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Algebra 2
Course: Algebra 2 > Unit 9
Lesson 3: Symmetry of functions- Function symmetry introduction
- Function symmetry introduction
- Even and odd functions: Graphs
- Even and odd functions: Tables
- Even and odd functions: Graphs and tables
- Even and odd functions: Equations
- Even and odd functions: Find the mistake
- Even & odd functions: Equations
- Symmetry of polynomials
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Symmetry of polynomials
CCSS.Math:
Learn how to determine if a polynomial function is even, odd, or neither.
What you should be familiar with before taking this lesson
A function is an even function if its graph is symmetric with respect to the y-axis.
Algebraically, f is an even function if f, left parenthesis, minus, x, right parenthesis, equals, f, left parenthesis, x, right parenthesis for all x.
A function is an odd function if its graph is symmetric with respect to the origin.
Algebraically, f is an odd function if f, left parenthesis, minus, x, right parenthesis, equals, minus, f, left parenthesis, x, right parenthesis for all x.
If this is new to you, we recommend that you check out our intro to symmetry of functions.
What you will learn in this lesson
You will learn how to determine whether a polynomial is even, odd, or neither, based on the polynomial's equation.
Investigation: Symmetry of monomials
A monomial is a one-termed polynomial. Monomials have the form f, left parenthesis, x, right parenthesis, equals, a, x, start superscript, n, end superscript where a is a real number and n is an integer greater than or equal to 0.
In this investigation, we will analyze the symmetry of several monomials to see if we can come up with general conditions for a monomial to be even or odd.
In general, to determine whether a function f is even, odd, or neither even nor odd, we analyze the expression for f, left parenthesis, minus, x, right parenthesis:
- If f, left parenthesis, minus, x, right parenthesis is the same as f, left parenthesis, x, right parenthesis, then we know f is even.
- If f, left parenthesis, minus, x, right parenthesis is the opposite of f, left parenthesis, x, right parenthesis, then we know f is odd.
- Otherwise, it is neither even nor odd.
As a first example, let's determine whether f, left parenthesis, x, right parenthesis, equals, 4, x, cubed is even, odd, or neither.
Here f, left parenthesis, minus, x, right parenthesis, equals, minus, f, left parenthesis, x, right parenthesis, and so function f is an odd function.
Now try some examples on your own to see if you can find a pattern.
Concluding the investigation
From the above problems, we see that if f is a monomial function of even degree, then function f is an even function. Similarly, if f is a monomial function of odd degree, then function f is an odd function.
| Even Function | Odd Function | |
|---|---|---|
| Examples | g, left parenthesis, x, right parenthesis, equals, 3, x, start superscript, start color #aa87ff, 2, end color #aa87ff, end superscript | h, left parenthesis, x, right parenthesis, equals, minus, 2, x, start superscript, start color #1fab54, 5, end color #1fab54, end superscript |
| In general | f, left parenthesis, x, right parenthesis, equals, a, x, start superscript, start color #aa87ff, n, end color #aa87ff, end superscript where n is start color #aa87ff, start text, e, v, e, n, end text, end color #aa87ff | f, left parenthesis, x, right parenthesis, equals, a, x, start superscript, start color #1fab54, n, end color #1fab54, end superscript where n is start color #1fab54, start text, o, d, d, end text, end color #1fab54 |
This is because left parenthesis, minus, x, right parenthesis, start superscript, n, end superscript, equals, x, start superscript, n, end superscript when n is even and left parenthesis, minus, x, right parenthesis, start superscript, n, end superscript, equals, minus, x, start superscript, n, end superscript when n is odd.
This is probably the reason why even and odd functions were named as such in the first place!
Investigation: Symmetry of polynomials
In this investigation, we will examine the symmetry of polynomials with more than one term.
Example 1: f, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 3, x, squared, minus, 5
To determine whether f is even, odd, or neither, we find f, left parenthesis, minus, x, right parenthesis.
Since f, left parenthesis, minus, x, right parenthesis, equals, f, left parenthesis, x, right parenthesis, function f is an even function.
Note that all the terms of f are of an even degree.
Example 2: g, left parenthesis, x, right parenthesis, equals, 5, x, start superscript, 7, end superscript, minus, 3, x, cubed, plus, x
Again, we start by finding g, left parenthesis, minus, x, right parenthesis.
