Even, odd, or neither: test the whole formula

Explain it like I am five

Think of folding a paper drawing along its vertical center line. If the two halves match, the graph is Symmetric about the y-axis, and its function is called an Even function. Another drawing may match after you turn the paper halfway around the center point. That graph is Symmetric about the origin, and its function is called an Odd function. The Origin is the point (0, 0), where the axes cross. These names describe the whole graph, not whether an answer is an even or odd integer. To decide from a formula, compare what happens at opposite inputs. A function may be Neither even nor odd.

Even: f(−x) = f(x)

Odd: f(−x) = −f(x)

Neither: neither whole-formula identity holds

First check that the domain allows opposite inputs

Compare complete expressions and check the domain before assigning a symmetry name.

Reminder

Powers of negative numbers. Parentheses include the negative in the power: (−2 $) squared$ = 4 and (−2 $) cubed$ = −8. Pair negative factors to see which sign remains.
Distributing an outside minus. Negate the entire output: −( $x^{3}$ + 1) = − $x^{3}$ − 1. The constant changes sign too.
Domain symmetry. Opposite inputs must both be allowed. [−3, 3] has opposite pairs; [0, 3] permits 1 but excludes −1.

Why it works. Equal heights at opposite inputs give the paired points (x, f(x)) and (−x, f(x)), which are mirrors across the y-axis. Opposite heights at opposite inputs give (x, f(x)) and (−x, −f(x)), which match after a half turn around the origin. Both comparisons require opposite inputs to be allowed. Algebra lets you test all inputs at once: replacing every x by (−x) produces a formula that can be compared with f(x) and with the negative of its entire output.

RuleOn a domain symmetric about zero: even means f(−x) = f(x); odd means f(−x) = −f(x), for every allowed x.
If neither identity holds for all allowed inputs, the function is neither. The zero-valued function on a symmetric domain satisfies both identities.

The same idea, five ways

Say it

An even function gives the same height at opposite addresses. An odd function gives opposite heights at opposite addresses. Neither means both required pairings fail.

Write it

On a domain symmetric about zero, compare f(−x) with f(x) and with −f(x) for every allowed input.

In math

Even: f(−x) = f(x); graph points (u, v) and (−u, v).
Odd: f(−x) = −f(x); graph points (u, v) and (−u, −v).
Neither: neither identity holds for every allowed input, or the domain lacks opposite input pairs.
Both: f(x) = 0 on a symmetric domain.

Even matches a fold along the vertical axis. Odd matches a half turn around the origin.

See it

Even: (2, 4) ↔ (−2, 4)

Odd: (2, 8) ↔ (−2, −8)

Neither: h(1) = 2, h(−1) = 0

Check opposite inputs in the domain first

The two symmetry tests require different partner points at opposite inputs.

The same idea, other ways

As two actions on a drawing

For $x^{2}$ , folding across the y-axis pairs (2, 4) with (−2, 4). For $x^{3}$ , a half turn pairs (2, 8) with (−2, −8). The first action keeps the height; the second changes both coordinate signs.

The equal heights at opposite inputs show the even pairing on .

As three complete expressions

For h(x) = $x^{3}$ + 1, write h(−x) = − $x^{3}$ + 1 beside h(x) = $x^{3}$ + 1 and −h(x) = − $x^{3}$ − 1. It matches neither. Comparing complete expressions prevents a constant from being overlooked.

As three complete expressions

Even: f(−x) = f(x); graph points (u, v) and (−u, v).

Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.

As a partner requirement

Before comparing heights, ask whether the partner address exists. The formula $x^{2}$ on x ≥ 0 accepts 1 but rejects −1, so no algebra can supply the missing partner. The zero function on [−3, 3] has every partner, and zero agrees with its own opposite, so both tests hold.

As a partner requirement

Even: f(−x) = f(x); graph points (u, v) and (−u, v).

Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.

Classification	Algebra test	Paired graph points	Picture
Even function	f(−x) = f(x)	(x, y) and (−x, y)	Mirror across the y-axis
Odd function	f(−x) = −f(x)	(x, y) and (−x, −y)	Half turn around the origin
Neither even nor odd	Neither identity holds for every input	Each symmetry has at least one missing or mismatched required pairing.	Neither symmetry
Both	f(x) = 0 on a symmetric domain	(x, 0) and (−x, 0)	The graph matches both actions

.1Even function

Imagine two seats the same distance left and right of a stage, both at the same height. An Even function pairs opposite horizontal positions with equal heights. Folding its graph along the y-axis would line up the points. The name concerns this pairing, not whether the function values are even whole numbers. The Absolute value function is another familiar example because opposite inputs have the same distance from zero.

