5.1 Graphs of Functions

Up until this point in the text, we have primarily focused on studying particular families of functions. These families and their relationships to one another provide useful examples of more abstract function structures and relationships. The notions introduced in this chapter will not only provide us a more formal vocabulary with which to describe the connections between the function families we have already studied, but, more importantly, give us additional lenses through which to view new families of functions that we'll encounter.

In this section, we review of the concepts associated with the graphs of functions. We introduced the notion of the graph of a function in Section 1.1, and the vast majority of the graphs we have encountered in this text were generated from an algebraic representation of a function. In this section, we define the functions geometrically from the outset and review the important concepts associated with the graphs of functions.

Recall the domain of a function is the set of inputs to the function and the range of a function is the set of outputs from the function. When graphing a function whose domain and range are subsets of real numbers, we plot the ordered pairs $(input, output)$ on the Cartesian plane. Hence, the domain values are found on the horizontal axis while the range values are found on the vertical axis.

Recall from Definition 1.3 that the largest output from the function (if there is one) is called the maximum or, when there may be some confusion, the absolute maximum of the function. Likewise, the smallest output from the function (again, if there is one) is called the minimum or absolute minimum.

A concept related to `absolute' maximum and minimum is the concept of `local' maximum and minimum as described in Definition 2.7. Here, a point $(a, b)$ on the graph of a function $f$ is a local maximum if $b$ is the maximum function value for some open interval in the domain containing $a$ . The notion of `local' here meaning instead of surveying the entire domain, we instead restrict our attention to inputs `local' or `near' the input $a$ . The concept of local minimum is defined similarly.

Next, we review the notions of increasing, decreasing, and constant as described in Definition 1.7. Recall a function is increasing over an interval if, as the inputs increase, do the outputs. This means that, geometrically, the graph of the function rises as we move left to right. Similarly, a function is decreasing over an interval if the outputs decrease as the inputs increase. Geometrically, a decreasing function falls as we move left to right. Finally, a function is constant over an interval if the output is the same regardless of the input. If a function is constant over an interval, its graph remains `flat' - a horizontal line.

Last, and according to some¹ least, we briefly review the notion of symmetry in the graphs of functions. Recall from Definition 2.2 that a function $f$ is called even if $f (- x) = f (x)$ for all $x$ in the domain of $f$ . The graphs of even functions are symmetric about the vertical (usually $y$ -) axis. In a similar manner, Definition 2.3 tells us a function $f$ is odd if $f (- x) = - f (x)$ for all $x$ in the domain of $f$ . Geometrically, the graphs of odd functions are symmetric about the origin.

The next example reviews all of the aforementioned concepts as well as many more.

Example 1

Given the graph of $y = f (x)$ below, answer all of the following questions.

Find the domain of $f$ .
Find the range of $f$ .
Find the maximum, if it exists.
Find the minimum, if it exists.
List the $x$ -intercepts, if any exist.
List the $y$ -intercepts, if any exist.
Find the zeros of $f$ .
Solve $f ⁢ open parenthesis x close parenthesis less than 0$ .
Determine $f ⁢ open parenthesis 2 close parenthesis$ .
Solve $f (x) = - 3$ .
Find the number of solutions to $f ⁢ open parenthesis x close parenthesis equals 1$ .
Does $f$ appear to be even, odd, or neither?
List the local maximums, if any exist.
List the local minimums, if any exist.
List the intervals on which $f$ is increasing.
List the intervals on which $f$ is decreasing.

