Review of Functions: Learn it 4

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Review of Functions: Learn it 4

Symmetry of Functions

Function graphs often exhibit symmetry, a feature that can simplify understanding their behavior.

Symmetry about the $y$ -axis means that mirroring the graph over the $y$ -axis results in the same graph, indicating an even function where $f(x)=f(−x)$ . For instance, $f(x)=x^4−2x^2−3$ is even because both sides of the $y$ -axis mirror each other.

Symmetry about the origin implies that rotating the graph $180$ degrees around the origin leaves the graph unchanged. This is characteristic of odd functions, satisfying $f(−x)=−f(x)$ . Take $f(x)=x^3−4x$ as an example; it’s odd because rotating its graph doesn’t alter it.

Algebraically, you can check for y-axis symmetry by seeing if $f(−x)$ equals $f(x)$ , and for origin symmetry by checking if $f(−x)$ equals $−f(x)$ .

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? It becomes straightforward to identify symmetry in functions once we determine if they are even or odd. Even functions are symmetric about the y-axis, whereas odd functions exhibit symmetry about the origin.

even and odd functions

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.

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

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

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 this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip using this link (opens in new window).

Absolute Value Function

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

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

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 can 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.

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

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 this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this video using this link (opens in new window).