Even, Odd, and Symmetry, precalculus lesson

Fold a sheet of paper down the middle and the two halves match. Some graphs fold like this along the $y$ -axis, and others match themselves after half a turn about the origin. When you spot such a symmetry you only need to study half of the graph, and calculus uses it the same way to shorten sketches and integrals. Checking for it takes one short algebra step.

Key terms

Symmetric domain

Definition A set $D$ of real numbers is symmetric about $0$ if $-x \in D$ for every $x \in D$ . The definitions of even and odd functions compare $f(x)$ with $f(-x)$ , so they make sense only when the domain of $f$ is symmetric about $0$ .

In plain words Whenever a number is in the set, so is its opposite. On the number line the set looks the same on both sides of $0$ .

Example Here are five sets, checked one by one.

Set	Symmetric about $0$	Reason
$[-3, 3]$	yes	if $x$ lies between $-3$ and $3$ , so does $-x$
$(-\infty, 0) \cup (0, \infty)$ , the domain of $1/x$	yes	the opposite of a nonzero number is nonzero
$[-2, 5]$	no	$5$ is in it but $-5$ is not
$[0, \infty)$ , the domain of $\sqrt{x}$	no	$1$ is in it but $-1$ is not
all $x \ne 1$ , the domain of $1/(x - 1)$	no	$-1$ is in it but $1$ is not

Even function

Definition A function $f$ with domain $D$ is even if $D$ is symmetric about $0$ and

f(-x) = f(x) \quad \text{for every } x \in D.

Examples on $\mathbb{R}$ are the constant functions, $\lvert x \rvert$ , and the powers $x^n$ with $n$ a positive even integer, such as $x^2$ and $x^4$ . The name comes from these even powers.

In plain words Opposite inputs give the same output. Once you know the outputs for positive inputs, you know them for negative inputs too.

Example For $f(x) = x^4 - 3x^2$ the domain is $\mathbb{R}$ , which is symmetric, and

\begin{aligned} f(-x) &= (-x)^4 - 3(-x)^2 \\[0.4em] &= x^4 - 3x^2 \\[0.4em] &= f(x), \end{aligned}

because an even power of $-x$ equals the same power of $x$ . So $f$ is even. For instance $f(2) = 16 - 12 = 4$ and $f(-2) = 4$ , and $f(1) = f(-1) = -2$ .

Symmetry about the y-axis

Definition A set of points in the plane is symmetric about the $y$ -axis if for every point $(x, y)$ in the set, the mirror point $(-x, y)$ is also in the set. The graph of a function $f$ is symmetric about the $y$ -axis exactly when $f$ is even, because the mirror point of $(x, f(x))$ is $(-x, f(x))$ , and that point is on the graph exactly when $-x$ is in the domain and $f(-x) = f(x)$ .

In plain words Fold the picture along the $y$ -axis and the left half lands exactly on the right half.

Example The function $g(x) = 8/(x^2 + 4)$ is even, since replacing $x$ by $-x$ leaves $x^2$ unchanged. Its graph pairs up points across the $y$ -axis.

$x$	$-2$	$-1$	$0$	$1$	$2$
$g(x)$	$1$	$1.6$	$2$	$1.6$	$1$

The point $(0, 2)$ lies on the axis itself and is its own mirror image.

Odd function

Definition A function $f$ with domain $D$ is odd if $D$ is symmetric about $0$ and

f(-x) = -f(x) \quad \text{for every } x \in D.

Examples are $1/x$ and the powers $x^n$ with $n$ a positive odd integer, such as $x$ and $x^3$ . If $0$ is in the domain of an odd function $f$ , then $f(0) = -f(0)$ , so $f(0) = 0$ .

In plain words Opposite inputs give opposite outputs. Change the sign of the input and the output changes sign too.

Example For $f(x) = x^3 - 4x$ on $\mathbb{R}$ ,

\begin{aligned} f(-x) &= (-x)^3 - 4(-x) \\[0.4em] &= -x^3 + 4x \\[0.4em] &= -(x^3 - 4x) \\[0.4em] &= -f(x), \end{aligned}

so $f$ is odd. Indeed $f(1) = -3$ and $f(-1) = 3$ , $f(3) = 15$ and $f(-3) = -15$ , and $f(0) = 0$ .

Symmetry about the origin

Definition A set of points in the plane is symmetric about the origin if for every point $(x, y)$ in the set, the point $(-x, -y)$ is also in the set. The origin is the midpoint of the segment from $(x, y)$ to $(-x, -y)$ , so moving every point $(x, y)$ to $(-x, -y)$ is a rotation by half a turn about the origin, and such a set looks the same after a half turn. The graph of a function $f$ is symmetric about the origin exactly when $f$ is odd, because the point $(-x, -f(x))$ is on the graph exactly when $-x$ is in the domain and $f(-x) = -f(x)$ .

In plain words Spin the picture half a turn around the origin and it matches itself. The same move can be done as a reflection across the $y$ -axis followed by a reflection across the $x$ -axis.