Figure 5.1

To find the domain of $f$ , we proceed as in Section 1.1. By projecting the graph to the $x$ -axis, we see that the portion of the $x$ -axis which corresponds to a point on the graph is everything from $minus 4$ to $4$ , inclusive. Hence, the domain is $open bracket minus 4, 4 close bracket$ .
To find the range, we project the graph to the $y$ -axis. We see that the $y$ values from $minus 3$ to $3$ , inclusive, constitute the range of $f$ . Hence, our answer is $open bracket minus 3, 3 close bracket$ .
The maximum value of $f$ is the largest $y$ -coordinate which is $3$ .
The minimum value of $f$ is the smallest $y$ -coordinate which is $minus 3$ .
The $x$ -intercepts are the points on the graph with $y$ -coordinate $0$ , namely $open parenthesis minus 2, 0 close parenthesis$ and $open parenthesis 2, 0 close parenthesis$ .
The $y$ -intercept is the point on the graph with $x$ -coordinate $0$ , namely $open parenthesis 0, 3 close parenthesis$ .
The zeros of $f$ are the $x$ -coordinates of the $x$ -intercepts of the graph of $y = f (x)$ which are $x = - 2, 2$ .
To solve $f ⁢ open parenthesis x close parenthesis less than 0$ , we look for the $x$ values of the points on the graph where the $y = f (x)$ is negative. Graphically, we are looking for where the graph is below the $x$ -axis. This happens for the $x$ values from $minus 4$ to $minus 2$ and again from $2$ to $4$ . So our answer is $[- 4, - 2) \cup (2, 4]$ .
Since the graph of $f$ is the graph of the equation $y = f (x)$ , $f ⁢ open parenthesis 2 close parenthesis$ is the $y$ -coordinate of the point which corresponds to $x equals 2$ . Since the point $open parenthesis 2, 0 close parenthesis$ is on the graph, we have $f ⁢ open parenthesis 2 close parenthesis equals 0$ .
To solve $f (x) = - 3$ , we look where $y = f (x) = - 3$ . We find two points with a $y$ -coordinate of $minus 3$ , namely $(- 4, - 3)$ and $open parenthesis 4, minus 3 close parenthesis$ . Hence, the solutions to $f (x) = - 3$ are $x = \pm 4$ .
As in the previous problem, to solve $f ⁢ open parenthesis x close parenthesis equals 1$ , we look for points on the graph where the $y$ -coordinate is $1$ . If we imagine the horizontal like $y equals 1$ superimposed over the graph of $f$ as sketched below, we get two intersections. Hence, even though these points aren't specified, we know there are two points on the graph of $f$ whose $y$ -coordinate is $1$ . Hence, there are two solutions to $f ⁢ open parenthesis x close parenthesis equals 1$ .
Figure 5.2
The graph appears to be symmetric about the $y$ -axis. This suggests² that $f$ is even.
The function has its only local maximum at $open parenthesis 0, 3 close parenthesis$ .
There are no local minimums. Why don't $(- 4, - 3)$ and $open parenthesis 4, minus 3 close parenthesis$ count? Let's consider the point $(- 4, - 3)$ for a moment. Recall that, in the definition of local minimum, there needs to be an open interval containing $x = - 4$ which is in the domain of $f$ . In this case, there is no open interval containing $x = - 4$ which lies entirely in the domain of $f$ , $open bracket minus 4, 4 close bracket$ . Because we are unable to fulfill the requirements of the definition for a local minimum, we cannot claim that $f$ has one at $(- 4, - 3)$ . The point $open parenthesis 4, minus 3 close parenthesis$ fails for the same reason no open interval around $x equals 4$ stays within the domain of $f$ .
As we move from left to right, the graph rises from $(- 4, - 3)$ to $open parenthesis 0, 3 close parenthesis$ . This means $f$ is increasing on the interval $open bracket minus 4, 0 close bracket$ . (Remember, the answer here is an interval on the $x$ -axis.)
As we move from left to right, the graph falls from $open parenthesis 0, 3 close parenthesis$ to $open parenthesis 4, minus 3 close parenthesis$ . This means $f$ is decreasing on the interval $open bracket 0, 4 close bracket$ . (Again, the answer here is an interval on the $x$ -axis.)

Our next example involves a more complicated function and asks more complicated questions.

Example 2

Consider the graph of the function $g$ below.

Figure 5.3 The graph of

y = g (t)

Find the domain of $g$ .
Find the range of $g$ .
Find the maximum, if it exists.
Find the minimum, if it exists.
List the local maximums, if any exist.
List the local minimums, if any exist.
Solve $(t^{2} - 25) g (t) = 0$ .
Solve $\frac{g (t)}{t^{2} + t - 30} \geq 0$ .

