Q 49. - Calculus

Question

 Determine the domains of the functions in Exercises 47–56, and find where the functions are continuous.

$f(x,y)=\sqrt{x^2+y}$ 

Step-by-Step Solution

Answer

The function $f(x,y)=\sqrt{x^2+y}$ is continuous on the set  $\left\{\left(x,y\right)\in f^2:y\ge -x^2\right\}$.

Step 1. Given information.

We have given expression: $f(x,y)=\sqrt{x^2+y}$

Step 2: Determine the domains of the functions.

Consider the function:$g:f^2\to f$Then the domain of the function is

$Domain \left(g\right)=\left\{\left(x,y\right)\in f^2:g(x,y) is defined\right\}$

Since the rational function $f(x,y)=\sqrt{x^2+y}$ assumes real values of all $\left(x,y\right)\in f^2$ such that.

$$\begin{gathered} x^2+y\ge 0 \\ y\ge -x^2 \end{gathered}$$

The domain of the function is $Domain \left(f\right)=\left\{\left(x,y\right)\in f^2:y\ge -x^2\right\}$.

Step 3. To find continuous of the function.

Since $x^2+y$  being a polynomial function of two variable is continuous for every point on  $f^2$.

The rational function is continuous for every point on $f^2$ where $\sqrt{x^2+y}$ is defined.

Since the square of the real number can never be negative, therefore $x^2\ne -y$.

Hence the function$f(x,y)=\sqrt{x^2+y}$ is continuous on the set$\left\{\left(x,y\right)\in f^2:y>-x^2\right\}$