Lesson Explainer: Even and Odd Functions Mathematics

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 𝑓(π‘₯) is

  • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
  • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),

for every π‘₯ in the function’s domain.

Note that the only function defined on the set of real numbers that is both even and odd is 𝑓(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 𝑓(π‘₯)=π‘₯+4 and 𝑔(π‘₯)=π‘₯.

We can check the parity of 𝑓(π‘₯) by evaluatin 𝑓(βˆ’π‘₯): 𝑓(βˆ’π‘₯)=(βˆ’π‘₯)+4=π‘₯+4=𝑓(π‘₯).

𝑓(π‘₯) is therefore an even function. Notice how the graph of 𝑓(π‘₯)=π‘₯+4 has reflectional symmetry with respect to the 𝑦-axis, or the line π‘₯=0. This is because the output of the function is the same if we input π‘₯ or βˆ’π‘₯. For instance, the points (2,8) and (βˆ’2,8) lie on the curve of 𝑦=𝑓(π‘₯).

In fact, 𝑓(βˆ’π‘₯)=𝑓(π‘₯) implies that the graph of the function will have reflectional symmetry with respect to the 𝑦-axis for every value of π‘₯ in the function’s domain. These functions are called even functions since a function 𝑓(π‘₯)=π‘₯ will have this property if 𝑛 is any even integer.

We now consider the function 𝑔(π‘₯)=π‘₯. To check the parity of this function, we will evaluate 𝑔(βˆ’π‘₯): 𝑔(βˆ’π‘₯)=(βˆ’π‘₯)=βˆ’π‘₯=βˆ’π‘”(π‘₯).

𝑔(π‘₯) is an odd function. This time, the graph of 𝑔(π‘₯) has rotational symmetry of order 2 about the origin, meaning that its graph remains unchanged after a rotation of 180∘ about (0,0). This is because if a point with coordinates (π‘₯,𝑦) lies on the curve, then since 𝑔(βˆ’π‘₯)=βˆ’π‘”(π‘₯), a corresponding point with coordinates (βˆ’π‘₯,βˆ’π‘¦) must also lie on the curve. For instance, since the point with coordinates (2,8) lies on the curve of 𝑦=𝑔(π‘₯), then a point with coordinates (βˆ’2,βˆ’8) must also lie on the curve.

𝑔(βˆ’π‘₯)=βˆ’π‘”(π‘₯) implies that the graph of the function will have rotational symmetry order 2 about the origin for every value of π‘₯ in the function’s domain. These functions are called odd functions since a function 𝑔(π‘₯)=π‘₯ will have this property if 𝑛 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, 𝑔(π‘₯)=βˆ’π‘”(π‘₯). We observe that 𝑔(0)=βˆ’π‘”(0), which corresponds to the rotational interpretation of an odd function.

Since, for an odd function, 𝑔(βˆ’π‘₯)=βˆ’π‘”(π‘₯), we can deduce that the absolute value of this function must in fact be even; for any odd function 𝑔(π‘₯), if β„Ž(π‘₯)=|𝑔(π‘₯)|, then β„Ž is even.

Definition: Graphs of Odd and Even Functions

The graph of any even function has reflectional symmetry with respect to the 𝑦-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 𝑓(π‘₯)=4π‘₯βˆ’3 even, odd, or neither?

Answer

We recall that a function 𝑓(π‘₯) is

  • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
  • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),

for every π‘₯ in the function’s domain.

Since 𝑓(π‘₯) is a linear function, its domain is ℝ. This is symmetric about 0, so we know that the symmetrical properties of even and odd functions apply. To test the parity of 𝑓(π‘₯), we will evaluate 𝑓(βˆ’π‘₯): 𝑓(βˆ’π‘₯)=4(βˆ’π‘₯)βˆ’3=βˆ’4π‘₯βˆ’3.

We notice that 𝑓(βˆ’π‘₯)≠𝑓(π‘₯), and nor does 𝑓(βˆ’π‘₯)=βˆ’π‘“(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 𝑦-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 function’s 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 [βˆ’8,8], not including π‘₯=0. Using set notation, the domain is given by [βˆ’8,8]βˆ’{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 𝑦-axis, or the line π‘₯=0. This means that, for any value of π‘₯ in the domain of the function, 𝑓(βˆ’π‘₯)=𝑓(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 𝑦-axis. It is important to realize that this must hold true for every value of π‘₯ in the function’s 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 β„βˆ’{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 𝑦-axis, and so this function cannot be even.