Projecting the graph of $g$ to the $t$ -axis, we see the domain contains values of $t$ from $minus 4$ up to, but not including $t equals 0$ and values greater than $t equals 0$ up to, but not including $t equals 7$ . Using interval notation, we write the domain as $open bracket minus 4, 0 close parenthesis union open parenthesis 0, 7 close parenthesis$ .
Projecting the graph of $g$ to the $y$ -axis, we see the range of $g$ contains all real numbers from $y = - 8$ up to, but not including, $y equals 5.5$ . Note that even though there is a hole in the graph at $open parenthesis 0, 0 close parenthesis$ , the points $open parenthesis minus 3, 0 close parenthesis$ and $open parenthesis 6, 0 close parenthesis$ put $y equals 0$ in the range of $g$ . Hence, the range of $g$ is $open bracket minus 8, 5.5 close parenthesis$ .
Owing to the hole in the graph at $open parenthesis 7, 5.5 close parenthesis$ , $g$ has no maximum.³
The minimum of $g$ is $minus 8$ which occurs at the point $open parenthesis 3, minus 8 close parenthesis$ .
The point $open parenthesis minus 2, 4.5 close parenthesis$ is clearly a local maximum, but there are actually infinitely many more. Per Definition 2.7, all points of the form $open parenthesis t, minus 6 close parenthesis$ for $4 less than or equal to t less than 5$ are also local maximums. For each of these points, we can find an open interval on the $t$ axis within which we produce no points on the graph higher than $open parenthesis t, minus 6 close parenthesis$ . (You may think about `zooming in' on the point $open parenthesis 4.5, minus 6 close parenthesis$ to see how this works.)
The local minimums of the graph are $open parenthesis 3, minus 8 close parenthesis$ along with points of the form $open parenthesis t, minus 6 close parenthesis$ for $4 less than t less than or equal to 5$ . Note the point $(- 4, - 3)$ is not a local minimum since there is no open interval containing $t = - 4$ which lies entirely within the domain of $g$ .
To solve $(t^{2} - 25) g (t) = 0$ , we use the zero product property of real numbers⁴ to conclude either $t^{2} - 25 = 0$ or $g ⁢ open parenthesis t close parenthesis equals 0$ .
From $t^{2} - 25 = 0$ , we get $t = \pm 5$ . However, since $t = - 5$ isn't in the domain of $g$ , it cannot be regarded as a solution to the equation $(t^{2} - 25) g (t) = 0$ . (If we substitute $t = - 5$ into the equation, we'd get $({(- 5)}^{2} - 25) g (- 5) = 0 \cdot g (- 5)$ . Since $g ⁢ open parenthesis minus 5 close parenthesis$ is undefined, so is $0 \cdot g (- 5)$ .)
To solve $g ⁢ open parenthesis t close parenthesis equals 0$ , we look for the zeros of $g$ which are $t = - 3$ and $t equals 6$ . (Again, there is a hole at $open parenthesis 0, 0 close parenthesis$ , so $t equals 0$ doesn't count as a zero.) Our final answer to $(t^{2} - 25) g (t) = 0$ is $t = - 3$ , $5$ , or $6$ .
To solve $\frac{g (t)}{t^{2} + t - 30} \geq 0$ , we employ a sign diagram as we (most recently) have done in Section 4.3.⁵ To that end, we define $F (t) = \frac{g (t)}{t^{2} + t - 30}$ and we set about finding the domain of $f$ .
First, we note that since $F$ is defined in terms of $g$ , the domain of $F$ is restricted to some subset of the domain of $g$ , namely $open bracket minus 4, 0 close parenthesis union open parenthesis 0, 7 close parenthesis$ . Since $t^{2} + t - 30$ is in the denominator of $F (t)$ , we must also exclude the values where $t^{2} + t - 30 = (t + 6) (t - 5) = 0$ . Hence, we must exclude $t = - 6$ (which isn't in the domain of $g$ in the first place) along with $t equals 5$ . Hence, the domain of $F$ is $open bracket minus 4, 0 close parenthesis union open parenthesis 0, 5 close parenthesis union open parenthesis 5, 7 close parenthesis$ .
Next, we find the zeros of $F$ . Setting $F (t) = \frac{g (t)}{t^{2} + t - 30} = 0$ amounts to solving $g ⁢ open parenthesis t close parenthesis equals 0$ . Graphically, we see this occurs when $t = - 3$ and $t equals 6$ . Hence, we need to select test values in each of the following intervals: $[- 4, - 3)$ , $open parenthesis minus 3, 0 close parenthesis$ , $open parenthesis 0, 5 close parenthesis$ , $open parenthesis 5, 6 close parenthesis$ and $open parenthesis 6, 7 close parenthesis$ .
For the interval $[- 4, - 3)$ , we may choose $t = - 4$ . $F (- 4) = \frac{g (- 4)}{{(- 4)}^{2} + (- 4) - 30} = \frac{- 3}{- 18} > 0$ so is $(+)$ . For the interval $open parenthesis minus 3, 0 close parenthesis$ we choose $t = - 2$ and get $F (- 2) = \frac{g (- 2)}{{(- 2)}^{2} + (- 2) - 30} = \frac{4.5}{- 28} < 0$ so is $(-)$ . For the interval $open parenthesis 0, 5 close parenthesis$ , we choose $t equals 3$ and find $F (3) = \frac{g (3)}{{(3)}^{2} + (3) - 30} = \frac{- 8}{- 18} > 0$ which is $(+)$ again.
For the last two intervals, $open parenthesis 5, 6 close parenthesis$ and $open parenthesis 6, 7 close parenthesis$ , we do not have specific function values for $g$ . However, all we are interested in is the sign of the function over these intervals, and we can get that information about $g$ graphically.
For the interval $open parenthesis 5, 6 close parenthesis$ , we choose $t equals 5.5$ as our test value. Since the graph of $y = g (t)$ is below the $t$ -axis when $t equals 5.5$ . we know $g ⁢ open parenthesis 5.5 close parenthesis$ is $(-)$ . Hence, $F (5.5) = \frac{g (5.5)}{{(5.5)}^{2} + (5.5) - 30} = \frac{(-)}{5.75} < 0$ so is $(-)$ . Similarly, when $t equals 6.5$ , the graph of $y = g (t)$ is above the $t$ -axis so $F (6.5) = \frac{g (6.5)}{{(6.5)}^{2} + (6.5) - 30} = \frac{(+)}{18.75} > 0$ so is $(+)$ . Putting all of this together, we get the sign diagram for $F (t) = \frac{g (t)}{t^{2} + t - 30}$ below:
Figure 5.4
Hence, $F ⁢ open parenthesis t close parenthesis greater than or equal to 0$ on $[- 4, - 3] \cup (0, 5) \cup [6, 7)$ .