Test: f(−x) = f(x) for every input, on a domain symmetric about zero.
Graph description: Symmetric about the y-axis. A Horizontal reflection reproduces the same graph.
Even-power terms and constants form an even polynomial. This is a useful recognition aid, while substitution gives the proof.

Opposite inputs −2 and 2 on have the same height 4.

Worked exampleProve that is even

Classify f(x) = $x^{2}$ on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Opposite inputs −2 and 2 have the same height 4.

Every real input and its opposite are allowed.Squaring is defined for every real number, so the domain is symmetric about zero.
f(−x) = (−x $) squared$ = (−x)(−x) = $x^{2}$ = f(x).The two minus signs multiply to a plus.
The function is even. It is not odd: f(1) = 1 while −f(1) = −1, and f(−1) = 1.The even identity holds for every input, but the odd identity fails at input 1.

Answer

f(x) = is even and is not odd.

Check The graph has the matching points (−2, 4) and (2, 4), which are mirror images across the y-axis.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: is odd because its graph includes the origin.

Passing through the origin does not establish opposite heights. Inputs 1 and −1 both give 1, while oddness would require one output to be −1.

✓ Instead: f(x) = is even and is not odd.

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.2Odd function

Imagine a seesaw centered at the origin, with one dot right and up and another the same distance left and down. A half turn exchanges the two dots. An Odd function has that pairing throughout its graph. Opposite inputs produce opposite signed heights. Reflecting its graph across both axes gives the original graph again, because each coordinate has changed to its opposite.

Test: f(−x) = −f(x) for every input, on a domain symmetric about zero.
Graph description: Symmetric about the origin. The point (u, v) has a matching point (−u, −v).
If zero is in the domain of an odd function, f(0) = 0. This follows from f(0) = −f(0).
A symmetric domain need not contain zero. The Reciprocal function $1 over x$ is odd on the domain x ≠ 0.

A half turn around (0, 0) exchanges the two marked cubic points.

Worked exampleProve that the cubic function is odd

Classify f(x) = $x^{3}$ on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

The marked points match after a half turn around the origin.

The domain is all real numbers, so it includes opposite inputs.Every real number can be multiplied by itself three times.
f(−x) = (−x $) cubed$ = (−x)(−x)(−x) = − $x^{3}$ = −f(x).Two negative factors give a positive; multiplying by the third negative factor makes the product negative.
The function is odd and is not even.The odd identity holds for all inputs. The even identity fails because f(−1) = −1 differs from f(1) = 1.

Answer

The Cubic function f(x) = is odd and is not even.

Check The points (2, 8) and (−2, −8) are opposite in both coordinates, so a half turn maps one to the other.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: is even because the same shape appears on both sides of the origin.

The heights must match for evenness. Inputs 1 and −1 give 1 and −1, which are opposites. A half turn, not a vertical mirror, matches the curve.

✓ Instead: The Cubic function f(x) = is odd and is not even.

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.3Neither even nor odd

A picture can fail both the folding test and the half-turn test. That is what Neither even nor odd means. A graph may look familiar and still lose its symmetry when you move it. For example, lifting an odd graph up by a nonzero amount keeps its shape but changes the center of its half-turn symmetry. The definition of odd specifically requires the center to be the origin.

Neither means the even identity and the odd identity each fail for at least one allowed input, or the domain does not allow opposite inputs.
A mixture of nonzero even-power and odd-power polynomial terms is neither. A nonzero constant added to a nonzero odd polynomial also makes it neither.
Symmetry about a different line or point does not satisfy symmetry about the y-axis or the origin.

Lifting by 1 makes the heights at opposite inputs neither equal nor opposites.

Worked exampleA constant breaks the odd test

Classify h(x) = $x^{3}$ + 1 on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Opposite inputs give heights 0 and 2, which are neither equal nor opposites.

h(−x) = (−x $) cubed$ + 1 = − $x^{3}$ + 1.Only x is replaced; the constant 1 stays 1.
This differs from h(x) = $x^{3}$ + 1, so h is not even.The coefficient of $x^{3}$ has changed sign.
Compute −h(x) = −( $x^{3}$ + 1) = − $x^{3}$ − 1. This differs from h(−x), so h is not odd.Negating the whole output must also negate the constant.

Answer

h(x) = + 1 is neither even nor odd.

Check h(1) = 2 and h(−1) = 0. The outputs are neither equal nor opposites. Also h(0) = 1 rules out oddness because an odd function defined at zero must output zero there.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: + 1 is odd because its cubic term changes sign.

