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Even Functions

- Calculus
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While seated and waiting for a train at the North Camp station headed towards Milford for a walkover, I introspected a few troubling questions. Why would the product between two or more odd numbers give an odd number? For instance, the product of 3 and 7 which are odd numbers would give 21, another odd number. Likewise, when even numbers are multiplied between themselves the result is always an even number; still lost in thoughts, I missed my train. However, we would not miss out on the purpose of this discussion, hereafter, we shall learn about **even functions**.

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

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

With respect to this concept, functions are generally classified as either being even, odd or neither.

Confirm that $f\left(x\right)$ is even when

$f\left(x\right)={x}^{4}-8$

**Solution:**

Since

$f\left(x\right)={x}^{4}-8$

To determine the nature of this function we find $f(-x)$ by substituting $-x.$ Hence,

$f(-x)={(-x)}^{4}-8\phantom{\rule{0ex}{0ex}}f(-x)={x}^{4}-8$

Therefore,

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

This proves that $f\left(x\right)$ is an even function for the expression ${x}^{4}-8.$

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:

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

Confirm that f(x) is odd when

$f\left(x\right)={x}^{3}+x$

**Solution:**

Since

$f\left(x\right)={x}^{3}+x$

To determine the nature of this function we find $f(-x)$ by substituting $-x.$ Hence

$f(-x)={(-x)}^{3}+(-x)\phantom{\rule{0ex}{0ex}}f(-x)=-{x}^{3}-x$

When factorized by -1 we get

$f(-x)=-1({x}^{3}+x)$

Does it ring a bell now?😁

Therefore,

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

This proves that f(x) is an odd function for the expression ${x}^{3}+x.$

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

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

Confirm that $f\left(x\right)$ is even when

$f\left(x\right)={x}^{4}+x-1$

**Solution:**

Since

$f\left(x\right)={x}^{4}+x-1$

To determine the nature of this function we find $f(-x)$ by substituting $-x.$ Hence,

$f(-x)={(-x)}^{4}+(-x)-1\phantom{\rule{0ex}{0ex}}f(-x)={x}^{4}-x-1$

The expression ${x}^{4}-x-1$ is not equivalent to ${x}^{4}+x-1$, hence, it is not an even function

When factorised by -1

$f(-x)=-1(-{x}^{4}+x+1)$

The expression above is not equivalent to $-1({x}^{4}+x-1)$, hence, it is not an odd function

Therefore,

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

and

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

This proves that $f\left(x\right)$ is neither function for the expression ${x}^{4}+x-1.$

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 2, an image used in proving nature of functions among trigonometric identities when θ is negative, StudySmarter Originals

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

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

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,

$\mathrm{cos}(-\theta )=\frac{b}{\sqrt{{(-a)}^{2}+{b}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}(-\theta )=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$

Hence,

$\mathrm{cos}\theta =\mathrm{cos}(-\theta )$

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

$f\left(x\right)=\mathrm{cos}\left(x\right)$

and

$f(-x)=\mathrm{cos}\left(-x\right)\phantom{\rule{0ex}{0ex}}f\left(-x\right)=\mathrm{cos}\left(x\right)$

Therefore,

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

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

**Cosine functions** without addition to other function(s) are even** functions**.

If you refer to Figure 1, you would infer that

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

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

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

If you factorize by -1, you would arrive at

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

Recall that

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

Hence,

$\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\left(\theta \right)$

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

$f\left(x\right)=\mathrm{sin}\left(x\right)$

and

$f(-x)=\mathrm{sin}(-x)\phantom{\rule{0ex}{0ex}}f(-x)=-\mathrm{sin}\left(x\right)$

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

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

Recall that

$f\left(x\right)=\mathrm{sin}\left(x\right)$

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**.

**Sine functions** without addition to any other function(s) **are odd functions**.

Why not play about with those 2 diagrams to determine if tangent functions are even, odd or neither functions?

If you did attempt to determine what functions are tangent functions, you would note that since

$\mathrm{tan}\theta =\frac{a}{b}$

From the diagrams, we also know that

$\mathrm{tan}(-\theta )=\frac{-a}{b}$

This implies that

$\mathrm{tan}(-\theta )=-\mathrm{tan}\left(\theta \right)$

Hence, tangent functions are odd functions.