Our last example focuses on symmetry. The reader is encouraged to review the notes about symmetry as summarized on page in Section A.3.

Example 3

Below are the partial graphs of functions $f$ and $g$ .

If possible, complete the graphs of $f$ and $g$ assuming both functions are even.
If possible, complete the graphs of $f$ and $g$ assuming both functions are odd.

Figure 5.5 Partial graph of

y = f (x)

Figure 5.6 Partial graph of

y = g (x)

If $f$ and $g$ are even then their graphs are symmetric about the $y$ -axis. Hence, to complete each graph, we reflect each point on the graphs of $f$ and $g$ about the $y$ -axis.
Figure 5.7 The graph of $f$ assuming $f$ is even.
Figure 5.8 The graph of $g$ assuming $g$ is even.
If $f$ and $g$ are odd then their graphs are symmetric about the origin. Hence, to complete each graph, we imagine reflecting each of the points on their graphs through the origin. We complete the process on the graph of $f$ with no issues.
However, when attempting to do the same with the graph of the function $g$ , we find the point $open parenthesis 0, minus 5 close parenthesis$ is reflected to the point $open parenthesis 0, 5 close parenthesis$ . Hence, this new graph doesn't pass the vertical line test and hence is not a function. Therefore, $g$ cannot be odd.⁶
Figure 5.9 The graph of $f$ assuming $f$ is odd.
Figure 5.10 This graph fails the vertical line test.