Example The graph of $y = 1/x$ is symmetric about the origin. The point $\left(2, \frac{1}{2}\right)$ on the right branch pairs with $\left(-2, -\frac{1}{2}\right)$ on the left branch, and $\left(\frac{1}{2}, 2\right)$ pairs with $\left(-\frac{1}{2}, -2\right)$ . The graph does not pass through the origin, because $0$ is not in the domain, so the graph of an odd function need not contain $(0, 0)$ .

The algebraic test

Definition To decide whether a function $f$ with domain $D$ is even, odd or neither, follow these steps.

Check that $D$ is symmetric about $0$ . If it is not, then $f$ is neither even nor odd.
Simplify $f(-x)$ and compare it with $f(x)$ and with $-f(x)$ . If $f(-x) = f(x)$ for every $x$ in $D$ , the function is even. If $f(-x) = -f(x)$ for every $x$ in $D$ , it is odd.
To show that $f$ is not even, one input $a$ with $f(-a) \ne f(a)$ is enough. To show that $f$ is not odd, one input $b$ with $f(-b) \ne -f(b)$ is enough.

Numbers alone can only rule symmetry out. Agreement at a few inputs is not proof, because being even or odd is a statement about every $x$ in the domain.

In plain words Replace $x$ by $-x$ , simplify, and see whether you get back $f(x)$ , its negative, or something else. A proof needs the algebra, while a disproof needs just one input that fails.

Example Take the function

f(x) = \frac{x^3}{x^2 - 4}.

Its domain is all real $x$ except $-2$ and $2$ , a set that is symmetric about $0$ . For every $x$ in it

\begin{aligned} f(-x) &= \frac{(-x)^3}{(-x)^2 - 4} \\[0.8em] &= \frac{-x^3}{x^2 - 4} \\[0.8em] &= -f(x), \end{aligned}

so $f$ is odd.

Neither even nor odd

Definition A function is neither even nor odd if it is not even and it is not odd. That happens when its domain is not symmetric about $0$ , or when the domain is symmetric but it contains inputs $a$ and $b$ , possibly equal, with $f(-a) \ne f(a)$ and $f(-b) \ne -f(b)$ . Most functions are neither, and “not even” does not mean “odd”.

In plain words Even and odd are special properties, not two boxes that every function falls into. A function can fail both tests.

Example For $f(x) = x^2 + x$ we get $f(1) = 2$ and $f(-1) = 0$ . Since $0 \ne 2$ , the function is not even, and since $0 \ne -2$ , it is not odd. So $f$ is neither. The function $\sqrt{x}$ is also neither, without any computing, because its domain $[0, \infty)$ is not symmetric.

The zero function on a symmetric domain

Definition Let $D$ be a set that is symmetric about $0$ . A function with domain $D$ is both even and odd exactly when it is the zero function on $D$ , that is, $f(x) = 0$ for every $x \in D$ . If $f$ is both, then for every $x$ in $D$

f(x) = f(-x) = -f(x),

so $2f(x) = 0$ and $f(x) = 0$ . Conversely, the zero function satisfies both conditions, since $0 = 0$ and $0 = -0$ .

In plain words The only graph of a function that is both its own mirror image across the $y$ -axis and its own half-turn image is a flat graph lying on the $x$ -axis. On each symmetric domain there is exactly one such function.

Example The function $f(x) = \sqrt{x^2} - \lvert x \rvert$ is defined for every real $x$ , and since $\sqrt{x^2} = \lvert x \rvert$ , it equals $0$ everywhere. So it is both even and odd. By contrast, the zero function with domain $[0, 1]$ is neither, because that domain is not symmetric about $0$ .

Check yourself

Classify each of $f(x) = x^5 - 2x^3$ , $g(x) = x^2 - 3\lvert x \rvert$ and $h(x) = x^3 + 1$ as even, odd or neither.

Show the answer

All three have domain $\mathbb{R}$ . For the first, $f(-x) = -x^5 + 2x^3 = -f(x)$ , so $f$ is odd. For the second,

\begin{aligned} g(-x) &= (-x)^2 - 3\lvert -x \rvert \\[0.4em] &= x^2 - 3\lvert x \rvert \\[0.4em] &= g(x), \end{aligned}

so $g$ is even. For the third, $h(1) = 2$ and $h(-1) = 0$ , and since $0 \ne 2$ and $0 \ne -2$ , the function $h$ is neither.

An odd function $f$ has $0$ in its domain, and its graph passes through $(2, -5)$ . Name two more points on the graph.

Show the answer

Oddness gives $f(-2) = -f(2) = 5$ , so $(-2, 5)$ is on the graph. Since $0$ is in the domain, $f(0) = 0$ , so the graph also contains $(0, 0)$ .

A student notices that $f(x) = x^3 + x^2 - x$ satisfies $f(-1) = f(1)$ and concludes that $f$ is even. Is that right?

Show the answer

No. Both values are $1$ , but one matching pair proves nothing. At $2$ the values differ, with $f(2) = 10$ and $f(-2) = -2$ , so $f$ is not even. It is not odd either, since $f(-1) = 1$ while $-f(1) = -1$ . The function is neither.

Even, Odd, and Symmetry

Key terms

Symmetric domain

Even function

Symmetry about the y-axis

Odd function

Symmetry about the origin

The algebraic test

Neither even nor odd

The zero function on a symmetric domain

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus

Resources for this lesson

Function transformation explorer