1.3: Introduction to Functions

Math 370 Learning Objectives

This entire section is provided for you to review on your own, if needed.

Functions

One of the core concepts in Precalculus is the function. There are many ways to describe a function and we begin by defining a function as a special kind of relation.

Definition: Function

A relation in which each \(x\)-coordinate is matched with only one \(y\)-coordinate is said to describe \(y\) as a function of \(x\).

Example \( \PageIndex{1} \)

Which of the following relations describe \(y\) as a function of \(x\)?

1. \(R_{1} = \{ (-2,1), (1,3), (1,4), (3,-1) \}\)
2. \(R_{2} = \{ (-2,1), (1,3), (2,3), (3,-1) \}\)

Solution

1. A quick scan of the points in \(R_{1}\) reveals that the \(x\)-coordinate \(1\) is matched with two different \(y\)-coordinates: namely \(3\) and \(4\). Hence in \(R_{1}\), \(y\) is not a function of \(x\).
2. On the other hand, every \(x\)-coordinate in \(R_{2}\) occurs only once which means each \(x\)-coordinate has only one corresponding \(y\)-coordinate. So, \(R_{2}\) does represent \(y\) as a function of \(x\).

Note that in the previous example, the relation \(R_{2}\) contained two different points with the same \(y\)-coordinates, namely \((1,3)\) and \((2,3)\). Remember, in order to say \(y\) is a function of \(x\), we just need to ensure the same \(x\)-coordinate isn’t used in more than one point.¹

Subsection Footnotes

¹ We will have occasion later in the text to concern ourselves with the concept of \(\ x\) being a function of \(\ y\). In this case, \(\ R_{1}\) represents \(x\) as a function of \(\ y\); \(\ R_{2}\) does not.

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The Vertical Line Test

To see what the function concept means geometrically, we graph \(R_{1}\) and \(R_{2}\) in the plane.

The fact that the \(x\)-coordinate, \(1\), is matched with two different \(y\)-coordinates in \(R_{1}\) presents itself graphically as the points \((1,3)\) and \((1,4)\) lying on the same vertical line, \(x=1\). If we turn our attention to the graph of \(R_{2}\), we see that no two points of the relation lie on the same vertical line. We can generalize this idea as follows:

Theorem \(\PageIndex{1}\): Vertical Line Test

A set of points in the plane represents \(y\) as a function of \(x\) if and only if no two points lie on the same vertical line.

It is worth taking some time to meditate on the Vertical Line Test; it will check to see how well you understand the concept of "function" as well as the concept of "graph."

Example \( \PageIndex{2} \)

Use the Vertical Line Test to determine which of the following relations describes \(y\) as a function of \(x\).

Solution

Looking at the graph of \(R\), we can easily imagine a vertical line crossing the graph more than once. Hence, \(R\) does not represent \(y\) as a function of \(x\). However, in the graph of \(S\), every vertical line crosses the graph at most once, so \(S\) does represent \(y\) as a function of \(x\).

In the previous test, we say that the graph of the relation \(R\) fails the Vertical Line Test, whereas the graph of \(S\) passes the Vertical Line Test. Note that in the graph of \(R\) there are infinitely many vertical lines which cross the graph more than once. However, to fail the Vertical Line Test, all you need is one vertical line that fits the bill, as the next example illustrates.

Example \( \PageIndex{3} \)

Use the Vertical Line Test to determine which of the following relations describes \(y\) as a function of \(x\).

Solution

Both \(S_{1}\) and \(S_{2}\) are slight modifications to the relation \(S\) in the previous example whose graph we determined passed the Vertical Line Test. In both \(S_{1}\) and \(S_{2}\), it is the addition of the point \((1,2)\) which threatens to cause trouble.

In \(S_{1}\), there is a point on the curve with \(x\)-coordinate \(1\) just below \((1,2)\), which means that both \((1,2)\) and this point on the curve lie on the vertical line \(x=1\) (see Figure \( \PageIndex{1}A \)). Hence, the graph of \(S_{1}\) fails the Vertical Line Test, so \(y\) is not a function of \(x\) here. However, in \(S_{2}\) notice that the point with \(x\)-coordinate \(1\) on the curve has been omitted, leaving an open circle there. Hence, the vertical line \(x=1\) crosses the graph of \(S_{2}\) only at the point \((1,2)\). Indeed, any vertical line will cross the graph at most once, so we have that the graph of \(S_{2}\) passes the Vertical Line Test. Thus it describes \(y\) as a function of \(x\).

Figures \( \PageIndex{1}A \) (left) and \( \PageIndex{1}B \) (right)

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The Domain and Range of a Function

Suppose a relation \(F\) describes \(y\) as a function of \(x\). The sets of \(x\)- and \(y\)-coordinates are given special names which we define below.

Definition: Domain and Range

Suppose \(F\) is a relation which describes \(y\) as a function of \(x\).

• The set of the \(x\)-coordinates of the points in \(F\) is called the domain of \(F\).
• The set of the \(y\)-coordinates of the points in \(F\) is called the range of \(F\).

We demonstrate finding the domain and range of functions given to us either graphically or via the roster method in the following example.