Exercises

In Exercises -, use the graph of $y = f (x)$ given below to answer the question.

Coordinate-plane figure. — **Figure 5.11** $y = f (x)$

Find the domain of $f$ .
Find the range of $f$ .
Find the maximum, if it exists.
Find the minimum, if it exists.
List the local maximums, if any exist.
List the local minimums, if any exist.
List the intervals where $f$ is increasing.
List the intervals where $f$ is decreasing.
Determine $f ⁢ open parenthesis minus 2 close parenthesis$ .
Solve $f ⁢ open parenthesis x close parenthesis equals 4$ .
List the $x$ -intercepts, if any exist.
List the $y$ -intercepts, if any exist.
Find the zeros of $f$ .
Solve $f ⁢ open parenthesis x close parenthesis greater than or equal to 0$ .
Find the number of solutions to $f ⁢ open parenthesis x close parenthesis equals 1$ .
Find the number of solutions to $vertical bar f ⁢ open parenthesis x close parenthesis vertical bar equals 1$ .
Solve $(x^{2} - x - 2) f (x) = 0$
Solve $(x^{2} - x - 2) f (x) > 0$

With help from your classmates:

Find the domain of $R (x) = \frac{1}{f (x)}$
Find the range of $R (x) = \frac{1}{f (x)}$

In Exercises -, use the graph of $y = g (t)$ given below to answer the question.

Find the domain of $g$ .
Find the range of $g$ .
Find the maximum, if it exists.
Find the minimum, if it exists.
List the local maximums, if any exist.
List the local minimums, if any exist.
List the intervals where $g$ is increasing.
List the intervals where $g$ is decreasing.
Determine $g ⁢ open parenthesis 2 close parenthesis$ .
Solve $g (t) = - 5$ .
List the $t$ -intercepts, if any exist.
List the $y$ -intercepts, if any exist.
Find the zeros of $g$ .
Solve $g ⁢ open parenthesis t close parenthesis less than or equal to 0$ .
Find the domain of $G (t) = \frac{g (t)}{t + 2}$ .
Solve $\frac{g (t)}{t + 2} \leq 0$ .
How many solutions are there to $open bracket g ⁢ open parenthesis t close parenthesis close bracket squared equals 9$ ?
Does $g$ appear to be even, odd, or neither?
Prove that if $f$ is an odd function and $0$ is in the domain of $f$ , then $f ⁢ open parenthesis 0 close parenthesis equals 0$ .
Let $R (x)$ be the function defined as: $R ⁢ open parenthesis x close parenthesis equals 1$ if $x$ is a rational number, $R ⁢ open parenthesis x close parenthesis equals 0$ if $x$ is an irrational number. With help from your classmates, try to graph $R$ . What difficulties do you encounter?
NOTE: Between every pair of real numbers, there is both a rational and an irrational number …
Consider the graph of the function $f$ given below.
Figure 5.13
1. Explain why $f$ has a local maximum but not a local minimum at the point $open parenthesis minus 1, 1 close parenthesis$ .
2. Explain why $f$ has a local minimum but not a local maximum at the point $open parenthesis 1, 1 close parenthesis$ .
3. Explain why $f$ has a local maximum AND a local minimum at the point $open parenthesis 0, 1 close parenthesis$ .
4. Explain why $f$ is constant on the interval $open bracket minus 1, 1 close bracket$ and thus has both a local maximum AND a local minimum at every point $(x, f (x))$ where $minus 1 less than x less than 1$ .
Explain why the function $g$ whose graph is given below does not have a local maximum at $open parenthesis minus 3, 5 close parenthesis$ nor does it have a local minimum at $open parenthesis 3, minus 3 close parenthesis$ . Find its extrema, both local and absolute and find the intervals on which $g$ is increasing and those on which $g$ is decreasing.
Figure 5.14
For each function below, find the local maximum or local minimum and list the interval over which the function is increasing and the interval over which the function is decreasing.
1. Figure 5.15 Function I
2. Figure 5.16 Function II
1. Figure 5.17 Function III
2. Figure 5.18 Function IV

