# Which of the Following is an Even Function

## How to Determine if a Office is Even, Odd or Neither

I take prepared eight (viii) worked examples to illustrate the procedure or steps on how to figure out if a given function is fifty-fifty, odd, or neither. The math involved in the calculation is easy as long every bit yous are careful in every stride of your solution.

To become into the “heart” of this topic, report the illustration below.

## How to Tell if a Function is Fifty-fifty, Odd, or Neither

Allow us talk nearly each case.

**CASE ane: Fifty-fifty Function**

Given some “starting” role

f\left( 10 \right):

- If nosotros evaluate or substitute

\color{ruby}-x

into

f\left( x \right) and get the original or “starting” role over again, this implies that

f\left( x \right) is an

**even office**.

**Instance 2: Odd Function**

Given some “starting” part

f\left( ten \right):

- However, if we evaluate or substitute

\color{red}-10 into

f\left( 10 \right)

and go the negative or opposite of the “starting” function, this implies thatf\left( 10 \right)

is an

**odd function**.

**CASE three: Neither Even nor Odd Function**

Given some “starting” function

f\left( x \correct):

- If we evaluate or substitute

\color{red}-x

into

f\left( ten \right) and we don’t obtain either Case one or Instance 2, that implies

f\left( x \right) is

**neither fifty-fifty nor odd**. In other words, information technology does non fall under the nomenclature of beingness fifty-fifty or odd.

### Examples of How to Make up one’s mind Algebraically if a Role is Fifty-fifty, Odd, or Neither

**Example 1**: Determine algebraically whether the given function is even, odd, or neither.

f\left( x \correct) = two{x^ii} – 3

I start with the given part

f\left( ten \right) = 2{x^2} – 3, plug in the value

\color{red}-x

and then simplify. What exercise I get? Let u.s. work it out algebraically.

Since

f\left( { {\colour{blood-red}- x}} \right) = f\left( ten \right), it means

f\left( x \right)

is an

**even function**!

The graph of an even part is symmetric with respect to the

y-axis or forth the vertical line

x = 0. Observe that the graph of the part is cut evenly at the

y-centrality and each half is an exact mirror of the some other. Another style of describing it is that each half of the role is a reflection across the

y-axis.

See the animated illustration.

**Example two**: Determine algebraically whether the given function is even, odd, or neither.

f\left( x \right) =\, – 3{ten^3} + 2x

I will substitute

\color{ruby}-x

into the function

f\left( x \right) = 3{10^iii} + 2x, and then simplify.

## How to Determine an Odd Function

Important Tips to Remember:

- If always yous arrive at a unlike role later evaluating

\color{red}–x

into the given

f\left( ten \correct), immediately try to factor out

−1

from information technology and observe if the original function shows upwardly. If it does, then we take an

**odd function**.

- The result of factoring out

−1

results in the switching of the signs of the terms inside the parenthesis. This is a key step to identify an odd function.

Now, since

f\left( { {\color{cerise}- 10}} \right) = – f\left( ten \right), it implies that the original function

f\left( 10 \right)

is an

**odd part**!

The graph of an odd function has rotational symmetry nigh the origin, or at the point

\left( {0,0} \right). That ways nosotros cut its graph along the

y-centrality and then reflect its fifty-fifty half in the

ten-centrality showtime followed by the reflection in the

y-axis.

Meet the blithe illustration.

**Example 3**: Determine algebraically whether if the part is even, odd, or neither:

f\left( x \right) = 3{ten^vi} – 5{x^iv} + half-dozen{10^2} – one

Here I observed that the exponents of variable

ten

are all fifty-fifty numbers, namely6,4, and

two. Equally for the constant term, I must add that it can also be expressed as

– one = – 1{\color{bluish}{x^0}}

which has an even power of zip.

This feature of a office containing but fifty-fifty powers can likely result in an even function. Nonetheless, we must testify it algebraically. And so here information technology goes.

Evaluating

\color{red}-x

into

f\left( x \right), we have the following adding.

Information technology is conspicuously an

**even role**!

**Instance 4**: Determine whether the given office is fifty-fifty, odd, or neither:

f\left( x \right) =\, – {ten^vii} + 8{x^5} – {x^3} + 6x

In contrast to example 3 where the office has even powers, this one has odd powers which are

vii,

v,

3, and

1. By now, I hope yous’re already seeing the blueprint. This is more likely an odd part but we will verify.

Substituting

\colour{blood-red}-x

into the given

f\left( 10 \right), and simplifying, we become:

After factoring out

−i, the polynomial inside the parenthesis equals the starting function. It shows that this is an

**odd function**!

**Example 5**: Determine whether the given function is fifty-fifty, odd, or neither:

This time I volition show y’all an case of a function that is neither even nor odd. Are you set up?

- First, check if it is fifty-fifty. Do we have the case

f\left( {\color{cherry-red}{ – x}} \right) = f\left( 10 \right)?

Definitely

**non an even function**

since

f\left( {\color{reddish}{ – x}} \right) \ne f\left( x \right).

- Secondly, check if it is odd by showing

f\left( {\color{red}{ – x}} \correct) = – f\left( 10 \correct).

Even after factoring out

−1, I notwithstanding don’t become the original function.

This is

**not an odd role**

since

f\left( {\color{red}{ – ten}} \right) \ne – f\left( 10 \right).

- Conclusion: Since we reached the case where

f\left( {\color{red}{ – x}} \correct) \ne f\left( 10 \right)

and

f\left( {\color{ruddy}{ – x}} \right) \ne – f\left( x \right), this function is

**neither even nor odd**!

**Instance 6**: Determine whether the given function is fifty-fifty, odd, or neither:

Solution:

Therefore, function

thou\left( x \correct)

is an

**odd role**!

**Example seven**: Make up one’s mind whether the given function is even, odd, or neither:

Solution:

Therefore, the role

h\left( x \right)

is

**neither**!

**Instance viii**: Decide whether the given function is even, odd, or neither:

Solution:

Therefore, office

k\left( x \right)

is an

**even role**!

## Which of the Following is an Even Function

Source: https://www.chilimath.com/lessons/intermediate-algebra/even-and-odd-functions/