Even and Odd Functions

Some graphs are perfectly balanced: the left half is a mirror image of the right, or the graph looks the same after a half-turn. Functions with these symmetries are called even and odd. Spotting them saves work, because you only need to graph half the function and reflect or rotate to get the rest.

Key ideas

Even functions

A function is even if

$$f(-x) = f(x) \quad \text{for every } x \text{ in its domain}$$

Opposite inputs give the same output. The graph is symmetric about the $y$-axis: if $(a, b)$ is on the graph, so is $(-a, b)$.

Odd functions

A function is odd if

$$f(-x) = -f(x) \quad \text{for every } x \text{ in its domain}$$

Opposite inputs give opposite outputs. The graph has point symmetry about the origin: if $(a, b)$ is on the graph, so is $(-a, -b)$. Rotating the graph $180^\circ$ about the origin leaves it unchanged.

An even function is symmetric about the $y$-axis; an odd function is symmetric about the origin.

Most functions are neither even nor odd.

Testing algebraically

1. Replace every $x$ with $(-x)$ and simplify $f(-x)$.
2. Compare with $f(x)$ and with $-f(x)$.
   • $f(-x) = f(x)$: even.
   • $f(-x) = -f(x)$: odd.
   • Neither matches: neither. To show this convincingly, give one value of $x$ where each test fails.

The key facts: $(-x)^n = x^n$ when $n$ is even, and $(-x)^n = -x^n$ when $n$ is odd.

Even and odd polynomials

That’s where the names come from:

• A polynomial with only even powers of $x$ is even. (A constant counts as an even power, since $5 = 5x^0$.) Example: $3x^4 - x^2 + 7$.
• A polynomial with only odd powers of $x$ is odd. It has no constant term. Example: $x^5 - 2x^3 + x$.
• A polynomial with a mixture is neither. Example: $x^3 + x^2$.

Write the polynomial in expanded form before you decide: $x(x - 1)$ looks like it has one power, but it’s $x^2 - x$.

Other functions

The tests work for any function, not just polynomials:

• $\cos(-x) = \cos x$, so cosine is even. Its graph is symmetric about the $y$-axis.
• $\sin(-x) = -\sin x$, so sine is odd. Its graph is symmetric about the origin.
• $y = |x|$ is even, and $y = \dfrac{1}{x}$ is odd.
• $y = 2^x$ is neither: $2^{-1} = \tfrac{1}{2}$, which is not $2$ or $-2$.

These facts are true whether $x$ is measured in degrees or radians.

Useful consequences

• An odd function that is defined at $0$ passes through the origin, because $f(0) = -f(0)$ forces $f(0) = 0$.
• The $x$-intercepts of an even function come in pairs, $a$ and $-a$. So if an even function has a limited number of $x$-intercepts, that number is odd exactly when $0$ is one of them, and even when it isn’t.

Sums and products

If you combine even and odd functions, the result follows a pattern, much like multiplying positive and negative numbers:

Combination	Result
even $\times$ even, or odd $\times$ odd	even
even $\times$ odd	odd
even $+$ even	even
odd $+$ odd	odd
even $+$ odd (both nonzero)	neither

For example, if $f$ and $g$ are both odd and $h(x) = f(x)g(x)$, then

$$h(-x) = f(-x)\,g(-x) = \big(-f(x)\big)\big(-g(x)\big) = f(x)g(x) = h(x)$$

so $h$ is even. The same patterns hold for quotients: odd $\div$ odd is even, and so on.

Worked examples

Example 1: Polynomials by their powers

Is each function even, odd, or neither?

• (a) $f(x) = 3x^4 - x^2 + 7$
• (b) $g(x) = -2x^5 + x^3 - x$
• (c) $h(x) = x^3 + x^2$

Solution.

(a) Powers $4$, $2$, $0$: all even, so $f$ is even.

(b) Powers $5$, $3$, $1$: all odd, so $g$ is odd.

