Chapter 14, Problem 1RE

What is the domain of $z=\frac{3x+2\sqrt{y}}{x^2+y^2}$?

To determine

To calculate: The domain of the function $z=\frac{3}{2x-y}$.

Answer to Problem 1RE

Solution:

The domain of the function $z=\frac{3}{2x-y}$ is.

Explanation of Solution

Given Information:

The provided function is $z=\frac{3}{2x-y}$.

Formula used:

The domain of a function of two variables $f\left(x,y\right)$ is the set of all ordered pairs $\left(x,y\right)$ for which the function is well-defined.

If the function maps each element of the domain to an element in the codomain, then it is said to be well defined.

To find the domain of a rational function, set the denominator of the function equals to zero. Then the domain of the function is the set of all ordered pairs except the values for which the denominator is zero.

Calculation:

Consider the provided function, $z=\frac{3}{2x-y}$.

Rewrite $z$ as $f\left(x,y\right)$ in $z=\frac{3}{2x-y}$.

Thus,

$f\left(x,y\right)=\frac{3}{2x-y}$

The function $f\left(x,y\right)$ is a rational function.

Recall that, to find the domain of a rational function, set the denominator of the function equals to zero. Then the domain of the function is the set of all ordered pairs except the values for which the denominator is zero.

The denominator of the function $f\left(x,y\right)=\frac{3}{2x-y}$ is $2x-y$.

Thus, the denominator $2x-y=0$ and solve.

$\begin{array}{c}2x-y=0\\ y=2x\end{array}$

Therefore, the domain of the function $f\left(x,y\right)=\frac{3}{2x-y}$ is the set of all ordered pairs $\left(x,y\right)$,

where x and y are real and $y\ne 2x$.

Hence, the domain of the function $z=\frac{3}{2x-y}$ is.