Even Functions

Odd functions

Odd functions are functions like $f\left(x\right)$ which have the negative equivalent when the negative independent variables like $f(-x)$ are substituted. Hence they are best expressed as:

$f\left(x\right)\stackrel{}{\to }f(-x)=-f\left(x\right)$

Neither functions

Neither functions are functions like $f\left(x\right)$ which do not have equivalent values when the negative independent variables like $f(-x)$ are substituted. This suggests that they are neither even nor odd functions. Hence they are best expressed as:

$f\left(x\right)\stackrel{}{\to }f(-x)\ne f\left(x\right)$

and

$f\left(x\right)\underset{}{\to }f(-x)\ne -f\left(x\right)$

Even functions in trigonometric identities

It is possible to determine the nature of the function (i.e. even, odd or neither) among trigonometric identities. We shall use to diagrams below to explain this.

Figure 1, an image used in proving nature of functions among trigonometric identities when θ is positive, StudySmarter Originals

From the first diagram, we can use SOHCAHTOA to determine the cosθ. If we do this, we find out that

$\cos \theta =\frac{b}{\sqrt{a^2+b^2}}$

But what happens when θ is negative? From the second diagram we note that although the opposite side (a) has changed (to -a) because the rotation of the angle is in the opposite direction, the adjacent side (b) remains constant. In that case,

$$\begin{gathered} \cos (-\theta )=\frac{b}{\sqrt{{(-a)}^2+b^2}} \\ \cos (-\theta )=\frac{b}{\sqrt{a^2+b^2}} \end{gathered}$$

Hence,

$\cos \theta =\cos (-\theta )$

How relevant is this to even functions? Now, if we express cosine as a function of x, so that we have cos(x) instead of cos(θ). Then, if

$f\left(x\right)=\cos \left(x\right)$

and

$$\begin{gathered} f(-x)=\cos \left(-x\right) \\ f\left(-x\right)=\cos \left(x\right) \end{gathered}$$

Therefore,

$f\left(x\right)=f(-x)$

In this case, this suggests that cos(x) is an even function.

How about sine functions?

If you refer to Figure 1, you would infer that

$\sin \left(\theta \right)=\frac{a}{\sqrt{a^2+b^2}}$

However, when the rotation on the cartesian plane goes in the opposite direction for the angle -θ (as displayed in Figure 2), we notice that the opposite side 'a' in Figure 1, changes to '-a' in Figure 2 because a is located in the negative y-axis. This implies that

$\sin \left(-\theta \right)=\frac{-a}{\sqrt{a^2+b^2}}$

If you factorize by -1, you would arrive at

$\sin \left(-\theta \right)=-1\left(\frac{a}{\sqrt{a^2+b^2}}\right)$

Hence,

$\sin \left(-\theta \right)=-\sin \left(\theta \right)$

But, how useful is this piece of detail? If we express sine as a function of x rather than θ, so that we know how sin(x) as well as sin(-x), then, when

$f\left(x\right)=\sin \left(x\right)$

and

$$\begin{gathered} f(-x)=\sin (-x) \\ f(-x)=-\sin \left(x\right) \end{gathered}$$

with the factorization by -1 on the right-hand side of the equation, we would arrive at

$f(-x)=-1\left(\sin \right(x\left)\right)$

It surely means

$f(-x)=-f\left(x\right)$

This brings to the submission that for the sine function,

$f\left(x\right)\ne f(-x)$

but,

$f(-x)=-f\left(x\right)$

and by implication, we can ergo conclude that sine functions are not even functions but odd functions.

What is the formula of even functions?

For us to determine the formula of even functions, the exponent of the independent variable, x, is always even with or without a constant. Thus for xⁿ, n is an even number such as 2, 4, 6...n. Where a, b, and c are constants such as 1, 2, 3... and n an even number, then, an even function is expressed as

$f\left(x\right)=a{x}^n+b{x}^{n\pm 2}+c$

or

$f\left(x\right)=a{x}^n+b{x}^{n\pm 2}$

Graphs of even function

When an even function is graphed, its graph is symmetric to the vertical axis (y-axis).

For example the even function

$f\left(x\right)=x^4-2$

is graphed below

Graphing an even function, f(x)=x⁴-2, StudySmarter Originals

From the above graph, we see that the two points (-1, -1) and (1, -1) of the graph prove symmetry to the y-axis for the even function $x^4-2.$

What is the difference between even functions and odd functions?

Even functions and odd functions differ in 2 major ways; in their graphs, and general expression.

Difference in graph

We just discussed that the graph of even functions is symmetric to the vertical axis (y-axis).

But for an odd function, its graph is symmetric to the origin. This means that if the curve were to be rotated by 180° at the origin (0, 0), the graph remains the same.

Note as earlier mentioned, in graphs of even function, if you select a given point (p, q) on the graph, you would surely have another point at the opposite horizontal side of the curve which would be (-p, q).

Meanwhile, in odd functions, if you select a point (p, q) you would have a point (-p, -q) in the opposite vertical and horizontal axis. For instance the odd function graph of

$f\left(x\right)=x^3$

Marks points (2, 8) upwards to the right as well as another point (-2, -8) which is downwards to the left.

General expression

Even functions also differ from odd functions in their general expression. Even functions are expressed to conform to the rule.

$f\left(x\right)=f(-x)$

However, odd functions do not obey this because in their case,

$f\left(x\right)\ne f(-x)$

Instead, their functions are generally expressed to conform to the rule

$f(-x)=-f\left(x\right)$

Examples of even functions

To have a better understanding of even functions, it is advisable to practice some problems.

For the function

$h\left(x\right)=6x^6-4x^4+2x^2-1$

Determine if it is an even function. Plot the graph and pick any two points to prove that it is or is not an even function.

Solution:

The first task is to determine if it is an even function. If you apply the even function formula explained earlier, by looking at the expression $6x^6-4x^4+2x^2-1,$ we can conclude that it is an even function since all the exponents of x i.e. 6, 4 and 2 are all even numbers. Nonetheless, so as to make further confirmation we would just apply the rule:

$f\left(x\right)=f(-x)$

By substituting -x into the expression, we get

$$\begin{gathered} f(-x)=6{(-x)}^6-4{(-x)}^4+2\left(x^2\right)-1 \\ f(-x)=6x^6-4x^4+2x^2-1 \end{gathered}$$

Thus,

$f\left(x\right)=f(-x)$

Hence, we can say that the above expression is indeed an even function.

The next task is to plot the graph and using two points, further prove that this expression is indeed an even function.

Using points on a graph to prove even functions, StudySmarter Originals

From the above graph of the expression, we chose two points, (-1, 3) and (1, 3). This further proves that the expression$6x^6-4x^4+2x^2-1,$ is an even function since the pair (-1, 3) and (1, 3) conforms with (p, q) and (-p, q).