Even and Odd Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 1.1: “Review of Functions” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Same Value, Opposite Sign, or Neither?
For $f(x) = x^2$: compare $f(3)$ and $f(-3)$. What is the relationship?
For $g(x) = x^3$: compare $g(3)$ and $g(-3)$. What is the relationship?
For $h(x) = x^2 + x$: compare $h(3)$ and $h(-3)$. What is the relationship?
Predict: which of these has a graph that is symmetric about the $y$-axis? About the origin?
Quick Reference
Even function. $f(-x) = f(x)$ for all $x$ in the domain.
- Algebraic test: substitute $-x$ and simplify; if result equals $f(x)$, it is even.
- Geometric meaning: the graph is symmetric about the $y$-axis.
Odd function. $f(-x) = -f(x)$ for all $x$ in the domain.
- Algebraic test: substitute $-x$ and simplify; if result equals $-f(x)$, it is odd.
- Geometric meaning: the graph is symmetric about the origin (180° rotational symmetry).
Neither. Most functions are neither even nor odd.
Key Concepts
1. The Algebraic Test
Example 1. Is $f(x) = x^4 - 3x^2 + 1$ even, odd, or neither?
$f(-x) = (-x)^4 - 3(-x)^2 + 1 = x^4 - 3x^2 + 1 = f(x)$.
$f(-x) = f(x)$: even. The graph is symmetric about the $y$-axis.
Example 2. Is $g(x) = 5x^3 - 2x$ even, odd, or neither?
$g(-x) = 5(-x)^3 - 2(-x) = -5x^3 + 2x = -(5x^3 - 2x) = -g(x)$.
$g(-x) = -g(x)$: odd. The graph has rotational symmetry about the origin.
Example 3. Is $h(x) = x^2 + x + 1$ even, odd, or neither?
$h(-x) = x^2 - x + 1$.
$h(-x) \neq h(x)$ (the $-x$ term differs). $h(-x) \neq -h(x)$ (would need $x^2 - x + 1 = -x^2 - x - 1$, false). Neither.
2. Two Representations: Symmetry
Even functions. Fold the graph along the $y$-axis: the two halves match. Examples: $y = \cos x$, $y = x^2$, $y = |x|$.
Odd functions. Rotate the graph 180° about the origin: it maps onto itself. Examples: $y = \sin x$, $y = x^3$, $y = x$.
Table: key examples:
| Function | $f(-x) = ?$ | Type |
|---|---|---|
| $x^2$ | $x^2 = f(x)$ | Even |
| $x^3$ | $-x^3 = -f(x)$ | Odd |
| $\cos x$ | $\cos x = f(x)$ | Even |
| $\sin x$ | $-\sin x = -f(x)$ | Odd |
| $e^x$ | $e^{-x} \neq \pm e^x$ | Neither |
| $x^2 + x$ | $x^2 - x \neq \pm(x^2 + x)$ | Neither |
Translation prompt. Even functions have $f(0)$ well-defined from symmetry: $f(0) = f(-0) = f(0)$ (tautology, no constraint). Odd functions with $0$ in the domain must satisfy $f(0) = -f(0)$, so $f(0) = 0$. Check: $\sin(0) = 0$ ✓; $x^3$ at 0 is 0 ✓. A non-zero value at the origin disqualifies an odd function.
3. Products and Sums
- Even $\times$ even = even.
- Odd $\times$ odd = even.
- Even $\times$ odd = odd.
- Sum of two even = even. Sum of two odd = odd.
- Sum of even and odd = neither (in general).
Example 4. Is $x^2 \sin x$ even, odd, or neither?
$x^2$ is even; $\sin x$ is odd. Product of even and odd: odd.
Check: $(-x)^2\sin(-x) = x^2(-\sin x) = -(x^2 \sin x)$. Odd. Confirmed.
4. Ask Why: Why Does an Odd Function Vanish at 0?
If $f(-x) = -f(x)$ for all $x$ in the domain, and if $0$ is in the domain, then:
$f(-0) = -f(0) \Rightarrow f(0) = -f(0) \Rightarrow 2f(0) = 0 \Rightarrow f(0) = 0$.
This is a necessary consequence of oddness. It is also useful: if you compute $f(0) \neq 0$ for a proposed odd function, you have found an error.
Named Misconception: iconic-graph
Students sometimes think “symmetric about the $y$-axis” means the graph looks like a specific shape (like a parabola). In fact, any function satisfying $f(-x) = f(x)$ has $y$-axis symmetry, regardless of its shape. The function $|x|$, the constant function 5, and $\cos x$ are all even -- they look very different.
Similarly, “symmetric about the origin” does not mean “looks like a cubic.” $\tan x$, $x^5$, and any odd polynomial all have this symmetry.
The symmetry is a property of the function-value relationship, not a visual template.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Forgetting to test both conditions | Checking $f(-x) \neq f(x)$ and concluding “odd” | Must also check $f(-x) = -f(x)$; if neither holds, the function is neither |
| Assuming all functions are even or odd | “Every function must be one of the three types” | Most functions are neither |
| Odd function at 0 | Claiming $f(x) = x^2 + x$ is odd | $f(0) = 0$ is necessary but not sufficient for oddness; check $f(-x) = -f(x)$ everywhere |
Leveled Practice
Level 1 -- Classify Each Function
Problem 1. Classify as even, odd, or neither: (a) $x^4 - x^2$, (b) $x^3 + x$, (c) $x^2 + 1$, (d) $x + 1$.
