In short

A function is even if f(-x) = f(x) for all x in its domain — its graph is symmetric about the y-axis. A function is odd if f(-x) = -f(x) — its graph is symmetric about the origin (180° rotation). Most functions are neither. The classification matters because even/odd symmetry cuts computation in half: integrals, series expansions, and equation-solving all simplify when symmetry is present.

Stand on a flat road and look at a distant cell tower. The two guy-wires anchoring the tower to the ground — one running left, one running right — form a symmetric V-shape. If you sketched the height of the wire as a function of horizontal distance from the tower, you would get something like f(x) = |x|: the same shape on the left as on the right, a perfect mirror image across the vertical centre line.

Now think of a different shape. A motorcycle ramp at a stunt show curves upward to the right and downward to the left — like the graph of f(x) = x^3. There is no left-right mirror symmetry here. But there is a different kind of symmetry: if you rotate the entire picture 180° around the centre point, it looks exactly the same. The upward curve on the right becomes the downward curve on the left, and vice versa.

These two types of symmetry — mirror symmetry about the y-axis, and rotational symmetry about the origin — are the defining properties of even and odd functions.

The definitions

Even function

A function f is even if its domain is symmetric about 0 and

f(-x) = f(x) \quad \text{for all } x \text{ in the domain}

Geometrically: the graph is unchanged when reflected across the y-axis.

Odd function

A function f is odd if its domain is symmetric about 0 and

f(-x) = -f(x) \quad \text{for all } x \text{ in the domain}

Geometrically: the graph is unchanged when rotated 180° about the origin.

"Domain symmetric about 0" means: whenever x is in the domain, -x is also in the domain. The function f(x) = \sqrt{x}, defined only for x \ge 0, cannot be even or odd — the domain [0, \infty) is not symmetric about 0.

Side-by-side comparison of an even function and an odd functionTwo graphs. The left graph shows y equals x squared, a parabola symmetric about the y-axis, illustrating an even function. The right graph shows y equals x cubed, an S-shaped curve symmetric about the origin, illustrating an odd function. even: y = x² x y −a a f(−a) = f(a) odd: y = x³ x y −a a f(−a) = −f(a)
Left: the parabola $y = x^2$ is even — the left half mirrors the right half across the $y$-axis. Right: the cubic $y = x^3$ is odd — rotating the graph $180°$ about the origin maps it onto itself. The dashed lines show that corresponding points have equal $y$-values (even) or opposite $y$-values (odd).

Testing a function: the three-step method

Given a function f(x), test whether it is even, odd, or neither:

  1. Check the domain. Is it symmetric about 0? If not, stop — the function is neither even nor odd.
  2. Compute f(-x). Replace every x with -x and simplify.
  3. Compare. If f(-x) = f(x), the function is even. If f(-x) = -f(x), it is odd. If neither, it is neither.

Examples of the test

f(x) = x^4 - 3x^2 + 7. The domain is \mathbb{R}, symmetric about 0. Compute:

f(-x) = (-x)^4 - 3(-x)^2 + 7 = x^4 - 3x^2 + 7 = f(x)

Since f(-x) = f(x), this is even.

f(x) = x^5 + 2x. The domain is \mathbb{R}. Compute:

f(-x) = (-x)^5 + 2(-x) = -x^5 - 2x = -(x^5 + 2x) = -f(x)

Since f(-x) = -f(x), this is odd.

f(x) = x^2 + x. The domain is \mathbb{R}. Compute:

f(-x) = (-x)^2 + (-x) = x^2 - x

Is this f(x) = x^2 + x? No. Is this -f(x) = -x^2 - x? No. So the function is neither even nor odd.

The pattern for polynomials: a polynomial with only even powers of x (including the constant term, which is x^0) is even. A polynomial with only odd powers is odd. A polynomial with a mix of even and odd powers is neither.

Graph of y equals x squared plus x showing it is neither even nor oddA parabola shifted off-centre. It is not symmetric about the y-axis (so not even) and not symmetric about the origin (so not odd). The vertex is at x equals negative one half. x y 1 −1 2 vertex (−½, −¼) y = x² + x not symmetric about y-axis or origin
The parabola $y = x^2 + x$ has its vertex at $(-\frac{1}{2}, -\frac{1}{4})$. The graph is not symmetric about the $y$-axis (not even) and not symmetric about the origin (not odd). Mixing even and odd powers breaks both symmetries.

Beyond polynomials

The test works for any function:

The only function that is both even and odd

Is any function both even and odd at the same time? If f(-x) = f(x) and f(-x) = -f(x), then f(x) = -f(x), which forces 2f(x) = 0, so f(x) = 0 for all x. The zero function f(x) = 0 is the only function that is both even and odd.

