Definition of Domain, Range, and Graph

By now, we have conducted a fairly detailed study on linear, quadratic, and similar functions. We also studied curves such as lines, parabolas, ellipses, and hyperbolas, as well as operations with polynomials and algebraic functions in general. With this done, it will now be much easier to delve into some more foundational aspects regarding functions in general, which we will begin reviewing by introducing the concepts of domain, range, and graph.

Let f be a function defined between sets A and B

\begin{matrix}f & : & A & \longrightarrow & B \\ & & x & \longmapsto & y=f(x) \end{matrix}

Sets A and B are referred to as the “input” and “output” sets, respectively. From these, the following sets are defined:

Dom(f) = \{x\in A\;|\; (\exists y \in B)(y=f(x))\}

Rec(f) = \{y\in B\;|\; (\exists ! x \in Dom(f))(y=f(x))\}

Graf(f) = \{(x,y)\in A\times B\;|\; x\in Dom(f) \wedge y=f(x) \}

Example Analysis

Although everything that can be learned about the concepts of domain, range, and graph is essentially theoretical, understanding comes more from practical examples, which we will now analyze by reviewing the following three cases:

Calculate the domain, range, and graph of: f(x) = \sqrt{1-x^2}

Let’s start this analysis by writing y=f(x). Doing so gives us the equation

y = \sqrt{1-x^2}

If we square this expression, we quickly arrive at an equation leading to things we already know

\begin{array}{rl} & y^2 = 1-x^2 \\ \equiv & x^2 + y^2 = 1 \end{array}

This is the equation of the unit circle.

However, we must be careful here, because by squaring we have “added some information”. Algebraically, two values satisfy the condition of “being the square root of”, but initially, the square root is specified as a function, and functions only admit a single result. We are talking about the principal root. For this reason, the original formulation refers only to the upper part of the circle, rather than the full figure.

From this figure, it is clear that

Dom(f) = \{x\in\mathbb{R}\;|\; |x|\leq 1\} = [-1,1]

Rec(f) = \{y\in\mathbb{R}\;|\; 0\leq y\leq 1\} = [0,1]

Graf(f) = \{(x,y)\in \mathbb{R}\times \mathbb{R}\;|\; x\in [-1,1] \wedge y=\sqrt{1-x^2}\}

Although I have conducted this analysis from a graphical perspective, it is also possible to do this from a more analytical approach by reviewing the operations involved.

f(x) = \color{red}{\sqrt{{1-x^2}}}

The part 1-x^2 is well-defined for all reals.

However, the square root only admits values greater than or equal to zero.

From this, we have:

\begin{array}{rlrl} x\in Dom(f) & \leftrightarrow & 0 &\leq 1-x^2 \\ {} & \leftrightarrow & x^2 &\leq 1 \\ & \leftrightarrow & |x| &\leq 1 \\ & \leftrightarrow & -1 &\leq x \leq 1 \\ \end{array}

Therefore:\; Dom(f) = \{x\in \mathbb{R}\;|x| \leq 1\} = [-1,1]

Analytical methods for determining the range are generally much more complicated; simpler cases are solved by finding the inverse function, but before reviewing this topic in detail, it is advisable to first study function composition and other simpler cases to have a solid foundation. Meanwhile, the graphical methods we will soon review will cover much of the difficulties involved in determining the range.

Analysis for: g(x) =\displaystyle \frac{x^2 - 1}{x^2 + 1}

One way to quickly find the domain of the function is by asking for the values of x that “spoil the function.” It is clear that the function only breaks when the denominator equals zero. That is:

\begin{array}{rl} & x^2 + 1 = 0 \\ \equiv & x^2 = -1 \\ \end{array}

As no real number can satisfy this condition, it is clear that

\color{blue}{Dom(g) = \mathbb{R}}

Determining the graph is generally the quickest way to determine the range of a function; and to achieve this, polynomial division will be a good tool.

By performing polynomial division, we arrive at:

y= \displaystyle\frac{x^2-1}{x^2+1} = 1 -\displaystyle\frac{2}{x^2 + 1}

Thus, we have separated the original function into two simpler parts that we call the “whole” and “fractional” parts. Plotting each of these parts separately is much easier than plotting the original function all at once.

Analysis for: h(x) =\displaystyle \frac{x - 1}{\sqrt{x+1}}

An algebraic analysis will help to quickly determine the domain of this function. It is enough to notice that it will be well-defined as long as

\begin{array}{rrl} & 0 & \lt x + 1 \\ \equiv & -1 & \lt x \\ \end{array}

Therefore, it is clear that Dom(h)=]-1,+\infty[.

To find the range, it is useful to sketch the graph, and to do this simply, we will use a sign table. The function h(x) consists of two parts

h(x)=\displaystyle\frac{\color{green}{x-1}}{\color{red}{\sqrt{x+1}}}

The top part is zero at x=1; the bottom part, besides becoming zero at x=-1, becomes undefined if x\lt-1. With this information, we can construct the following sign table:

With the information displayed in this table, it is now very simple to graph the function.

And with this, determining the domain and range is now a trivial matter:

Dom(h)=]-1,+\infty[

Rec(h)=\mathbb{R}

Proposed Exercise

Using the tools we just reviewed, find the domain, range, and graph of the following function

F(x) = \displaystyle\frac{4x^3 + 6x^2 -2x + 1}{x^2-4}