(c) Powers $3$ and $2$: a mixture, so $h$ is neither. Check with $x = 1$: $h(1) = 2$ and $h(-1) = -1 + 1 = 0$. That’s neither $h(1)$ nor $-h(1)$.

Example 2: The algebraic test

Determine whether each function is even, odd, or neither: (a) $f(x) = x(x^2 - 9)$ (b) $g(x) = (x - 1)^2$.

Solution.

(a) Replace $x$ with $-x$:

$$f(-x) = (-x)\big((-x)^2 - 9\big) = -x(x^2 - 9) = -f(x)$$

So $f$ is odd.

(b)

$$g(-x) = (-x - 1)^2 = (x + 1)^2$$

This isn’t the same as $(x - 1)^2$ or $-(x - 1)^2$. To be sure, test $x = 1$: $g(1) = 0$, but $g(-1) = 4$. Since $g(-1) \ne g(1)$ and $g(-1) \ne -g(1)$, $g$ is neither. (Its graph is a parabola with vertex $(1, 0)$, which isn’t symmetric about the $y$-axis.)

Example 3: Beyond polynomials

Is each function even, odd, or neither? (a) $f(x) = x^2 + \cos x$ (b) $g(x) = x + \sin x$ (c) $h(x) = x \sin x$

Solution. Use $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$.

(a) $f(-x) = (-x)^2 + \cos(-x) = x^2 + \cos x = f(x)$: even (even $+$ even).

(b) $g(-x) = -x + \sin(-x) = -x - \sin x = -(x + \sin x) = -g(x)$: odd (odd $+$ odd).

(c) $h(-x) = (-x)\sin(-x) = (-x)(-\sin x) = x \sin x = h(x)$: even (odd $\times$ odd).

Example 4: Using symmetry to build a polynomial

An even polynomial function of degree $4$ has zeros at $1$ and $3$, and $f(0) = 18$. Find $f(x)$.

Solution. Because $f$ is even, its zeros come in pairs: $\pm 1$ and $\pm 3$. That’s four zeros for a quartic, so (as in families of polynomials)

$$f(x) = k(x - 1)(x + 1)(x - 3)(x + 3) = k(x^2 - 1)(x^2 - 9)$$ $$f(0) = k(-1)(-9) = 9k = 18 \quad\Rightarrow\quad k = 2$$

$f(x) = 2(x^2 - 1)(x^2 - 9) = 2x^4 - 20x^2 + 18$. Check: only even powers, so it’s even. ✓

Common mistakes

Testing only one value and concluding “even” or “odd”. One pair like $f(2) = f(-2)$ doesn’t prove $f$ is even; $f(-x) = f(x)$ has to hold for every $x$. Use algebra to prove even or odd. (One value is enough to prove “neither”, though.)

Forgetting that a constant term is an even power. $f(x) = x^3 + 1$ is not odd: the $1$ is like $x^0$. Check: $f(-1) = 0$, but $-f(1) = -2$.

Deciding from the factored form. $f(x) = x(x - 1)$ has only odd-looking pieces, but it expands to $x^2 - x$, which is neither. Expand first, or use the algebraic test.

Thinking “odd degree” means “odd function”. $x^3 + x^2$ has odd degree but isn’t odd. Every power must be odd.

Mixing up the symmetries. Even means a mirror image in the $y$-axis. Odd means a half-turn about the origin, not a mirror image in the $x$-axis (which wouldn’t even be a function).

Practice

1. (Warm-up) Is each function even, odd, or neither? (a) $y = 5x^6 - 2x^2$ (b) $y = x^7 - 3x$ (c) $y = x^4 + x$ (d) $y = 4x^3 + 1$

Solution

(a) Even: powers $6$ and $2$.

(b) Odd: powers $7$ and $1$.

(c) Neither: powers $4$ and $1$ are mixed.

(d) Neither: the constant $1$ is an even power, mixed with the odd power $3$.

2. (Warm-up) Suppose $f(4) = -7$. Find $f(-4)$ if (a) $f$ is odd, (b) $f$ is even.