The constant stays +1 under input substitution, but the negative of the entire original output has constant −1. The two expressions disagree.

✓ Instead: h(x) = + 1 is neither even nor odd.

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.4Zero function and domain symmetry

Imagine drawing only along the horizontal axis, with a dot at each allowed address and every height zero. Zero is its own opposite. If the allowed addresses come in opposite pairs, that drawing matches both the fold and the half turn. The zero function can therefore be both even and odd. But choosing only addresses to the right removes the needed left-hand partners, even though the formula still says zero.

Domain symmetry: whenever x is allowed, −x must be allowed too. Intervals such as [−3, 3] are symmetric; [0, 3] is not.
On a symmetric domain, z(x) = 0 is both even and odd, because z(−x) = 0 = z(x) and z(−x) = 0 = −z(x).
Only a zero-valued function can be both: if f(x) = f(−x) and f(−x) = −f(x), then f(x) = −f(x), so 2f(x) = 0 and f(x) = 0 at every allowed input.
A nonzero Constant function on a symmetric domain is even and is not odd.

z(x) = 0 on [−3, 3]

(−2, 0) and (2, 0)

Mirror: (2, 0) → (−2, 0)

Half turn: (2, 0) → (−2, −0) = (−2, 0)

A zero-height graph on a symmetric domain survives both a reflection across the y-axis and a half turn.

Worked exampleCheck zero on a symmetric domain

Classify z(x) = 0 with domain [−3, 3].
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

On the stated interval [−3, 3], the zero graph is unchanged by either symmetry.

If −3 ≤ x ≤ 3, then −3 ≤ −x ≤ 3 too.Negating a number in this interval keeps it between the same two opposite endpoints.
z(−x) = 0 = z(x), so z is even.Every allowed input returns the same zero height.
z(−x) = 0 = −0 = −z(x), so z is also odd.Zero is its own negative.

Answer

z is both even and odd on [−3, 3].

Check Its points (2, 0) and (−2, 0) satisfy both required pairings, and the algebra establishes the same result for every input in the domain.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The formula z(x) = 0 makes the function both even and odd on domain [0, 3].

Input 1 lacks its opposite −1 in that domain. Both definitions also require a symmetric domain. The example uses [−3, 3], which does have opposite partners.

✓ Instead: z is both even and odd on [−3, 3].

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

Strategy: step by step

1. Check that every allowed input x has its opposite −x in the domain. A domain that fails this condition rules out both symmetries.
2. Replace every x in the formula with (−x), including x inside powers and multiplied terms.
3. Simplify. For an even power, the minus signs cancel in pairs. For an odd power, one minus sign remains.
4. Compare the entire result with f(x). If they match for every allowed input, the function is even.
5. Separately write −f(x), distributing its minus to every term. If f(−x) matches this expression for every allowed input, the function is odd. If only the even comparison succeeded, it is even. If neither comparison succeeded, it is neither. If both succeed, it is both; the zero function on a symmetric domain does this.

Strategy

Classify a function by checking both symmetries

Does every allowed input have its opposite in the Domain?

YesCompute f(−x), f(x), and −f(x).

NoClassify the function as neither; its graph lacks a required opposite-input partner.

↓

Does f(−x) equal f(x) for every allowed input?

YesRecord the even property, then check the odd property too.

NoThe function is not even. Continue to the odd test.

↓

Does f(−x) equal −f(x) for every allowed input?

YesRecord the odd property. If the even test also passed, the function is both and every output is zero.

NoIf the even test passed, classify it as even. If it also failed, classify it as neither.

1. Inspect the Domain to see whether every allowed x has an allowed −x. This establishes whether the symmetry tests can apply at all.
2. Replace every x by (−x), then simplify. This computes the height at the opposite address using the same rule.
3. Write −f(x), changing every term's sign. This gives the entire opposite output needed for the odd test.
4. Compare f(−x) with f(x) and with −f(x) as identities. Matching only the first means even; matching only the second means odd; matching neither means neither; matching both means both.
5. Check the result with opposite inputs, while keeping the identity as the proof. For $x^{3}$ + 2x, inputs 1 and −1 give 3 and −3, confirming the odd pairing.

Worked exampleClassify the three draft functions

On their full real domains, classify f(x) = $x^{3}$ + 2x, g(x) = x⁴ + 3 $x^{2}$ + 7, and h(x) = $x^{3}$ + 1.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

f(x) = + 2x: odd.

g(x) = x⁴ + 3 + 7: even.

h(x) = + 1: neither.

The card records the exact result of the worked coordinate or symmetry calculation.