Show answer
(a) $f(-x) = x^4 - x^2 = f(x)$. Even.
(b) $g(-x) = -x^3 - x = -(x^3+x) = -g(x)$. Odd.
(c) $h(-x) = x^2 + 1 = h(x)$. Even.
(d) $k(-x) = -x+1$. $k(-x) \neq k(x)$ and $k(-x) \neq -k(x) = x - 1$. Neither.
Problem 2. If $f$ is odd and $g$ is even, classify $f(x)^2$, $g(f(x))$, and $f(x) + g(x)$.
Show answer
$f(x)^2$: $f(-x)^2 = (-f(x))^2 = f(x)^2$. Even.
$g(f(x))$: $g(f(-x)) = g(-f(x)) = g(f(x))$ (since $g$ is even). Even.
$f(x) + g(x)$: $(f+g)(-x) = f(-x) + g(-x) = -f(x) + g(x)$. Not equal to $f(x)+g(x)$ or $-(f(x)+g(x))$ in general. Neither.
Level 2 -- Using Symmetry
Problem 3. Suppose $f$ is even and $f(2) = 5$. What is $f(-2)$?
Show answer
$f(-2) = f(2) = 5$.
Problem 4. Show that every function $f$ can be written as the sum of an even function and an odd function: $f(x) = E(x) + O(x)$.
Show answer
Define $E(x) = \dfrac{f(x)+f(-x)}{2}$ and $O(x) = \dfrac{f(x)-f(-x)}{2}$.
$E(-x) = \frac{f(-x)+f(x)}{2} = E(x)$: even.
$O(-x) = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -O(x)$: odd.
$E(x) + O(x) = \frac{f(x)+f(-x)+f(x)-f(-x)}{2} = f(x)$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 5 (Extension).
(a) (Floor) Sketch or describe a function that is both even and odd. Is such a function possible?
(b) (Mid) Let $f$ be any function defined on a domain symmetric about 0 (i.e., if $x$ is in the domain, so is $-x$). Using the decomposition in Problem 4, prove that the decomposition $f = E + O$ is unique.
(c) (Ceiling) Show that the only function that is both even and odd (with a symmetric domain) is the zero function $f(x) = 0$.
Show answer
(a) A function both even ($f(-x) = f(x)$) and odd ($f(-x) = -f(x)$) satisfies $f(x) = -f(x)$, so $2f(x) = 0$, $f(x) = 0$. Only the zero function.
(b) Suppose $f = E_1 + O_1 = E_2 + O_2$ with $E_1, E_2$ even and $O_1, O_2$ odd. Then $E_1 - E_2 = O_2 - O_1$. Left side is even; right side is odd. A function that is both even and odd is zero (part c). So $E_1 = E_2$ and $O_1 = O_2$. Unique.
(c) If $f$ is both even and odd: for all $x$, $f(-x) = f(x)$ and $f(-x) = -f(x)$. So $f(x) = -f(x)$, giving $f(x) = 0$ for all $x$.
Common Misconceptions
even and odd refer to whether the exponents in the formula are even or odd. The names come from the behavior of the simplest power functions ($x^2$ is even, $x^3$ is odd), but the definition is algebraic: $f(-x) = f(x)$ for even, $f(-x) = -f(x)$ for odd. The function $f(x) = \cos x$ is even despite having no polynomial exponents. The function $g(x) = x^2 + x$ has only even and odd exponents but is neither even nor odd because the mixed terms break both symmetry conditions.
$f(-x) = -f(x)$ means the graph is “flipped.” The odd condition $f(-x) = -f(x)$ means the graph has 180-degree rotational symmetry about the origin, not a simple reflection across an axis. Reflecting across the $y$-axis (replacing $x$ with $-x$) produces the same graph only for even functions. For odd functions, the reflection across the $y$-axis is the same as a reflection across the $x$-axis, which together is a rotation.
Mastery Checklist
Mental Model
Even functions “ignore the sign” of the input: $f(3) = f(-3)$. The positive and negative sides are mirror images. Odd functions “flip the sign” of the output when the input sign flips: $f(-3) = -f(3)$. A 180° rotation maps the graph onto itself.
Most functions are neither even nor odd -- they have no such symmetry. But when symmetry is present, it halves the work: knowing $f(x)$ for $x > 0$ gives you $f(x)$ for $x < 0$ for free.
Connections
Within MATH161
- Integration (MATH162): $\int_{-a}^{a} f(x)\,dx = 0$ if $f$ is odd, and $= 2\int_0^a f(x)\,dx$ if $f$ is even. Symmetry drastically simplifies definite integrals over symmetric intervals.
- Taylor series: The Taylor series of an even function has only even-degree terms; the series of an odd function has only odd-degree terms. This explains the structure of the series for $\cos x$ and $\sin x$.