Answers

$open bracket minus 5, 3 close bracket$
$open bracket minus 5, 4 close bracket$
$f (- 3) = 4$
$f (- 5) = - 5$
$open parenthesis minus 3, 4 close parenthesis$ , $open parenthesis 2, 3 close parenthesis$
$open parenthesis 0, minus 1 close parenthesis$
$[- 5, - 3]$ , $open bracket 0, 2 close bracket$
$open bracket minus 3, 0 close bracket$ , $open bracket 2, 3 close bracket$
$f (- 2) = 2$
$x = - 3$
$open parenthesis minus 4, 0 close parenthesis$ , $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$
$open parenthesis 0, minus 1 close parenthesis$
$minus 4$ , $minus 1$ , $1$
$[- 4, - 1]$ , $open bracket 1, 3 close bracket$
$4$
$6$
$x = - 4, - 1, 1, 2$
$(- 4, - 1) \cup (- 1, 1) \cup (2, 3)$
To find the domain of $R (x) = \frac{1}{f (x)}$ , we start with the domain of $f$ and exclude values where $f ⁢ open parenthesis x close parenthesis equals 0$ . Hence, the domain of $R$ is $[- 5, - 4) \cup (- 4, - 1) \cup (- 1, 1) \cup (1, 3]$ .
To find the range of $R (x) = \frac{1}{f (x)}$ , we start with the range of $f$ (excluding $0$ ) and take reciprocals. If $minus 5 less than or equal to y less than 0$ , then $\frac{1}{y} \leq - \frac{1}{5}$ . If $0 less than y less than or equal to 4$ , then $\frac{1}{y} \geq \frac{1}{4}$ . Hence the range of $R$ is $(- \infty, - \frac{1}{5}] \cup [\frac{1}{4}, \infty)$ .
$open bracket minus 4, 4 close bracket$
$open bracket minus 5, 5 close parenthesis$
none
$g (- 2) = - 5$
none
$(- 2, - 5)$ , $open parenthesis 2, 3 close parenthesis$
$open bracket minus 2, 2 close parenthesis$
$[- 4, - 2]$ , $open parenthesis 2, 4 close bracket$
$g ⁢ open parenthesis 2 close parenthesis equals 3$
$t = - 2$
$open parenthesis minus 4, 0 close parenthesis$ , $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 4, 0 close parenthesis$
$open parenthesis 0, 0 close parenthesis$
$minus 4$ , $0$ , $4$
$open bracket minus 4, 0 close bracket union open brace 4 close brace$
$[- 4, - 2) \cup (- 2.4]$
${- 4} \cup (- 2, 0] \cup {4}$
$5$
Neither.
1. Local maximum: $open parenthesis 0, 1 close parenthesis$ , no local minimum. Increasing: $open parenthesis 0, 2 close bracket$ , decreasing: $open bracket minus 2, 0 close parenthesis$ .
2. No local maximum, local minimum: $open parenthesis 0, 1 close parenthesis$ . Increasing: $open bracket minus 2, 0 close parenthesis$ , decreasing: $open parenthesis 0, 2 close bracket$ .
3. No local maximum, local minimum: $open parenthesis 0, minus 1 close parenthesis$ . Increasing: $open bracket 0, 2 close bracket$ , decreasing: $open bracket minus 2, 0 close bracket$ .
4. Local maximum: $open parenthesis 0, 5 close parenthesis$ , no local minimum. Increasing: $open bracket minus 2, 0 close bracket$ , decreasing: $open bracket 0, 2 close bracket$ .

Jeff ↩
but does not prove ↩
There is no real number `right before' $5.5$ … ↩
see Section A.2, the related section ↩
Note that $g$ is continuous on its domain, and hence, it follows that $\frac{g (t)}{t^{2} + t - 30}$ is, too. (Thank Calculus!) This means the Intermediate Value Theorem applies so a Sign Diagram approach is valid. ↩
We leave it as an exercise to show that if a function $f$ is odd and $0$ is in the domain of $f$ , then, necessarily, $f ⁢ open parenthesis 0 close parenthesis equals 0$ . ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.