All three domains are (−∞, ∞), so opposite inputs are allowed.These polynomials use sums, products, and whole-number powers, which are defined at every real number.
f(−x) = (−x $) cubed$ + 2(−x) = − $x^{3}$ − 2x = −( $x^{3}$ + 2x) = −f(x). Thus f is odd.Three negative factors leave a negative product, and multiplying −x by 2 keeps its minus. The result is the negative of the whole original expression.
g(−x) = (−x)⁴ + 3(−x $) squared$ + 7 = x⁴ + 3 $x^{2}$ + 7 = g(x). Thus g is even.Pairs of negative factors multiply to positives, and the constant 7 has no x to replace.
h(−x) = − $x^{3}$ + 1. It differs from h(x) = $x^{3}$ + 1 and from −h(x) = − $x^{3}$ − 1. Thus h is neither.The inside substitution changes the cubic term but keeps the constant. Negating the entire output changes both terms.

Answer

f(x) = $x^{3}$ + 2x: odd.
g(x) = x⁴ + 3 $x^{2}$ + 7: even.
h(x) = $x^{3}$ + 1: neither.

Check At opposite inputs 1 and −1: f gives 3 and −3; g gives 11 and 11; h gives 2 and 0. These point pairs match the classifications. The formula identities above establish the classifications for every input; these numerical checks only confirm them.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: an even power

Classify p(x) = $x^{2}$ on its full domain.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

The paired points at −3 and 3 both have height 9.

The full domain is all real numbers, which allows opposite inputs.A square is defined for any real number.
p(−x) = (−x $) squared$ = $x^{2}$ = p(x).A negative multiplied by a negative is positive.
p is even and not odd.The even identity holds for all x, while p(−1) = 1 differs from −p(1) = −1.

Answer

Even.

Check At −3 and 3 the outputs are both 9, matching the mirror-image point pairing.

Rung 2Rung 2: negate every occurrence of x

Classify f(x) = $x^{3}$ + 2x on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Odd.

The card records the exact result of the worked coordinate or symmetry calculation.

f(−x) = (−x $) cubed$ + 2(−x).Both occurrences of x must be replaced; changing only the cubic term would test a different function.
f(−x) = − $x^{3}$ − 2x = −( $x^{3}$ + 2x) = −f(x).The odd power and the linear term both change sign. Factoring out −1 negates the whole original expression.
The function is odd on its symmetric real domain.The identity f(−x) = −f(x) holds for every real x.

Answer

Odd.

Check f(2) = 8 + 4 = 12 and f(−2) = −8 − 4 = −12. Their points are exchanged by a half turn around the origin.

Rung 3Rung 3: even powers and a constant

Classify g(x) = x⁴ + 3 $x^{2}$ + 7 on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Even.

The card records the exact result of the worked coordinate or symmetry calculation.

g(−x) = (−x)⁴ + 3(−x $) squared$ + 7.Replace each x and keep the constant 7 unchanged.
g(−x) = x⁴ + 3 $x^{2}$ + 7 = g(x).Both even powers pair their minus signs, so those signs cancel. At x = 0 the powers are zero; they are nonnegative, not always positive.
The function is even and not odd.The even identity holds on its full symmetric domain. At zero, g(0) = 7, so it cannot equal −g(0) = −7 as oddness would require.

Answer

Even.

Check g(2) = 16 + 12 + 7 = 35 and g(−2) = 16 + 12 + 7 = 35.

Rung 4Rung 4: compare with the negative of the whole function

Classify h(x) = $x^{3}$ + 1 on all real inputs.
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Adding 1 keeps the constant positive during input substitution.

h(−x) = − $x^{3}$ + 1, while h(x) = $x^{3}$ + 1.The cubic term changes sign under substitution and the constant does not. These are not identical expressions.
−h(x) = − $x^{3}$ − 1, which also differs from h(−x).An outside minus must distribute to both the cubic term and the constant.
h is neither even nor odd.It satisfies neither whole-formula identity, even though its domain is symmetric.

Answer

Neither even nor odd.

Check h(2) = 9 and h(−2) = −7. The second value is neither 9 nor −9.

Rung 5Rung 5: the formula does not erase a domain restriction

A function has formula r(x) = $x^{2}$ but accepts only x ≥ 0. Is it even, odd, or neither?
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

The stated domain includes only the right half of .

The domain is [0, ∞), so 1 is allowed and −1 is not.The restriction is part of the function, not an optional detail of its formula.
The domain is not symmetric about zero.At least one allowed input lacks its opposite partner.
The function is neither even nor odd on this domain.Both definitions require the paired opposite input. Writing (−x $) squared$ = $x^{2}$ does not make a forbidden input allowed.