Solution

(a) $f(-4) = -f(4) = 7$.

(b) $f(-4) = f(4) = -7$.

3. (Warm-up) The points $(2, 5)$ and $(-3, 1)$ are on the graph of an odd function $f$. Name two more points that must be on the graph, and give $f(0)$.

Solution

$(-2, -5)$ and $(3, -1)$. Since an odd function defined at $0$ passes through the origin, $f(0) = 0$.

4. (Core) Show algebraically that $f(x) = -2x^5 + 3x^3$ is odd.

Solution

$$\begin{aligned} f(-x) &= -2(-x)^5 + 3(-x)^3 \\ &= -2(-x^5) + 3(-x^3) \\ &= 2x^5 - 3x^3 \\ &= -(-2x^5 + 3x^3) = -f(x) \end{aligned}$$

So $f$ is odd.

5. (Core) Is each function even, odd, or neither? (a) $f(x) = x^2(x - 2)(x + 2)$ (b) $g(x) = x(x - 2)(x + 2)$ (c) $h(x) = (x + 2)^2(x - 2)$

Solution

(a) $f(x) = x^2(x^2 - 4) = x^4 - 4x^2$: only even powers, so even.

(b) $g(x) = x(x^2 - 4) = x^3 - 4x$: only odd powers, so odd.

(c) $h(x) = (x^2 + 4x + 4)(x - 2) = x^3 + 2x^2 - 4x - 8$: mixed powers, so neither. (Check: $h(1) = 9(-1) = -9$, while $h(-1) = 1(-3) = -3$, which is neither $-9$ nor $9$.)

6. (Core) Is each function even, odd, or neither? (a) $f(x) = x^2 \sin x$ (b) $g(x) = \cos x + x^4$ (c) $h(x) = \sin x + \cos x$

Solution

(a) $f(-x) = (-x)^2 \sin(-x) = x^2(-\sin x) = -f(x)$: odd (even $\times$ odd).

(b) $g(-x) = \cos(-x) + (-x)^4 = \cos x + x^4 = g(x)$: even.

(c) $h(-x) = -\sin x + \cos x$. Test $x = 90^\circ$: $h(90^\circ) = 1 + 0 = 1$, but $h(-90^\circ) = -1 + 0 = -1$, so $h$ is not even. Test $x = 0$: $h(0) = 0 + 1 = 1 \ne 0$, so $h$ is not odd (an odd function would pass through the origin). Neither.

7. (Core) Suppose $f$ is even and $g$ is odd, with the same domain. Prove that $p(x) = f(x)g(x)$ is odd.

Solution

$$p(-x) = f(-x)\,g(-x) = f(x)\big(-g(x)\big) = -f(x)g(x) = -p(x)$$

Since $p(-x) = -p(x)$ for every $x$, $p$ is odd.

8. (Challenge) An odd polynomial function of degree $5$ has zeros at $0$, $2$, and $3$, and passes through $(1, 48)$. Find its equation.

Solution

Since $f$ is odd, the zeros $2$ and $3$ come with $-2$ and $-3$. That’s five zeros: $0, \pm 2, \pm 3$.

$$f(x) = kx(x - 2)(x + 2)(x - 3)(x + 3) = kx(x^2 - 4)(x^2 - 9)$$$$48 = k(1)(-3)(-8) = 24k \quad\Rightarrow\quad k = 2$$

$f(x) = 2x(x^2 - 4)(x^2 - 9) = 2x^5 - 26x^3 + 72x$. Check: only odd powers. ✓

9. (Challenge) Show that the only function that is both even and odd (on a domain of all real numbers) is $f(x) = 0$.

Solution

If $f$ is both even and odd, then for every $x$:

$$f(-x) = f(x) \quad \text{and} \quad f(-x) = -f(x)$$

So $f(x) = -f(x)$, which gives $2f(x) = 0$, so $f(x) = 0$ for every $x$. And $f(x) = 0$ does satisfy both tests, since $0 = 0$ and $0 = -0$.