The graph does, however, remain unchanged after a rotation of 180∘ 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 𝑦-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 𝑦-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 function’s 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 2≀π‘₯≀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

𝑓(π‘₯)=π‘₯cos and 𝑓(π‘₯)=π‘₯sec are even functions.

𝑓(π‘₯)=π‘₯sin, 𝑓(π‘₯)=π‘₯csc, 𝑓(π‘₯)=π‘₯tan, and 𝑓(π‘₯)=π‘₯cot are odd functions.

Example 5: Identifying the Parity of a Function

Is the function 𝑓(π‘₯)=π‘₯6π‘₯οŠͺtan even, odd, or neither?

Answer

A function 𝑦=𝑓(π‘₯) is

  • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
  • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),

for every π‘₯ in the function’s 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‘₯οŠͺtan is the product of two functions, so its domain will be the intersection of the domains of each function.

Since π‘₯ 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 tanοŠͺ6π‘₯ is the set of real numbers except those that make cosοŠͺ6π‘₯=0. The values of π‘₯ that make cosοŠͺ6π‘₯=0 are π‘₯=πœ‹12,3πœ‹12,βˆ’πœ‹12,βˆ’3πœ‹12, and so on. These values are symmetrical with respect to the 𝑦-axis, meaning the domain of tanοŠͺ6π‘₯ 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 𝑓(βˆ’π‘₯): 𝑓(βˆ’π‘₯)=(βˆ’π‘₯)(βˆ’6π‘₯).οŠͺtan

And we rewrite (βˆ’π‘₯) as (βˆ’π‘₯)=(βˆ’1Γ—π‘₯)=(βˆ’1)Γ—π‘₯=βˆ’π‘₯.

To evaluate tanοŠͺ(βˆ’6π‘₯), we can consider the graph of the function tan6π‘₯; this is a horizontal stretch of the graph of 𝑦=(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(βˆ’6π‘₯)=βˆ’(6π‘₯) and we can write 𝑓(βˆ’π‘₯) as 𝑓(βˆ’π‘₯)=(βˆ’π‘₯)Γ—(βˆ’6π‘₯)=βˆ’π‘₯Γ—6π‘₯=βˆ’π‘₯6π‘₯=βˆ’π‘“(π‘₯).οŠͺοŠͺοŠͺtantantan

We can now see, for every π‘₯ in the domain of 𝑓, 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯).

Hence, the function 𝑓(π‘₯)=π‘₯6π‘₯οŠͺtan is odd.

In example 5, we multiplied an odd function, π‘₯, by an even function, tanοŠͺ(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 π‘“οŠ§ and π‘“οŠ¨ be even functions and π‘”οŠ§ and π‘”οŠ¨ be odd functions:

  • π‘“Β±π‘“οŠ§οŠ¨ is even and π‘”Β±π‘”οŠ§οŠ¨ is odd,
  • π‘“Β±π‘”οŠ§οŠ§ is neither even nor odd,
  • 𝑓⋅𝑓,𝑓𝑓,π‘”β‹…π‘”οŠ§οŠ¨οŠ§οŠ¨οŠ§οŠ¨, and π‘”π‘”οŠ§οŠ¨ are even,
  • π‘“β‹…π‘”οŠ§οŠ§ and π‘“π‘”οŠ§οŠ§ 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 𝑓 is even, odd, or neither given that 𝑓(π‘₯)=ο­βˆ’9π‘₯βˆ’8π‘₯<0,9π‘₯βˆ’8π‘₯β‰₯0.ifif

Answer

A function 𝑓(π‘₯) is

  • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
  • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),

for every π‘₯ in the function’s domain.

We do need to ensure that the domain of the function is symmetric about 0; otherwise, the symmetrical properties of even and odd functions will not apply.

The domain of a piecewise defined function is the union of the subdomains of the various subfunctions. In this question, we have a subfunction, βˆ’9π‘₯–8, defined over the interval ]βˆ’βˆž,0[ and another, 9π‘₯–8, defined over the interval [0,∞[. Both subfunctions are linear and so they are defined over their entire subdomain. Therefore, the union of these intervals is the set of real numbers. The domain of 𝑓(π‘₯) can be written as ℝ.

This is symmetric about 0, so we can now test the parity of the function by evaluating 𝑓(βˆ’π‘₯). We will need to do this for negative and positive inputs separately to determine whether the function displays reflectional symmetry with respect to the 𝑦-axis.