Answer

Neither, because the domain is not symmetric about zero.

Check The graph includes (1, 1) but no point at input −1. It therefore lacks both the mirror partner (−1, 1) and the half-turn partner (−1, −1).

Rung 6Rung 6: show which function can be both

Suppose f is both even and odd on a domain symmetric about zero. What must every output be?
This asks whether opposite allowed inputs always give equal heights, opposite heights, or neither pattern.

Every height is 0, so both required point pairings coincide.

Evenness gives f(−x) = f(x), and oddness gives f(−x) = −f(x).Both identities apply to every allowed input.
Set the two equal expressions equal: f(x) = −f(x). Add f(x) to both sides to get 2f(x) = 0.Both expressions equal f(−x). Adding the same amount to both sides preserves equality.
Divide by 2: f(x) = 0 at every allowed input.Zero divided by a nonzero number is zero.
Conversely, a zero output satisfies both tests: 0 = 0 and 0 = −0.This confirms the candidate actually meets both definitions on the symmetric domain.

Answer

Every output must be zero. The zero-valued function on a symmetric domain is both even and odd.

Check For z(x) = 0 on all real inputs, the graph is the x-axis. Reflecting it across the y-axis or rotating it halfway around the origin reproduces the same graph.

Rung 7Rung 7: read three complete graphs without starting from formulas

Classify the three supplied complete graphs A, B, and C as even, odd, or neither. Their curves continue with the same shapes outside the viewing window, and all real inputs are allowed. This asks you to test the mirror and half-turn pairings on the entire supplied curves.

Graph A has the same height on opposite sides of the y-axis.

Graph B has opposite heights at opposite inputs throughout its complete curve.

Graph C fails both pairings at inputs −2 and 2.

Graph A has matching arms across the y-axis. Its points at inputs −2 and 2 both have height 4; the same mirror pairing holds along both complete arms.An even function has equal heights at opposite inputs. Here the full V is mirrored about the y-axis.
Graph B matches after a half turn around (0, 0). Its point (1, 2) has partner (−1, −2), and the entire rising curve has this opposite-coordinate pairing.An odd function pairs opposite inputs with opposite heights; the complete cubic-shaped drawing matches that half turn.
Graph C has point (2, 1) but its point at the opposite input −2 has height 9.The heights 1 and 9 are neither equal nor opposites, so this one failing pair disproves each required symmetry.

Answer

Graph A: even.
Graph B: odd.
Graph C: neither.

Check Reflect the complete V left to right: it matches. Rotate the complete S halfway about the origin: it matches. For C, its vertex is at (1, 0), and the marked heights at ±2 already fail both tests. A few matching dots alone would not prove a symmetry.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: h(x) = + 1 is odd because the term changes sign.

The entire output must change sign. Substitution gives h(−x) = − + 1, while negating the output gives −h(x) = − − 1. The constants disagree.

✓ Instead: h is neither. For example, h(1) = 2 and h(−1) = 0. Those outputs are neither equal nor opposites.

✗ Not this: The formula guarantees an Even function even when the Domain is x ≥ 0.

The domain is part of the function. Input 1 is allowed but −1 is forbidden, so the required partner point is missing.

✓ Instead: On [0, ∞), this function is neither. On all real inputs, is even because opposite inputs exist and (−x = .

✗ Not this: An even result at one pair of opposite inputs proves the function is even.

One pair checks only those inputs. For h(x) = + 1, the pair at x = 0 matches itself, but h(1) = 2 and h(−1) = 0 fail both symmetry tests.

✓ Instead: Use a formula identity to cover every allowed input. One failing pair can disprove a proposed symmetry. Also test both identities for the zero function, which is both even and odd on a symmetric domain.

Tips and tricks

Remember even agrees, odd opposes. The agreement concerns outputs at opposite inputs, not whether a value is an even or odd integer.
Write three separate expressions on the exam: f(x), f(−x), and −f(x). Parenthesize every substituted −x and negate every term in −f(x).
Check the Domain before doing algebra. A missing opposite input rules out both symmetries.
A graph-only check uses point partners. For an even graph, (2, 4) needs (−2, 4). For an odd graph, (2, 8) needs (−2, −8). Check the whole drawing, because a few matching dots cannot establish its symmetry.

Trap. Calling a function odd because some terms change sign. Odd means the entire output changes sign, including its constant term. Compare with −f(x) written in full. A few matching number pairs can check a proof, but cannot replace an identity that covers every input.