Were you correct ?

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 x^{n}, 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

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

For odd functions, the exponent of the independent variable, x, is always odd and a constant must not be present. Thus for x^{n}, n is an odd number such as 1, 3, 5...n. Where a and b are constants such as 1, 2, 3... and n, an odd number, then, an odd function is expressed as

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

For neither function, the exponent of the independent variable, x, is both even and odd with or without the presence of a constant. Thus for x^{n}, n is an odd and or even number such as 1, 2, 3, 4, 5...n. Where a, b and c are constants such as 1, 2, 3... and n, both even and odd numbers, then, neither function is expressed as

$f\left(x\right)=a{x}^{n}+b{x}^{n\pm 1}+c$

or

$f\left(x\right)=a{x}^{n}+b{x}^{n\pm 1}$

or in the event that all exponents value the independent variable, x, are odd with a constant. Neither function is expressed as

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

where n is an odd number.

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

When a graph is symmetric to an axis, if rotated around a point or reflected over a line, the graph remains the same, although, the point on that axis is the same, the point on the other axis would carry an opposite sign because they are a reflection just like a mirror image.

Thus, when a graph is symmetric to the vertical axis, the given points (p, q) on that graph would have points (-p, q) on that graph. Note that the value of y (point q on the vertical axis) is unchanged while the value of x on the first point (p) has an opposite value for the second point (-p).

For example the even function

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

is graphed below

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 ${x}^{4}-2.$

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

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

Did you just forget that ?

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.

This rotation can be achieved by choosing points (b, 0) and (0, b) on the graph if you drag point (b, 0) horizontally to the point (-b, 0) and drag point (0, b) vertically to point (0, -b) you would confirm that the graph is just the same. Isn't that amazing ?

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).

Kindly refer to the graph of x^{4}-2 as an example.

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.

Graphing an odd function, f(x)=x^{3}, StudySmarter Originals

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

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

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

For the function

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

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 $6{x}^{6}-4{x}^{4}+2{x}^{2}-1,$ 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

$f(-x)=6{(-x)}^{6}-4{(-x)}^{4}+2\left({x}^{2}\right)-1\phantom{\rule{0ex}{0ex}}f(-x)=6{x}^{6}-4{x}^{4}+2{x}^{2}-1$

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.

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

If

$f\left(x\right)=3{x}^{2}$

and

$g\left(x\right)={x}^{4}$

Determine the class of the sum of both functions.

**Solution:**

$f\left(x\right)+g\left(x\right)=3{x}^{2}+{x}^{4}$

Now let's determine the nature of the sum. Let h(x) be the sum, so that,

$h\left(x\right)=3{x}^{2}+{x}^{4}\phantom{\rule{0ex}{0ex}}h(-x)=3{(-x)}^{2}+{(-x)}^{4}\phantom{\rule{0ex}{0ex}}h(-x)=3{x}^{2}+{x}^{4}\phantom{\rule{0ex}{0ex}}h\left(x\right)=h(-x)$

Hence the sum of f(x) and g(x) which are both even functions gives us h(x) which is another even function.

- Even functions are functions like f(x) which have the same values when the negative independent variables like f(-x) are substituted.
- Odd functions are functions like f(x) which have the negative equivalent when the negative independent variables like f(-x) are substituted.
- Neither functions are functions like f(x) which do not have equivalent values when the negative independent variables like f(-x) are substituted.
- It is possible to determine the nature of the function (i.e. even, odd or neither) among trigonometric identities. Cosine functions are even while sine functions are odd.
- For us to determine the formula of even functions, the exponent of the independent variable, x, is always even with or without a constant.
- When an even function is graphed, its graph is symmetric to the vertical axis (y-axis).
- Even functions and odd functions differ in 2 major ways; in their graphs, and general expression.

A function, f(x) is an even function if f(x) = f(-x)

An even function is symmetric about the y-axis.

A graph symmetric about the y-axis represents an even function.

More about Even Functions

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