Even and Odd Functions
A function is even if its graph is symmetric about the y-axis, and odd if its graph is symmetric about the origin. Both properties are checked with a single algebraic test and have important consequences in calculus, physics, and engineering.
Quick Reference
| Property | Even Function | Odd Function |
|---|---|---|
| Algebraic test | $f(-x) = f(x)$ for all $x$ in domain | $f(-x) = -f(x)$ for all $x$ in domain |
| Symmetry | About the $y$-axis | About the origin |
| Power function $x^n$ | Even when $n$ is even | Odd when $n$ is odd |
| Simple examples | $x^2,\ x^4,\ \lvert x \rvert$ | $x,\ x^3,\ \dfrac{1}{x}$ |
Even Functions
A function $f$ is even if
$f(-x) = f(x) \quad \text{for every } x \in \operatorname{Dom}(f).$A function $y = f(x)$ defined on a symmetric interval $(-a, a)$ is called even if changing the sign of any $x$ in that interval leaves the function value unchanged:
$f(-x) = f(x) \qquad \text{for every } x \text{ in } (-a,\, a).$What does y-axis symmetry mean geometrically?
If a function is even, its graph is symmetric about the y-axis. That is, if we draw a horizontal segment from any point on the graph to the $y$-axis and continue the same distance on the other side, we reach another point on the graph. Equivalently, folding the coordinate plane along the $y$-axis makes the left and right halves of the graph coincide exactly.
Examples of even functions are $f(x) = x^{2}$ and $g(x) = \dfrac{1}{x^{2}} - 1$.
Odd Functions
A function $f$ is odd if
$f(-x) = -f(x) \quad \text{for every } x \in \operatorname{Dom}(f).$A function $y = f(x)$ defined on a symmetric interval $(-a, a)$ is called odd if changing the sign of any $x$ in that interval changes only the sign, not the absolute value, of the function:
$f(-x) = -f(x) \qquad \text{for every } x \text{ in } (-a,\, a).$What does origin symmetry mean geometrically?
If a function is odd, its graph is symmetric about the origin. That is, if we draw a segment from any point on the graph through the origin and continue the same distance on the other side, we reach another point on the graph. Equivalently, for any point $(x, y)$ on the graph, the point $(-x, -y)$ also lies on the graph. Rotating the graph 180° about the origin produces the same curve.
Examples of odd functions are $f(x) = x^{3}$ and $g(x) = \dfrac{1}{x}$.
Worked Example
Determine whether each function is even, odd, or neither:
- $f(x) = x^3 - 3x$
- $g(x) = x^4 + 2x^2$
- $h(x) = 1 - \dfrac{1}{x}$
Solution
(a) Compute $f(-x)$: $f(-x) = (-x)^3 - 3(-x) = -x^3 + 3x = -(x^3 - 3x) = -f(x).$ Since $f(-x) = -f(x)$, the function $f$ is odd. Note that $f$ has only odd-degree terms ($x^3$ and $x$), which confirms the result.
Algebraic Properties: Sums and Products
When two functions with known parity are combined, the parity of the result follows fixed rules. These rules hold for all algebraic functions (polynomials, rational functions, power functions) and are proved directly from the definitions.
Sums and Products Table
| Operation | Result | Why |
|---|---|---|
| even $+$ even | even | $f(-x)+g(-x) = f(x)+g(x)$ |
| odd $+$ odd | odd | $f(-x)+g(-x) = -f(x)-g(x) = -(f+g)(x)$ |
| odd $+$ even | neither (in general) | no symmetry is preserved |
| even $\times$ even | even | $f(-x)\cdot g(-x) = f(x)\cdot g(x)$ |
| odd $\times$ odd | even | $(-f(x))\cdot(-g(x)) = f(x)\cdot g(x)$ |
| odd $\times$ even | odd | $(-f(x))\cdot g(x) = -(f\cdot g)(x)$ |
Proofs of all six rules
In each proof below, $f$ and $g$ are functions with the stated parity, defined on a common symmetric domain. We verify the definition directly.- even + even = even
Let $f(-x) = f(x)$ and $g(-x) = g(x)$. Set $h = f + g$. Then: $ \begin{aligned} (f+g)(-x) &= f(-x) + g(-x) \\ &= f(x) + g(x) \\ &= (f+g)(x). \checkmark \end{aligned} $ - odd + odd = odd
Let $f(-x) = -f(x)$ and $g(-x) = -g(x)$. Set $h = f + g$. Then: $ \begin{aligned} (f+g)(-x) &= f(-x) + g(-x) \\ &= -f(x) + (-g(x)) \\ &= -(f(x)+g(x)) \\ &= -(f+g)(x). \checkmark \end{aligned} $ - odd + even = neither (in general)
A general argument: suppose $h = f + g$ with $f$ odd and $g$ even. If $h$ were even, then $h(-x) = h(x)$ would give $-f(x)+g(x) = f(x)+g(x)$, forcing $f(x) = 0$ for all $x$. If $h$ were odd, then $h(-x) = -h(x)$ would give $-f(x)+g(x) = -f(x)-g(x)$, forcing $g(x) = 0$ for all $x$. So unless $f$ or $g$ is identically zero, $h$ is neither.