At this point, notice that each term in g, left parenthesis, minus, x, right parenthesis is the opposite of each term in g, left parenthesis, x, right parenthesis. In other words, g, left parenthesis, minus, x, right parenthesis, equals, minus, g, left parenthesis, x, right parenthesis, and so g is an odd function.
Note that all the terms of g are of an odd degree.
Example 3: h, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 7, x, cubed
Let's find h, left parenthesis, minus, x, right parenthesis.
2, x, start superscript, 4, end superscript, plus, 7, x, cubed is not the same as h, left parenthesis, x, right parenthesis nor is it the opposite of h, left parenthesis, x, right parenthesis.
Mathematically, h, left parenthesis, minus, x, right parenthesis, does not equal, h, left parenthesis, x, right parenthesis and h, left parenthesis, minus, x, right parenthesis, does not equal, minus, h, left parenthesis, x, right parenthesis, and so h is neither even nor odd.
Note that h has one even-degree term and one odd-degree term.
Concluding the investigation
In general, we can determine whether a polynomial is even, odd, or neither by examining each individual term.
| empty space | General rule | Example polynomial |
|---|---|---|
| Even | A polynomial is even if each term is an even function. | f, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 3, x, squared, minus, 5 |
| Odd | A polynomial is odd if each term is an odd function. | g, left parenthesis, x, right parenthesis, equals, 5, x, start superscript, 7, end superscript, minus, 3, x, cubed, plus, x |
| Neither | A polynomial is neither even nor odd if it is made up of both even and odd functions. | h, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 7, x, cubed |
Check your understanding
Want to join the conversation?
Is a constant considered an "even" term?(36 votes)
Yes. Constant functions are symmetric about the y axis and are even functions. (You can think of the power of the x variable as zero, which is an even number, to help you remember this.)(59 votes)
Why would you use the F(x)=F(-x)/-F(x) rule when you can simply check whether the exponents are odd or even? Why check for points on the graph when you clearly know that if it's not symmetrical, it's not even? Is it just to figure out whether a graph is perfectly even or perfectly odd?(2 votes)
There is no such thing as "perfectly even or odd". There is only even or odd or neither.
Not all functions we wish to classify are polynomials. For instance, how would you tell me if the sine function was even or odd or neither? Good observation though – all polynomial functions only have parity if all exponents have the same parity.(6 votes)
- why does (-x)^2=x^2 in the explanation of the 2nd question?(2 votes)
(-x)^2 = (-x)(-x)
a negative times a negative = a positive
So, the final results is +x^2
Hope this helps.(4 votes)
Now that I know the "general rule" I know how to correctly solve these type of problems. But just abstractly solving without using a real number for x, the test still confuses me. I think my confusion has to do with f(-x) and -f(-x). So the question is who do I put that idea more concretely in my mind?(2 votes)
So look, the symmetry of any function basically depends on f(-x), where x is some real number within the domain. So, the article says that if f(-x) is the same as the function (i.e. f(-x) = f(x)), the function is even. Conversely, if f(-x) is the opposite of the function (i.e. f(-x) = -f(x), NOT -(f(-x))), the function is odd.
I'm surprised they didn't use a proper number example here, but here you go:
So, let's use the same example given. So, let y = 5x^7−3x^3+x. From the proof in Example 3, we know this function is odd. Let's test it.
Let's first find f(2). So, we have y = 5(2)^(7) - 3(2)^(3) + 2. This gives us 640 - 24 + 2 = 618.
Now, let's find f(-2). So, we have y = 5(-2)^(7) - 3(-2)^(3) + (-2). This gives us -640 + 24 - 2 = -618.
Now, see that f(x) isn't equal to f(-x). So, the function isn't even. Now, let's find -f(2). We know f(2) = 618. So, -f(2) is simply -618, which is exactly what f(-2) is. Hence, f(-2) = -f(2) and thus, the function is odd.
Hope this helped!(4 votes)
- i want to know about absolute value if it's even or odd(2 votes)
The absolute value function would be an even function.
If f(x)=|x| then f(-x)=|-x|=x therefore f(-x)= f(x)(2 votes)
- i want to know about absolute value if it´s even or odd(2 votes)
It is even since F(x) = F(-x)(2 votes)
- Is every function an odd function if the degree is odd and the function goes through the origin?(2 votes)
Not necessarily. something like x(x-1)(x-2) has a degree of 3 but doesn't fit the definition of an odd function.
