10.2: Symmetry

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Review: Symmetry - Even and Odd Functions (level = college algebra)

Definitions and Theorems

Throughout this section, \(f\) denotes a real function whose domain is symmetric about the origin; that is, whenever \(x\) is in the domain, \(-x\) is also in the domain.

Definition: Even and Odd Functions

The function \( f \) is called an even function if for every input \( x \) in the domain of \( f \)\[f ( -x ) = f ( x ). \nonumber \]The function \( f \) is called an odd function if for every input \( x \) in the domain of \( f \)\[ f ( -x ) = − f ( x ). \nonumber \]

Neither Even nor Odd

A function that fails to satisfy \(f(-x)=f(x)\) for at least one \(x\), and also fails to satisfy \(f(-x)=-f(x)\) for at least one \(x\), is said to be neither even nor odd. Most functions fall into this category.

Theorem: Symmetry of Even and Odd Functions

Let \(f\) be a real function.

• \(f\) is even if and only if its graph is symmetric with respect to the \(y\)-axis; that is, the graph is unchanged under reflection across the \(y\)-axis.
• \(f\) is odd if and only if its graph is symmetric with respect to the origin; that is, the graph is unchanged under a \(180^{\circ}\) rotation about the origin.

Equivalently, \(f\) is even when the point \((-x,y)\) lies on the graph whenever \((x,y)\) does, and \(f\) is odd when the point \((-x,-y)\) lies on the graph whenever \((x,y)\) does.

Theorem: Parity of a Polynomial

Let \(f\) be a polynomial function. Then \(f\) is even if and only if every term has even degree, and \(f\) is odd if and only if every term has odd degree. A nonzero constant term is treated as having even degree.

The Only Function That Is Both

The only function that is both even and odd is the function that equals \(0\) everywhere on its domain. Satisfying \(f(-x)=f(x)\) and \(f(-x)=-f(x)\) simultaneously forces \(f(x)=-f(x)\), hence \(f(x)=0\).

Examples

Example \(\PageIndex{1}\): An Even Polynomial

Classify \(f(x)=2x^4-x^2+5\) as even, odd, or neither.

Solution

Replace \(x\) with \(-x\) and simplify, using the fact that an even power removes the sign:\[f(-x)=2(-x)^4-(-x)^2+5=2x^4-x^2+5.\nonumber\]This is identical to \(f(x)\), so \(f(-x)=f(x)\) and \(f\) is even. Its graph is symmetric about the \(y\)-axis. (Every term has even degree, which confirms the result.)

Example \(\PageIndex{2}\): An Odd Polynomial

Classify \(f(x)=x^3-4x\) as even, odd, or neither.

Solution

Compute \(f(-x)\), noting that an odd power keeps the sign:\[f(-x)=(-x)^3-4(-x)=-x^3+4x.\nonumber\]Factor out \(-1\) to compare with \(f(x)\):\[-x^3+4x=-\left(x^3-4x\right)=-f(x).\nonumber\]Since \(f(-x)=-f(x)\), the function is odd, and its graph is symmetric about the origin.

Example \(\PageIndex{3}\): Neither

Classify \(f(x)=x^3+x^2\) as even, odd, or neither.

Solution

Compute \(f(-x)\):\[f(-x)=(-x)^3+(-x)^2=-x^3+x^2.\nonumber\]Compare with \(f(x)=x^3+x^2\): the two are not identical, so \(f\) is not even. Compare with\[-f(x)=-x^3-x^2;\nonumber\]this does not match \(f(-x)=-x^3+x^2\) either, so \(f\) is not odd. The function is neither. (The mixed even and odd degrees signal this outcome directly.)

Example \(\PageIndex{4}\): A Rational Function

Classify \(f(x)=\dfrac{x}{x^2+1}\) as even, odd, or neither.

Solution

Substitute \(-x\) and simplify each part. The numerator becomes \(-x\); the denominator is unchanged because \((-x)^2=x^2\):\[f(-x)=\dfrac{-x}{(-x)^2+1}=\dfrac{-x}{x^2+1}=-\dfrac{x}{x^2+1}=-f(x).\nonumber\]Since \(f(-x)=-f(x)\), the function is odd.

Example \(\PageIndex{5}\): Reading Symmetry from a Graph

The graph of a function \(f\) is shown in Figure \(\PageIndex{1}\). Determine whether \(f\) is even, odd, or neither.

Figure \(\PageIndex{1}\): The graph of a function \(f\).

Solution

Test the graph against the two symmetry conditions. A reflection across the \(y\)-axis does not return the same curve, so \(f\) is not even. However, rotating the graph \(180^{\circ}\) about the origin leaves it unchanged: for each plotted point \((x,y)\), the opposite point \((-x,-y)\) also lies on the curve. This is exactly the condition for symmetry about the origin, so \(f\) is odd.

Example \(\PageIndex{6}\): A Combination Involving a Trigonometric Term

Classify \(f(x)=x^2+\cos x\) as even, odd, or neither.

Solution

Compute \(f(-x)\), using \((-x)^2=x^2\) and the fact that cosine is even, so \(\cos(-x)=\cos x\):\[f(-x)=(-x)^2+\cos(-x)=x^2+\cos x.\nonumber\]This equals \(f(x)\), so \(f\) is even. (The result also follows because a sum of two even functions is even, and both \(x^2\) and \(\cos x\) are even.)

MyOpenMath Practice

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Sources

Several parts of this text use modifications from the following source:

• Wikipedia article: "Even and odd functions"

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