Symmetry of Functions

Learning Outcomes

Describe the symmetry properties of a function

The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph. For example, consider the function $f(x)=x^4-2x^2-3$ shown in Figure 13(a). If we take the part of the curve that lies to the right of the $y$ -axis and flip it over the $y$ -axis, it lays exactly on top of the curve to the left of the $y$ -axis. In this case, we say the function has symmetry about the $y$ -axis. On the other hand, consider the function $f(x)=x^3-4x$ shown in Figure 13(b). If we take the graph and rotate it 180° about the origin, the new graph will look exactly the same. In this case, we say the function has symmetry about the origin.

An image of two graphs. The first graph is labeled “(a), symmetry about the y-axis” and is of the curved function “f(x) = (x to the 4th) - 2(x squared) - 3”. The x axis runs from -3 to 4 and the y axis runs from -4 to 5. This function decreases until it hits the point (-1, -4), which is minimum of the function. Then the graph increases to the point (0,3), which is a local maximum. Then the the graph decreases until it hits the point (1, -4), before it increases again. The second graph is labeled “(b), symmetry about the origin” and is of the curved function “f(x) = x cubed - 4x”. The x axis runs from -3 to 4 and the y axis runs from -4 to 5. The graph of the function starts at the x intercept at (-2, 0) and increases until the approximate point of (-1.2, 3.1). The function then decreases, passing through the origin, until it hits the approximate point of (1.2, -3.1). The function then begins to increase again and has another x intercept at (2, 0).

Figure 13. (a) A graph that is symmetric about the $y$ -axis. (b) A graph that is symmetric about the origin.

If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without a graph, how can we determine algebraically whether a function $f$ has symmetry? Looking at Figure 13(a) again, we see that since $f$ is symmetric about the $y$ -axis, if the point $(x,y)$ is on the graph, the point $(−x,y)$ is on the graph. In other words, $f(−x)=f(x)$ . If a function $f$ has this property, we say $f$ is an even function, which has symmetry about the $y$ -axis. For example, $f(x)=x^2$ is even because

f (- x) = (- x)^{2} = x^{2} = f (x)

$f(−x)=(−x)^2=x^2=f(x)$ .

In contrast, looking at Figure 13(b) again, if a function $f$ is symmetric about the origin, then whenever the point $(x,y)$ is on the graph, the point $(−x,−y)$ is also on the graph. In other words, $f(−x)=−f(x)$ . If $f$ has this property, we say $f$ is an odd function, which has symmetry about the origin. For example, $f(x)=x^3$ is odd because

f (- x) = (- x)^{3} = - x^{3} = - f (x)

$f(−x)=(−x)^3=−x^3=−f(x)$ .

Definition

If $f(-x)=f(x)$ for all $x$ in the domain of $f$ , then $f$ is an even function. An even function is symmetric about the $y$ -axis.

If $f(−x)=−f(x)$ for all $x$ in the domain of $f$ , then $f$ is an odd function. An odd function is symmetric about the origin.

Example: Even and Odd Functions

Determine whether each of the following functions is even, odd, or neither.

$f(x)=-5x^4+7x^2-2$
$f(x)=2x^5-4x+5$
$f(x)=\dfrac{3x}{x^2+1}$

Show Solution

To determine whether a function is even or odd, we evaluate $f(−x)$ and compare it to $f(x)$ and $−f(x)$ .

$f(−x)=-5(−x)^4+7(−x)^2-2=-5x^4+7x^2-2=f(x)$ . Therefore, $f$ is even.
$f(−x)=2(−x)^5-4(−x)+5=-2x^5+4x+5$ . Now, $f(−x)\ne f(x)$ . Furthermore, noting that $−f(x)=-2x^5+4x-5$ , we see that $f(−x)\ne −f(x)$ . Therefore, $f$ is neither even nor odd.
$f(−x)=\frac{3(−x)}{((−x)^2+1)}=\frac{-3x}{(x^2+1)}=−\left[\frac{3x}{(x^2+1)}\right]=−f(x)$ . Therefore, $f$ is odd.

Watch the following video to see the worked solution to Example: Even and Odd Functions

Closed Captioning and Transcript Information for Video

Try It

Determine whether $f(x)=4x^3-5x$ is even, odd, or neither.

Hint

Compare $f(−x)$ with $f(x)$ and $−f(x)$ .

Show Solution

$f(x)$ is odd.

One symmetric function that arises frequently is the absolute value function, written as $|x|$ . The absolute value function is defined as

f (x) = {\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}

$f(x)=\begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$

Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value function, we see that if $x<0$ , then $|x|=−x>0$ , and if $x>0$ , then $|x|=x>0$ . However, for $x=0, \, |x|=0$ . Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if $x=0$ , the output $|x|=0$ . We conclude that the range of the absolute value function is $\{y|y\ge 0\}$ . In Figure 14, we see that the absolute value function is symmetric about the $y$ -axis and is therefore an even function.

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -4 to 4. The graph is of the function “f(x) = absolute value of x”. The graph starts at the point (-3, 3) and decreases in a straight line until it hits the origin. Then the graph increases in a straight line until it hits the point (3, 3).

Figure 14. The graph of $f(x)=|x|$ is symmetric about the $y$ -axis.

Example: Working with the Absolute Value Function

Find the domain and range of the function $f(x)=2|x-3|+4$ .

Show Solution

Since the absolute value function is defined for all real numbers, the domain of this function is $(−\infty ,\infty )$ . Since $|x-3|\ge 0$ for all $x$ , the function $f(x)=2|x-3|+4\ge 4$ . Therefore, the range is, at most, the set $\{y|y\ge 4\}$ . To see that the range is, in fact, this whole set, we need to show that for $y\ge 4$ there exists a real number $x$ such that

$2|x-3|+4=y$ .

A real number $x$ satisfies this equation as long as

$|x-3|=\frac{1}{2}(y-4)$ .

Since $y\ge 4$ , we know $y-4\ge 0$ , and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore,

$|x-3|= \begin{cases} x-3, & \text{ if } \, x \ge 3 \\ -(x-3), & \text{ if } \, x < 3 \end{cases}$

Therefore, we see there are two solutions:

$x=\pm\frac{1}{2}(y-4)+3$ .

The range of this function is $\{y|y\ge 4\}$ .

Watch the following video to see the worked solution to Example: Working with the Absolute Value Function

Closed Captioning and Transcript Information for Video

Try It

For the function $f(x)=|x+2|-4$ , find the domain and range.

Hint

$|x+2|\ge 0$ for all real numbers $x$ .

Show Solution

Domain = $(−\infty ,\infty )$ , range = $\{y|y\ge -4\}$ .

Module 1: Functions and Graphs