Also worth noting: if f is odd, then setting x = 0 gives f(-0) = -f(0), i.e., f(0) = -f(0), so f(0) = 0. Every odd function (whose domain includes 0) must pass through the origin.

An odd function must pass through the originThe graph of an odd function passing through the origin. The curve rises to the right and dips to the left in a rotationally symmetric pattern. The origin is marked with a red dot and a label explaining that f of 0 must equal 0. x y f(0) = 0 every odd function passes through the origin
Any odd function whose domain includes $0$ must satisfy $f(0) = 0$. The origin is always on the graph. If you find that $f(0) \neq 0$, the function is immediately not odd — no further checking needed.

Properties under operations

What happens when you add, subtract, multiply, or compose even and odd functions? The table below summarises the results. Let E stand for even and O for odd.

Operation Result
E + E Even
O + O Odd
E + O Neither (in general)
E \cdot E Even
O \cdot O Even
E \cdot O Odd
E \circ E Even
O \circ O Odd
E \circ O Even
O \circ E Even

Each of these can be verified directly from the definitions. Take the product rule "O \cdot O is even" as an example. Let f and g both be odd. Then

(f \cdot g)(-x) = f(-x) \cdot g(-x) = (-f(x)) \cdot (-g(x)) = f(x) \cdot g(x) = (f \cdot g)(x)

The two negatives cancel, giving an even product. The same pattern — replacing -x, using the symmetry condition, simplifying — proves every row in the table.

And "E \cdot O is odd": let f be even and g be odd. Then

(f \cdot g)(-x) = f(-x) \cdot g(-x) = f(x) \cdot (-g(x)) = -(f(x) \cdot g(x)) = -(f \cdot g)(x)

One negative survives, making the product odd.

Every function can be split into even and odd parts

Here is a surprising and beautiful fact. Take any function f whose domain is symmetric about 0. Define:

f_e(x) = \frac{f(x) + f(-x)}{2}, \qquad f_o(x) = \frac{f(x) - f(-x)}{2}

Then f_e is even, f_o is odd, and f(x) = f_e(x) + f_o(x).

Check that f_e is even: f_e(-x) = \frac{f(-x) + f(x)}{2} = f_e(x).

Check that f_o is odd: f_o(-x) = \frac{f(-x) - f(x)}{2} = -\frac{f(x) - f(-x)}{2} = -f_o(x).

Check the sum: f_e(x) + f_o(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2} = f(x).

Example. Take f(x) = e^x.

f_e(x) = \frac{e^x + e^{-x}}{2} = \cosh x, \qquad f_o(x) = \frac{e^x - e^{-x}}{2} = \sinh x

The exponential function is neither even nor odd, but it splits cleanly into hyperbolic cosine (even) and hyperbolic sine (odd). This decomposition appears throughout physics and engineering.

Decomposition of e to the x into its even part cosh x and odd part sinh xThree curves on one set of axes. The exponential function e to the x rises steeply. Its even part cosh x is a symmetric U-shape. Its odd part sinh x is an odd S-shape. The three curves pass through the point (0,1) for the exponential, (0,1) for cosh, and (0,0) for sinh. x y 1 2 −2 eˣ cosh x sinh x
The exponential $e^x$ (solid black) decomposes into its even part $\cosh x$ (solid red, symmetric U-shape) and its odd part $\sinh x$ (dashed, passing through the origin). At every $x$, $e^x = \cosh x + \sinh x$.

Two worked examples

Example 1: Determine whether $f(x) = \frac{x^2}{1 + |x|}$ is even, odd, or neither

Step 1. Check the domain. The denominator is 1 + |x|, which is always positive (since |x| \ge 0). So the domain is all of \mathbb{R}, which is symmetric about 0.

Why: a domain check is always the first step. If the domain is not symmetric, the question is over immediately.

Step 2. Compute f(-x).

f(-x) = \frac{(-x)^2}{1 + |-x|} = \frac{x^2}{1 + |x|}

Why: (-x)^2 = x^2 and |-x| = |x|, so every piece is unchanged.

Step 3. Compare. f(-x) = \frac{x^2}{1 + |x|} = f(x).

Why: the expressions are identical, confirming even symmetry.

Step 4. The function is even.

Result: f(x) = \frac{x^2}{1 + |x|} is even.

Graph of x squared over 1 plus absolute value of x showing y-axis symmetryA smooth curve symmetric about the y-axis. It passes through the origin, rises on both sides, and eventually behaves like a straight line for large x values. The left and right halves are mirror images. x y 1 2 −1 −2 f(−1) = ½ f(1) = ½ y = x²/(1+|x|)
The graph of $f(x) = \frac{x^2}{1 + |x|}$ is symmetric about the $y$-axis. The marked points at $x = \pm 1$ both give $y = \frac{1}{2}$, confirming the mirror symmetry.

