Lesson Explainer: Even And Odd Functions - Nagwa
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In this explainer, we will learn how to decide whether a function is even, odd, or neither both from a graph of the function and from its rule.
The parity of a function describes whether the function is even or odd.
Definition: Odd and Even Functions
A function π(π₯) is
- an even function if π(βπ₯)=π(π₯),
- an odd function if π(βπ₯)=βπ(π₯),
for every π₯ in the functionβs domain.
Note that the only function defined on the set of real numbers that is both even and odd is π(π₯)=0; thus, once we have determined the parity of a function, we do not need to test again.
The graphs of even and odd functions also have some key properties that can make them easy to identify. Consider the graphs of the functions π(π₯)=π₯+4ο¨ and π(π₯)=π₯ο©.
We can check the parity of π(π₯) by evaluatin π(βπ₯): π(βπ₯)=(βπ₯)+4=π₯+4=π(π₯).ο¨ο¨
π(π₯) is therefore an even function. Notice how the graph of π(π₯)=π₯+4ο¨ has reflectional symmetry with respect to the π¦-axis, or the line π₯=0. This is because the output of the function is the same if we input π₯ or βπ₯. For instance, the points (2,8) and (β2,8) lie on the curve of π¦=π(π₯).
In fact, π(βπ₯)=π(π₯) implies that the graph of the function will have reflectional symmetry with respect to the π¦-axis for every value of π₯ in the functionβs domain. These functions are called even functions since a function π(π₯)=π₯ο will have this property if π is any even integer.
We now consider the function π(π₯)=π₯ο©. To check the parity of this function, we will evaluate π(βπ₯): π(βπ₯)=(βπ₯)=βπ₯=βπ(π₯).ο©ο©
π(π₯) is an odd function. This time, the graph of π(π₯) has rotational symmetry of order 2 about the origin, meaning that its graph remains unchanged after a rotation of 180β about (0,0). This is because if a point with coordinates (π₯,π¦) lies on the curve, then since π(βπ₯)=βπ(π₯), a corresponding point with coordinates (βπ₯,βπ¦) must also lie on the curve. For instance, since the point with coordinates (2,8) lies on the curve of π¦=π(π₯), then a point with coordinates (β2,β8) must also lie on the curve.
π(βπ₯)=βπ(π₯) implies that the graph of the function will have rotational symmetry order 2 about the origin for every value of π₯ in the functionβs domain. These functions are called odd functions since a function π(π₯)=π₯ο will have this property if π is any odd integer.
If an odd function is defined at zero, then its graph must pass through the origin. We can demonstrate this by letting π₯=0 in the definition for an odd function, π(π₯)=βπ(π₯). We observe that π(0)=βπ(0), which corresponds to the rotational interpretation of an odd function.
Since, for an odd function, π(βπ₯)=βπ(π₯), we can deduce that the absolute value of this function must in fact be even; for any odd function π(π₯), if β(π₯)=|π(π₯)|, then β is even.
Definition: Graphs of Odd and Even Functions
The graph of any even function has reflectional symmetry with respect to the π¦-axis.
The graph of any odd function has rotational symmetry of order 2 about the origin.
We can use both the definition of the function and its graph to help determine the parity of the function. In our first example, we will demonstrate how to use the definition of the function to determine whether a function is even, odd, or neither.
Example 1: Stating the Parity of a Linear Function
Is the function π(π₯)=4π₯β3 even, odd, or neither?
Answer
We recall that a function π(π₯) is
- an even function if π(βπ₯)=π(π₯),
- an odd function if π(βπ₯)=βπ(π₯),
for every π₯ in the functionβs domain.
Since π(π₯) is a linear function, its domain is β. This is symmetric about 0, so we know that the symmetrical properties of even and odd functions apply. To test the parity of π(π₯), we will evaluate π(βπ₯): π(βπ₯)=4(βπ₯)β3=β4π₯β3.
We notice that π(βπ₯)β π(π₯), and nor does π(βπ₯)=βπ(π₯).
The function is neither even nor odd.
In our next two examples, we will look at how the definition of even and odd functions (with respect to the symmetry of their graphs) can help us to determine the parity of the function.
Example 2: Determining If a Graphed Function Is Even, Odd, or Neither
Determine whether the function represented by the following figure is even, odd, or neither.
Answer
We recall that the graph of an even function has reflectional symmetry with respect to the π¦-axis while the graph of an odd function has rotational symmetry of order 2 about the origin. It is important to realize that this must hold true for every value of π₯ in the functionβs domain, and as such, we must ensure that the domain of the function is symmetric about 0.
The domain of a function is the set of possible π₯-values that can be substituted into the function; this can be deduced from the graph of a function by looking at the spread of π₯-values from left to right.
The domain of this function is values of π₯ in the interval [β8,8], not including π₯=0. Using set notation, the domain is given by [β8,8]β{0}.
Since this domain is symmetric about 0, we can now check whether the function is even, odd, or neither.
We observe the graph to have reflectional symmetry with respect to the π¦-axis, or the line π₯=0. This means that, for any value of π₯ in the domain of the function, π(βπ₯)=π(π₯).
The function is even.
In our previous example, we demonstrated how to determine the parity of a function defined over a bounded domain from its graph. We will see in example 3 how this process can be applied to functions defined over an unbounded domain.