For π‘₯<0, βˆ’π‘₯ will be positive: 𝑓(βˆ’π‘₯)=9Γ—(βˆ’π‘₯)βˆ’8=βˆ’9π‘₯βˆ’8.

This is equal to the other part of the piecewise function, the subfunction used for negative values of π‘₯.

Then, for π‘₯>0, βˆ’π‘₯ will be negative: 𝑓(βˆ’π‘₯)=βˆ’9Γ—(βˆ’π‘₯)βˆ’8=9π‘₯βˆ’8.

Again, this is equal to the other part of the piecewise function, the subfunction used for positive values of π‘₯.

We can confirm our findings, and check what happens at π‘₯=0, by drawing a sketch of the graph.

The graph has reflectional symmetry with respect to the 𝑦-axis.

Since 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯) for all π‘₯ in the domain of 𝑓, the function is even.

We will now investigate how the parity of a function can be affected by its domain.

Example 7: Identifying the Parity of Functions

Determine whether the function 𝑓(π‘₯)=9π‘₯ is even, odd, or neither given that π‘“βˆΆ]βˆ’7,7]→ℝ.

Answer

A function 𝑓(π‘₯) is

  • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
  • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),

for every π‘₯ in the function’s domain.

We need to ensure that the domain of the function is symmetric about 0; otherwise, the symmetrical properties of even and odd functions will not apply.

We are given that π‘“βˆΆ]βˆ’7,7]→ℝ. We can read this as β€œthe function 𝑓 maps numbers from the left open, right closed interval from βˆ’7 to 7 onto the set of real numbers.” The domain is the interval ]βˆ’7,7], while the codomain is the set of real numbers.

It might seem like this domain is symmetric about 0; however, we are told that π‘₯ can be equal to 7 but cannot be equal to βˆ’7. This means it is not symmetric about 0.

Since the domain of 𝑓(π‘₯) is not symmetric about 0, the function is neither even nor odd.

In our final example, we will demonstrate how knowing the parity of a function can help us to infer information about its variables.

Example 8: Finding an Unknown in a Rational Function given Its Parity

Find the value of π‘Ž given 𝑓 is an even function where 𝑓(π‘₯)=68π‘₯+π‘Žπ‘₯βˆ’3 and π‘₯β‰ 0.

Answer

We know that if π‘“οŠ§ and π‘“οŠ¨ are even functions, their quotient π‘“π‘“οŠ§οŠ¨ is also even. Similarly, a function 𝑓(π‘₯) is said to be even if 𝑓(βˆ’π‘₯)=𝑓(π‘₯) for every π‘₯ in the function’s domain.

Since the function on the numerator is independent of π‘₯, it is even. This means that the function on the denominator must also be even. Let the function 𝑓(π‘₯)=8π‘₯+π‘Žπ‘₯βˆ’3 so that 𝑓(βˆ’π‘₯)=8(βˆ’π‘₯)+π‘Ž(βˆ’π‘₯)βˆ’3=8π‘₯βˆ’π‘Žπ‘₯βˆ’3.

For the function to be even, 𝑓(βˆ’π‘₯)=𝑓(π‘₯) for every value of π‘₯ in the domain of 𝑓: 8π‘₯βˆ’π‘Žπ‘₯βˆ’3=8π‘₯+π‘Žπ‘₯βˆ’3.

Subtracting 8π‘₯ and adding 3 to both sides, this equation simplifies to βˆ’π‘Žπ‘₯=π‘Žπ‘₯.

Since π‘₯β‰ 0, we can divide through by π‘₯: βˆ’π‘Ž=π‘Ž.

The only way for this equation to be true is if π‘Ž=0.

For 𝑓 to be an even function, π‘Ž=0.

We will now recap the key points from this explainer.

Key Points

  • A function 𝑓(π‘₯) is
    • an even function if 𝑓(βˆ’π‘₯)=𝑓(π‘₯),
    • an odd function if 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯),
    for every π‘₯ in the function’s domain.
  • The graph of any even function has reflectional symmetry with respect to the 𝑦-axis. Similarly, a function whose graph has reflectional symmetry with respect to the 𝑦-axis is an even function.
  • The graph of any odd function has rotational symmetry of order 2 about the origin. Similarly, a function whose graph has rotational symmetry of order 2 about the origin is an odd function.
  • A function whose domain is not symmetric about 0 is neither even nor odd.

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