The graph tells the same story as the algebra. At x = -1, f(-1) = \frac{1}{2}. At x = 1, f(1) = \frac{1}{2}. Every pair of opposite inputs gives the same output — the hallmark of an even function.

Example 2: Determine whether $f(x) = x^3 - \sin x$ is even, odd, or neither

Step 1. The domain is \mathbb{R}, symmetric about 0.

Why: both x^3 and \sin x are defined for all real numbers.

Step 2. Compute f(-x).

f(-x) = (-x)^3 - \sin(-x) = -x^3 - (-\sin x) = -x^3 + \sin x

Why: (-x)^3 = -x^3 (cube preserves the sign change) and \sin(-x) = -\sin x (sine is odd).

Step 3. Compare with -f(x).

-f(x) = -(x^3 - \sin x) = -x^3 + \sin x

Why: distribute the negative sign across both terms.

Step 4. Since f(-x) = -x^3 + \sin x = -f(x), the function is odd.

Result: f(x) = x^3 - \sin x is odd.

Graph of x cubed minus sin x showing rotational symmetry about the originA curve through the origin that rises steeply to the right and falls steeply to the left, with slight undulations from the sine term. The curve has 180-degree rotational symmetry about the origin. x y 1 −1 2 −2 f(0) = 0 f(1) ≈ 0.16 f(−1) ≈ −0.16 y = x³ − sin x
The graph of $y = x^3 - \sin x$ passes through the origin and has 180° rotational symmetry. The values at $x = 1$ and $x = -1$ are equal in magnitude but opposite in sign: $f(1) = 1 - \sin 1 \approx 0.16$ and $f(-1) = -1 + \sin 1 \approx -0.16$.

Both x^3 (odd) and \sin x (odd) are odd functions. The rule "O - O is odd" (which follows from "O + O is odd" since subtraction is addition of a negative, and the negative of an odd function is odd) correctly predicts the result. The graph confirms it: the curve through the origin has the characteristic 180° rotational symmetry of an odd function.

Common confusions

Going deeper

If you can test any function for even/odd symmetry, apply the operation rules, and decompose a function into its even and odd parts, you have what you need for the next chapter. The material below explores extensions and applications.

Even and odd extensions

Suppose a function g is defined only for x \ge 0. You can extend it to all of \mathbb{R} in two natural ways:

Even extension:

f(x) = \begin{cases} g(x) & \text{if } x \ge 0 \\ g(-x) & \text{if } x < 0 \end{cases}

This makes f even by copying the right-half graph onto the left half as a mirror image.

Odd extension:

f(x) = \begin{cases} g(x) & \text{if } x \ge 0 \\ -g(-x) & \text{if } x < 0 \end{cases}

This makes f odd by copying the right half, flipping it vertically, and placing it on the left. For the odd extension to be continuous at x = 0, you need g(0) = 0.

Even and odd extensions of a function defined on the positive halfThree panels. Top: the original function g defined for x greater than or equal to 0, a curve rising from the origin. Middle: the even extension, which mirrors the curve across the y-axis. Bottom: the odd extension, which reflects the curve through the origin. original g(x), x ≥ 0 even extension odd extension must pass through origin
Top: a function $g$ defined only for $x \ge 0$. Middle: its even extension mirrors the graph across the $y$-axis. Bottom: its odd extension rotates the graph $180°$ through the origin — this requires $g(0) = 0$ for continuity.

Even and odd extensions are central in Fourier analysis. When you expand a function defined on [0, L] in terms of sines (odd functions) or cosines (even functions), you are implicitly working with the odd or even extension of the original function.

An interactive test

Drag the slider to blend between f(x) = x^2 (even) and f(x) = x^3 (odd). At the extremes, the symmetry is clear; in between, the function is neither.

Interactive blend between x squared and x cubedA graph that can be smoothly adjusted between the even function x squared and the odd function x cubed using a draggable point. At one extreme the curve is a symmetric parabola; at the other it is an S-shaped cubic. x y 1 −1 drag to blend
At blend $= 0$, the curve is $y = x^2$ (even, symmetric about the $y$-axis). At blend $= 1$, it is $y = x^3$ (odd, symmetric about the origin). In between, the function $(1-t)x^2 + tx^3$ is neither even nor odd.

Where this leads next

Symmetry is one of the most powerful ideas in mathematics. Even and odd functions are the simplest case — symmetry of a function under the transformation x \mapsto -x. The next articles extend this idea.