Question

Is domain of f(x,y) = 1 + (4 -y^2)^1/2 open, closed or neither is it bounded or unbounded?

Answer 1

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:

$f(x,y) = 1 + (4 -y^2)^{\frac{1}{2}} \:id\:est\:f(x,y)=1+\sqrt{(4 -y^2)}$

Since every quadratic root are defined for values $\geq$ 0  then, this help us to understand that we need calculate what interval this Domain is:

$4-y^{2} \geq 0\\4-y^{2}-4\geq -4\\-y^{2}\geq-4 \\y^{2}\leq4\\-2\leq y\leq 2$

$D=[-2,2]$

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.