Example \( \PageIndex{4} \)

Find the domain and range of the function \(F = \{ (-3, 2), (0, 1), (4, 2), (5, 2) \}\) and of the function \(G\) whose graph is Figure \( \PageIndex{1}B \).

Solution

The domain of \(F\) is the set of the \(x\)-coordinates of the points in \(F\), namely \(\{ -3, 0, 4, 5 \}\) and the range of \(F\) is the set of the \(y\)-coordinates, namely \(\{ 1,2 \}.\)

To determine the domain and range of \(G\), we need to determine which \(x\) and \(y\) values occur as coordinates of points on the given graph. To find the domain, it may be helpful to imagine collapsing the curve to the \(x\)-axis and determining the portion of the \(x\)-axis that gets covered. This is called projecting the curve to the \(x\)-axis. Before we start projecting, we need to pay attention to two subtle notations on the graph: the arrowhead on the lower left corner of the graph indicates that the graph continues to curve downwards to the left forever more; and the open circle at \((1,3)\) indicates that the point \((1,3)\) isn’t on the graph, but all points on the curve leading up to that point are.

Figure \( \PageIndex{2} \)

We see from Figure \( \PageIndex{2} \) that if we project the graph of \(G\) to the \(x\)-axis, we get all real numbers less than \(1\). Using interval notation, we write the domain of \(G\) as \((-\infty, 1)\). To determine the range of \(G\), we project the curve to the \(y\)-axis as follows:

Figure \( \PageIndex{3} \)

Note that even though there is an open circle at \((1,3)\), we still include the \(y\) value of \(3\) in our range, since the point \((-1,3)\) is on the graph of \(G\). We see that the range of \(G\) is all real numbers less than or equal to \(4\), or, in interval notation, \((-\infty, 4]\).

All functions are relations, but, as we have seen, not all relations are functions. Thus the equations which described the relations in Section 1.2 may or may not describe \(y\) as a function of \(x\). The algebraic representation of functions is possibly the most important way to view them so we need a process for determining whether or not an equation of a relation represents a function. We delay the discussion of finding the domain of a function given algebraically until Section 1.4.

Example \( \PageIndex{5} \)

Determine which equations represent \(y\) as a function of \(x\).

1. \(x^3 + y^2 = 1\)
2. \(x^2 + y^3 = 1\)
3. \(x^2y = 1 - 3y\)

Solution

For each of these equations, we solve for \(y\) and determine whether each choice of \(x\) will determine only one corresponding value of \(y\).

1.  

\[ \begin{array}{crcl}
& x^3 + y^2 & = & 1 \\
\implies & y^2 & = & 1 - x^3 \\
\implies & \sqrt{y^2} & = & \sqrt{1 - x^3} \\
\implies & y & = & \pm \sqrt{1 - x^3} \\
\end{array} \nonumber \]

If we allow \( x = 0 \), we see this relation has two values for \( y \): \( +1 \) and \( -1 \). Thus, \( y \) is not a function of \( x \).

2.  

\[ \begin{array}{crcl}
& x^2 + y^3 & = & 1 \\
\implies & y^3 & = & 1 - x^2 \\
\implies & \sqrt[3]{y^3} & = & \sqrt[3]{1 - x^3} \\
\implies & y & = & \sqrt[3]{1 - x^2} \\
\end{array} \nonumber \]

The defining difference between this example and the previous one is that the previous example resulted in \( y = \pm \ldots \). This can be read as \( y \) can be either positive \( \ldots \) or its opposite. That is the primary reason the previous example was not a function of \( x \); however, in this case, we have no such issue. No matter what choice is made for \( x \), there is only one possible value for \( y \). Thus, \( y \) is a function of \( x \).

3.  

\[\begin{array}{crclr}
& x^2y & = & 1 - 3y & \\
\implies & x^2y + 3y & = & 1 & \\
\implies & y \left(x^2 + 3\right) & = & 1 & \\
\implies & y & = & \dfrac{1}{x^2 + 3} & \left( \text{Dividing both sides by the same non-zero value} \right) \\
\end{array} \nonumber \]

For each choice of \(x\), there is only one value for \(y\), so this equation describes \(y\) as a function of \(x\).

In part 3 of Example \( \PageIndex{5} \), we ended up dividing both sides of the equation by \( x^2 + 3 \). It is incredibly important to note that you should try to avoid doing this in most cases. Why? Suppose you were given \( y \left( x + 3 \right) = 1 \). Dividing both sides by the variable expression \( \left( x + 3 \right) \) leads to \( y = \frac{1}{x + 3} \); however, if \( x \) is ever \( -3 \), we have a major issue. In that case, we would have been dividing by zero! The only reason we got away with the division by a variable expression in Example \( \PageIndex{5} \) is that we divided by \( \left( x^2 + 3 \right) \). Since \( x^2 \ge 0 \), we can add \( 3 \) to both sides to find that \( x^3 + 3 \ge 3 \gt 0 \). Hence, there is no possibility for the variable expression \( \left( x^2 + 3 \right) \) to ever be zero.

If you think that last paragraph was too much detail for you, I have some bad news - the rest of this course, Calculus, and all of your Science classes will require that level of thought-process and meticulousness.