Example 3: Determining the Parity of a Graphed Rational Function
Is the function represented by the figure even, odd, or neither?
Answer
We recall that the graph of an odd function has rotational symmetry of order 2 about the origin, while the graph of an even function has reflectional symmetry with respect to the π¦-axis. It is important to realize that this must hold true for every value of π₯ in the functionβs domain, and as such, we must ensure that the domain of the function is symmetric about 0.
The graph of the function has a vertical asymptote given at π₯=0. This is the only value of π₯ where the function is not defined; hence, its domain is given by ββ{0}.
Since this domain is symmetric about 0, we can now check whether the function is even, odd, or neither.
We can see that the graph does not have reflectional symmetry given by the π¦-axis, and so this function cannot be even.
The graph does, however, remain unchanged after a rotation of 180β about the origin.
Therefore, the function is odd.
In our previous two examples, we began by checking that the domain of the function was symmetric about 0. Since the parity of a function is dependent on its symmetrical properties with respect to either the π¦-axis or the origin, it follows that a function whose domain is not symmetric about 0 will be neither even nor odd.
In our next example, we will see how acknowledging this element of the definition can save us some time when determining whether a function is even, odd, or neither.
Example 4: Determining If a Graphed Function Is Even, Odd, or Neither
Is the function represented by the figure even, odd, or neither?
Answer
The graph of an even function has reflectional symmetry with respect to the π¦-axis while the graph of an odd function has rotational symmetry of order 2 about the origin. It is important to realize that this must hold true for every value of π₯ in the functionβs domain, and as such, we must ensure that the domain of the function is symmetric about 0.
The domain of a function is the set of possible inputs, or π₯-values, that we can substitute into that function.
The domain of this function is the interval 2β€π₯β€6. This domain is not symmetric about 0.
Since the domain of this function is not symmetric about 0, we can deduce that the function is neither even nor odd.
In our next example, we will look at how to determine the parity of a trigonometric function from its equation using the following definitions.
Definition: Parity of Trigonometric Functions
π(π₯)=π₯cos and π(π₯)=π₯sec are even functions.
π(π₯)=π₯sin, π(π₯)=π₯csc, π(π₯)=π₯tan, and π(π₯)=π₯cot are odd functions.
Example 5: Identifying the Parity of a Function
Is the function π(π₯)=π₯6π₯ο«οͺtan even, odd, or neither?
Answer
A function π¦=π(π₯) is
- an even function if π(βπ₯)=π(π₯),
- an odd function if π(βπ₯)=βπ(π₯),
for every π₯ in the functionβs domain.
We begin by finding the domain of the function. We need to ensure that it is symmetric about 0; otherwise, the symmetrical properties of even and odd functions will not apply.
π₯6π₯ο«οͺtan is the product of two functions, so its domain will be the intersection of the domains of each function.
Since π₯ο« is a polynomial, we know its domain is the set of real numbers.
The domain of the tangent function is the set of real numbers except those where cos(π₯)=0. This means that the domain of the function tanοͺ6π₯ is the set of real numbers except those that make cosοͺ6π₯=0. The values of π₯ that make cosοͺ6π₯=0 are π₯=π12,3π12,βπ12,β3π12, and so on. These values are symmetrical with respect to the π¦-axis, meaning the domain of tanοͺ6π₯ must be symmetric about 0.
The intersection of the two domains is therefore also symmetric about 0, so we can now test for parity by evaluating π(βπ₯): π(βπ₯)=(βπ₯)(β6π₯).ο«οͺtan
And we rewrite (βπ₯)ο« as (βπ₯)=(β1Γπ₯)=(β1)Γπ₯=βπ₯.ο«ο«ο«ο«ο«
To evaluate tanοͺ(β6π₯), we can consider the graph of the function tan6π₯; this is a horizontal stretch of the graph of π¦=(π₯)tan by a scale factor of 16.
We can see that tan6π₯ is odd, since the graph of an odd function has rotational symmetry of order 2 about the origin.
Therefore, tantan(β6π₯)=β(6π₯) and we can write π(βπ₯) as π(βπ₯)=(βπ₯)Γ(β6π₯)=βπ₯Γ6π₯=βπ₯6π₯=βπ(π₯).ο«οͺο«οͺο«οͺtantantan
We can now see, for every π₯ in the domain of π, π(βπ₯)=βπ(π₯).
Hence, the function π(π₯)=π₯6π₯ο«οͺtan is odd.
In example 5, we multiplied an odd function, π₯ο«, by an even function, tanοͺ(6π₯), which resulted in an odd function. In fact, the product of an even and an odd function will always be odd. We can generalize this result alongside some further properties of combining functions.
Definition: Combining Even and Odd Functions
Let πο§ and πο¨ be even functions and πο§ and πο¨ be odd functions:
- πΒ±πο§ο¨ is even and πΒ±πο§ο¨ is odd,
- πΒ±πο§ο§ is neither even nor odd,
- πβ π,ππ,πβ πο§ο¨ο§ο¨ο§ο¨, and ππο§ο¨ are even,
- πβ πο§ο§ and ππο§ο§ are odd.
We will now learn how to apply this concept to determine the parity of a piecewise-defined function.
Example 6: Determining the Parity of a Piecewise-Defined Function
Determine whether the function π is even, odd, or neither given that π(π₯)=οβ9π₯β8π₯
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