values [tex]\geq[/tex] 0 then, this help us to understand that we need calculate what interval this Domain is:
[tex]4-y^{2} \geq 0\\4-y^{2}-4\geq -4\\-y^{2}\geq-4 \\y^{2}\leq4\\-2\leq y\leq 2[/tex]
[tex]D=[-2,2][/tex]
2) Graphically speaking, the domain is closed. For the values -2 and 2 are included, and bounded.
Bounded functions have all of their points contained by some circle origin centered. Check it out below.
