Q 48. - Calculus

Question

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

$f(x,y,z)=\frac{x{y}^2}{x+y-z}$

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

Step 2: Determine the domains of the functions.

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

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

Since the rational function $f(x,y,z)=\frac{x{y}^2}{x+y-z}$is defined for all $\left(x,y,z\right)\in f^2$ such that

$$\begin{gathered} x+y-z\ne 0 \\ x+y\ne z \end{gathered}$$

The domain of the function is $Domain\left(f\right)=\left\{\left(x,y,z\right)\in f^3:x+y\ne z\right\}$

Step 3. To find continuous of the function.

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

The rational function is continuous where all those points where $f(x,y,z)=\frac{x{y}^2}{x+y-z}$ defined.

The rational function is discontinuous only at the points where $x+y-z=0$ that is $x+y=z$.

Hence, the given function$f(x,y,z)=\frac{x{y}^2}{x+y-z}$ is continuous on the set $\left\{\left(x,y,z\right)\in f^3:x+y\ne z\right\}$.