Domain and Range of Algebraic Functions

Recall the definition of domain and range for the one variable function y = f(x):

Domain (of a one variable function): The set of all values of the independent variable for which the function is defined.

Range(of a one variable function): The set of all possible values which the dependent variable can take, based on the domain.

Similar definitions hold for functions of many variables. However, instead of the domain consisting of a portion of the real number line, we must now consider what portion of n-dimensional space the domain of a function of n variables inhabits. Points in n-dimensional space are called ordered n-tuples because they must be written down in a specific order. You are already familiar with this concept. To graph a point in three-space, you need to specify its coordinates as an ordered triple-(x,y,z). Thus, the following definitions hold:

Domain: The domain of a function of n variables is the set of all n-tuples in n-dimensional space (also written as ${\mathbb R}^n$) for which the function is defined.

Range: The range of a function of n variables is the set of all possible values of the dependent variable.

For example, the function z = f(x,y) = x² + y² has a domain consisting of all points in two dimensional space. In other words, any point in ${\mathbb R}^2$, any ordered pair (x,y) is an element of the domain. The range is a different matter, though. Regardless of the choice of x and y, the quantity $x^2 + y^2 \ge 0$. Thus, the range of this function is $z \ge 0$.

As another example, consider the function $z = \sqrt{x^2 - y^2}$. This function is only defined if the quantity $x^2 - y^2 \ge 0$. Thus, either $x \ge y$ or $x \le -y$. The domain is then all (x,y) such that either $x \ge y$ or $x \le -y$. Since the square root is positive, the range is again $z \ge 0$.

What about the function z = e^{-(x2 + y2)}? Clearly the domain is all ordered pairs in ${\mathbb R}^2$ but what about the range? To determine this, let's rewrite the function as z = 1/e^{x2 + y2}. Since $x^2 + y^2 \ge 0$ and e raised to any power is positive, we see that z > 0. Further, z = 0 only in the limit as x and y go to infinity. Thus, z > 0 is part of the range. Is there an upper bound to z? The function obviously decreases in value as x and y increase, so the maximum value must occur at x = y = 0. Here, z = 1. The range is then $0 < z \le 1$.