# Math 22 Functions of Several Variables

## Definition of a Function of Two Variables

 Let ${\displaystyle D}$ be a set of ordered pairs of real numbers.
If to each ordered pair ${\displaystyle (x,y)}$ in ${\displaystyle D}$ there corresponds a unique real number ${\displaystyle f(x,y)}$, then ${\displaystyle f}$ is a function of ${\displaystyle x}$ and ${\displaystyle y}$.
The set ${\displaystyle D}$ is the domain of ${\displaystyle f}$, and the corresponding set of values for ${\displaystyle f(x,y)}$ is the range of ${\displaystyle f}$. Functions of three, four, or more variables are defined similarly.


Exercises 1 Given ${\displaystyle f(x,y)=2x+y-3}$. Evaluate:

1) ${\displaystyle f(0,2)}$

Solution:
${\displaystyle f(x,y)=2x+y-3}$
So, ${\displaystyle f(0,2)=2(0)+2-3=-1}$

2) ${\displaystyle f(5,20)}$

Solution:
${\displaystyle f(x,y)=2x+y-3}$
So, ${\displaystyle f(5,20)=2(5)+20-3=27}$

3) ${\displaystyle f(-1,2)}$

Solution:
${\displaystyle f(x,y)=2x+y-3}$
So, ${\displaystyle f(-1,2)=2(-2)+2-3=-5}$

4) ${\displaystyle f(4,2)}$

Solution:
${\displaystyle f(x,y)=2x+y-3}$
So, ${\displaystyle f(4,2)=2(3)+2-3=5}$

## The Domain and Range of a Function of Two Variables

Example: Find the domain of ${\displaystyle f(x,y)={\sqrt {9-x^{2}-y^{2}}}}$

Notice that : The radicand should be non-negative. So, ${\displaystyle 9-x^{2}-y^{2}\geq 0}$, hence the domain is ${\displaystyle x^{2}+y^{2}\leq 9}$.

Notice: ${\displaystyle x^{2}+y^{2}=9}$ is the circle center at ${\displaystyle (0,0)}$, radius 3.

Since the point ${\displaystyle (0,0)}$ satisfies the inequality ${\displaystyle x^{2}+y^{2}\leq 9}$. Hence the range is ${\displaystyle 0\leq x\leq 3}$