Also there are functions that are not polynomials that can be even or odd.(2 votes)
- Can terms be functions?(1 vote)
Yes, terms can be anything you can add. That includes functions, vectors, matrices, and numbers.(3 votes)
- What do I do if the function is in a factored form and its degree is, so high that it will be hard to deduce its original form?(0 votes)
- You could just get the explicit definition of f(-x) and/or -f(-x) and see if they are equal to the definition of f(x) to determine if f(x) is odd, even, or neither.
EDIT:
I've thought about your question for a while. Another scenario would be is what if there were literally so many factors that replacing x's with -x's, negating, and then simplifying the whole expression was too cumbersome? So, I've devised a method that comes directly from the basic arithmetic properties of odd and even functions.
Let P(x) = a ⋅ x₁ ⋅ x₂ ⋅ ... ⋅ xₙ ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ) where P(x) is a polynomial in factored form and each factor is linear—this is possible for any polynomial according to The Linear Factorization Theorem.
a is an even function because a = a ⋅ x⁰. Factors of the form xₖ are odd functions because xₖ = xₖ¹. And factors of the form (x - cⱼ) are neither odd nor even (let's call them noden for short) since x - cⱼ = x¹ - cⱼ ⋅ x⁰.
Here are some properties of odd, even, and noden functions (each function is strictly of that parity). For this section, I'm going to use even to denote an even function, odd to denote an odd function, and noden to denote a function that is neither odd nor even—all of which are polynomials. When I use any of those terms multiple times, they do not necessarily refer to the same function.
even ⋅ even = even
odd ⋅ even = odd
noden ⋅ even = noden
odd ⋅ odd = even
Or more generally,
odd₁ ⋅ odd₂ ⋅ ... ⋅ oddₙ = even ; iff. n is even
odd₁ ⋅ odd₂ ⋅ ... ⋅ oddₙ = odd ; iff. n is odd
odd ⋅ noden = noden
noden ⋅ noden = even ; if both nodens are linear of the form (x - c) and are conjugates of each other. The simplest example is (x + a) and (x - a), they are conjugates of each other (I'll refer to them as a conjugate pair). Their product is x² - a², which is an even function. The product of any conjugate pair is always an even function.
From those properties, we can see that multiplying by an even function doesn't affect the parity of the product. When we only care about parity, we can ignore the even functions/factors in a factored polynomial. I'm going to use p as a variable in the context of algorithms and := as an assignment operator. Let's assign p as P(x) without a, we can ignore it for the sake of the overall parity of P(x).
p := x₁ ⋅ x₂ ⋅ ... ⋅ xₙ ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)
p has the same parity as P(x).
Now, let's determine what n is, which tells us how many factors of the form xₖ there are. Remember, factors of the form xₖ are odd functions, so when there are an even number of them, their product is an even function, otherwise (if the n is odd), the product is an odd function.
If n is even,
p := (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)
otherwise, if n is odd,
p := x ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)
p and P(x) still have the same parity.
Now, let's remove all the conjugate pairs. Remember that the product of any conjugate pair is always an even function. I'm going to denote that operation as a method, RemoveConjugatePairs.
p := RemoveConjugatePairs(p)
p and P(x) still have the same parity.
if p = 1, P(x) is an even function.
; we removed all the even functions, and we are left with 1, that means P(x) was an even function all along. Another way interpret this is that 1 is an even function, so P(x) is an even function.
if p = (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ), P(x) is a noden function.
; the product of noden functions of the form (x - c) is a noden function if there exists a noden factor which is not part of a conjugate pair—all instances of conjugate pairs have already been removed of course.
if p = x, then P(x) is an odd function.
; x is an odd function so P(x) is an odd function.
if p = x ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ), P(x) is a noden function.
; we already know that (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ) is a noden function so if we multiply that by an odd function, the product will be a noden function.
So now we know the parity of P(x). Hopefully you'll have less of a headache determining the parity of extremely large polynomials in factored form.(4 votes)
- I am not sure i understand how to determine the odd or even graphs.(1 vote)
Even graphs are symmetric over the y-axis. y=x^2 is a even graph because it is symmetric over the y-axis. Odd graphs are symmetric over the origin. y = x^3 is an odd graph because it is symmetric over the origin.(2 votes)