Concrete example: Let $f(x) = x$ (odd) and $g(x) = x^2$ (even), so $h(x) = x + x^2$. $h(-1) = -1 + 1 = 0, \quad h(1) = 1 + 1 = 2.$ Since $h(-1) \neq h(1)$, $h$ is not even. Since $h(-1) \neq -h(1) = -2$, $h$ is not odd. $\checkmark$ - even × even = even
Let $f(-x) = f(x)$ and $g(-x) = g(x)$. Then: $ \begin{aligned} (f \cdot g)(-x) &= f(-x)\cdot g(-x) \\ &= f(x)\cdot g(x) \\ &= (f\cdot g)(x). \checkmark \end{aligned} $ - odd × odd = even
Let $f(-x) = -f(x)$ and $g(-x) = -g(x)$. Then: $ \begin{aligned} (f \cdot g)(-x) &= f(-x)\cdot g(-x) \\ &= (-f(x))\cdot(-g(x)) \\ &= f(x)\cdot g(x) \\ &= (f\cdot g)(x). \checkmark \end{aligned} $ The two minus signs cancel, making the product satisfy the even condition. - odd × even = odd
Let $f(-x) = -f(x)$ (odd) and $g(-x) = g(x)$ (even). Then: $\begin{aligned} (f \cdot g)(-x) &= f(-x)\cdot g(-x) \\ &= (-f(x))\cdot g(x) \\ &= -(f(x)\cdot g(x)) \\ &= -(f\cdot g)(x). \checkmark \end{aligned} $
Key surprise: The product of two odd functions is even, not odd. The two negatives cancel. For example, $x \cdot x = x^2$ (even), and $x^3 \cdot x = x^4$ (even). Think of it like multiplying signs: odd $\times$ odd behaves like $(-)\times(-) = (+)$.
For quotients the same sign rules apply, since $f/g = f \cdot (1/g)$ and $1/g$ has the same parity as $g$. So odd/even is odd, even/odd is odd, even/even is even, and odd/odd is even.
Which Algebraic Functions Are Even or Odd?
Power functions $x^n$ ($n$ a positive integer): even when $n$ is even, odd when $n$ is odd. This is the origin of the terminology.
Polynomials follow from the addition rules above:
- A polynomial with only even-degree terms, such as $3x^4 - 5x^2 + 7$, is even.
- A polynomial with only odd-degree terms, such as $2x^5 - x^3 + 4x$, is odd.
- A polynomial mixing even and odd degrees, such as $x^2 + x$, is neither.
Rational functions $p(x)/q(x)$: combine the parities of numerator and denominator using the product/quotient rules in the table above.
Decomposing Any Function Into Even and Odd Parts
Every function defined on a symmetric domain can be written uniquely as the sum of an even function and an odd function. This is called the even–odd decomposition (or parity decomposition) of $f$.
Theorem (Even–Odd Decomposition). Let $f$ be any function whose domain $D$ is symmetric about $0$ (i.e., $x \in D \Rightarrow -x \in D$). Define
$E(x) = \frac{f(x)+f(-x)}{2}, \qquad O(x) = \frac{f(x)-f(-x)}{2}.$Then $E$ is even, $O$ is odd, and
$f(x) = E(x) + O(x) \quad \text{for every } x \in D.$Moreover, this decomposition is unique: there is only one way to write $f$ as a sum of an even function and an odd function.
Proof
$E$ is even: $ \begin{aligned} E(-x) = \frac{f(-x)+f(-(-x))}{2} \\ &= \frac{f(-x)+f(x)}{2} \\ &= E(x). \checkmark \end{aligned} $ $O$ is odd: $ \begin{aligned} O(-x) &= \frac{f(-x)-f(-(-x))}{2} \\ &= \frac{f(-x)-f(x)}{2} \\ &= -\frac{f(x)-f(-x)}{2} \\ &= -O(x). \checkmark \end{aligned} $ $\boldsymbol{E + O = f}$: $ \begin{aligned} E(x)+O(x) &= \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} \\ &= \frac{2f(x)}{2} \\ &= f(x). \checkmark \end{aligned} $ Uniqueness. Suppose $f = E_1 + O_1 = E_2 + O_2$ with $E_1, E_2$ even and $O_1, O_2$ odd. Then $E_1 - E_2 = O_2 - O_1.$ The left side is even (difference of even functions) and the right side is odd (difference of odd functions). A function that is both even and odd satisfies $h(x) = h(-x)$ and $h(x) = -h(-x)$, so $h(x) = -h(x)$, giving $h(x) = 0$ for all $x$. Therefore $E_1 = E_2$ and $O_1 = O_2$. $\blacksquare$Find the even and odd parts of $f(x) = x^3 + 2x^2 - x + 3$.
