Functions
Even and Odd Functions
Two symmetry conditions, how to test for them in one substitution, and why most functions are neither.
What this page covers
- Test a function for evenness and oddness
- Connect each condition to a geometric symmetry
- Recognise that most functions are neither
- Use the parity of exponents as a quick predictor
The two conditions
The words for all x carry weight. A function that happens to satisfy the condition at one or two points is not even; the identity must hold throughout the domain.
| Condition | Symmetry | Source wording |
|---|---|---|
| f(-x) = f(x) | About the y-axis | Mirror symmetry about the y-axis |
| f(-x) = -f(x) | About the origin | Rotational symmetry about the origin |
Mirror symmetry means folding the plane along the y-axis leaves the graph unchanged. Rotational symmetry means turning it through half a revolution about the origin does.
The test
- Compute f(-x) by substituting -x everywhere x appears.
- Simplify, using (-x)n = xn for even n and (-x)n = -xn for odd n.
- Compare with f(x). Identical means even.
- Compare with -f(x). Identical means odd.
- Neither? Then the function is neither, which is the usual case.
(-x)2 = x2, not -x2. Squaring a negative gives a positive, and the bracket is what makes that visible. Writing -x2 instead means -(x2), which is a different quantity.
Even functions, worked
The source's examples
f(x) = x2 + 2:
g(x) = x4 + 3x2:
g(x) = 3x6 - x2 + 4:
Every exponent in all three is even, counting the constant as x0. That is the pattern.
Odd functions, worked
The source's examples
F(x) = x3:
G(x) = 5x5 - 3x:
The factoring step at the end is what makes the comparison visible. -5x5 + 3x does not obviously equal -G(x) until the minus is taken out.
An odd function defined at zero must satisfy f(0) = -f(0), hence f(0) = 0. Every odd function therefore passes through the origin, which is a quick sanity check: a function with a non-zero constant term cannot be odd.
Neither
The source's mixed case
f(x) = 2x2 - 3x + 5:
| Quantity | Value |
|---|---|
| f(x) | 2x2 - 3x + 5 |
| f(-x) | 2x2 + 3x + 5 |
| -f(x) | -2x2 + 3x - 5 |
f(-x) matches neither. The source concludes: neither even nor odd. The -3x term has odd parity while 2x2 and 5 have even parity, and the mixture destroys both symmetries.
The source poses one more, y = x3 - x1/5, and marks it odd? check. Both exponents are odd, so the answer is yes: (-x)3 = -x3 and (-x)1/5 = -x1/5, giving -(x3 - x1/5).
The parity shortcut
All exponents even (with a constant counted as x0) gives an even function. All exponents odd gives an odd one. A mixture gives neither.
| Function | Exponents | Verdict |
|---|---|---|
| x4 + 3x2 | 4, 2 — all even | Even |
| 3x6 - x2 + 4 | 6, 2, 0 — all even | Even |
| 5x5 - 3x | 5, 1 — all odd | Odd |
| 2x2 - 3x + 5 | 2, 1, 0 — mixed | Neither |
| x3 + 1 | 3, 0 — mixed | Neither |
It says nothing about sin x, cos x, |x| or 1x. Those must be tested directly — and they are worth knowing: cos x and |x| are even, while sin x, tan x and 1x are odd.
Why the classification is useful
Halves the sketching
One half of the graph determines the other by reflection or half-turn, exactly as in the source's symmetry tests.
Predicts intercepts
An odd function passes through the origin. An even function has a horizontal tangent there if it is smooth.
Simplifies later work
Integrals of odd functions over symmetric intervals vanish; those of even functions double. This is where the classification pays off most.
Checks trigonometric identities
cos(-t) = cos t and sin(-t) = -sin t are the parity statements, and they are used constantly in deriving the addition formulae.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| (-x)2 = -x2 | = x2 | At x = 3: 9, not -9 |
| (-x)3 = x3 | = -x3 | At x = 2: -8 |
| Concluding odd without factoring out the minus | Compare with -f(x) explicitly | -5x5 + 3x = -(5x5 - 3x) |
| Assuming every function is one or the other | Most are neither | 2x2 - 3x + 5 |
| Applying the parity shortcut to sin x | It is for polynomials | Test directly |
| Calling x3 + 1 odd | Neither | An odd function must satisfy f(0) = 0 |
Frequently asked questions
Can a function be both even and odd?
Only f(x) = 0. Both conditions together force f(x) = -f(x), hence f(x) = 0 for every x.
What if it is neither?
That is the usual case, and it is not a failure. The source works f(x) = 2x2 - 3x + 5 and concludes 'neither even nor odd'. Most functions with mixed-parity terms are neither.
Is there a shortcut?
For a polynomial, yes. All-even exponents (counting a constant as x0) means even; all-odd exponents means odd; a mixture means neither. It is only a shortcut for polynomials, though.
Why does it matter?
It halves the work of sketching, since one half of the graph determines the other. It also simplifies integration over symmetric intervals, which is where the classification earns its keep in later work.
Source. Handwritten teaching notes, Week 5, pages 3-5.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.