Even and odd functions — graphical and algebraic tests: Worked examples

This page is the "no surprises" drill for the parent topic. We take the algebraic test $f (- x)$ vs $f (x)$ vs $- f (x)$ and run it through every kind of input a problem can hand you: clean polynomials, sign traps, functions that break because their domain is lopsided, functions that are secretly neither, a real-world word problem, and an exam twist. If a scenario can appear, it appears below.

Figure — Even and odd functions — graphical and algebraic tests

The scenario matrix

Before working anything, here is the full list of case-classes this topic can throw at you. Each row is a distinct trap or shape. The right column names the example that clears it.

Cell	Case class	What makes it tricky	Cleared by
A	Pure even power(s)	baseline sanity	Ex 1
B	Pure odd power(s) mixed	"different powers ⇒ neither" trap	Ex 2
C	Even $+$ odd term	genuinely neither	Ex 3
D	Constant / the zero function	degenerate: is a flat line even? odd? both?	Ex 4
E	Quotient (rational)	sign travels through division	Ex 5
F	Composition with $∣ x ∣$ or square	inner symmetry forces outer	Ex 6
G	Lopsided domain (e.g. $x$ )	fails before you even compute	Ex 7
H	Piecewise / spot-check trap	one point lies to you	Ex 8
I	Word problem (physics)	translate reality → symmetry	Ex 9
J	Exam twist: decompose + integrate	odd part vanishes on $[- a, a]$	Ex 10

We now hit every cell.

Worked example Example 1 — Cell A: pure even

Statement. Classify $f (x) = 3 x^{4} - 2 x^{2} + 7$ .

Forecast: every power is even, and $7 = 7 x^{0}$ (power $0$ , also even). Guess: even.

Domain check. All terms are polynomials, defined for every real number, so the domain is $R$ . It contains $- x$ whenever it contains $x$ . ✓ Why this step? Symmetry compares $f (x)$ with $f (- x)$ ; if $- x$ isn't allowed, the question is meaningless (see Ex 7).
Compute $f (- x)$ . Replace every $x$ with $- x$ : $f (- x) = 3 (- x)^{4} - 2 (- x)^{2} + 7 = 3 x^{4} - 2 x^{2} + 7.$ Why this step? $(- x)^{even} = x^{even}$ — an even number of sign-flips cancels.
Compare. $f (- x) = 3 x^{4} - 2 x^{2} + 7 = f (x)$ . Matches the even condition $f (- x) = f (x)$ . Why this step? This is the definition of even.

Verify: test one concrete point as a smell-check: $f (2) = 3 (16) - 2 (4) + 7 = 48 - 8 + 7 = 47$ and $f (- 2) = 47$ too. Equal, consistent with even.

Worked example Example 4 — Cell D: the degenerate constant / zero function

Statement. (a) Classify the constant $c (x) = 5$ . (b) Classify the zero function $z (x) = 0$ .

Forecast: a flat line is its own mirror image (even), but $z (x) = 0$ might be special. Guess: (a) even; (b) both even and odd.

Constant, domain check. $c (x) = 5$ defined on $R$ . ✓
Constant, compute. $c (- x) = 5 = c (x)$ . Matches even. Why this step? A constant has no $x$ to flip, so nothing changes → even. Is it odd? Odd needs $c (- x) = - c (x)$ , i.e. $5 = - 5$ — false. So $c$ is even only.
Zero function, compute. $z (- x) = 0$ . Then $z (- x) = 0 = z (x)$ (even) and $z (- x) = 0 = - 0 = - z (x)$ (odd). Why this step? $0$ is the unique fixed point of sign-flip: $0 = - 0$ . So the zero function satisfies both conditions simultaneously.

Conclusion: the only function that is both even and odd is the zero function (on a symmetric domain). Non-zero constants are even, never odd.

Verify: for evenness of $c$ : $c (3) = 5 = c (- 3)$ ✓. For $z$ being odd: $z (- 3) = 0 = - z (3) = - 0$ ✓.

Worked example Example 6 — Cell F: composition forcing symmetry

Statement. Classify $p (x) = cos (x^{2})$ and separately $q (x) = sin (x^{2})$ .

Forecast: the inner $x^{2}$ is even, so both inputs "forget the sign" before the outer function ever sees them. Guess: both even.

Domain check. $x^{2}$ defined on $R$ ; sine and cosine defined everywhere. Domain $R$ . ✓
Compute $p (- x)$ . $p (- x) = cos ((- x)^{2}) = cos (x^{2}) = p (x) .$ Why this step? $(- x)^{2} = x^{2}$ : the even inner layer kills the sign before the outer function acts, so the outer function's own parity is irrelevant.
Compute $q (- x)$ . $q (- x) = sin ((- x)^{2}) = sin (x^{2}) = q (x) .$

Conclusion: both even, even though $sin$ on its own is odd. Lesson: an even inner function makes the whole composite even, regardless of the outer function.

Verify: $q (1) = sin (1) \approx 0.8415$ ; $q (- 1) = sin (1) \approx 0.8415$ , equal ⇒ even. ✓

Worked example Example 7 — Cell G: domain kills it before you start

Statement. Classify $s (x) = x + x$ ... let's take $s (x) = x$ .

Forecast: the graph only lives for $x \geq 0$ . If half the mirror is missing, symmetry can't even be asked. Guess: neither — and for a domain reason, not an algebra reason.

Domain check. $x$ requires $x \geq 0$ , so domain $= [0, \infty)$ . Why this step? Even/odd both require: whenever $x$ is in the domain, $- x$ must be too.
Test the requirement. Take $x = 4$ (in domain). Then $- x = - 4$ , and $- 4 \in / [0, \infty)$ . The domain is not symmetric. Why this step? The definition $f (- x) = f (x)$ can't be evaluated when $f (- 4)$ doesn't exist.
Conclusion. Since the definition can't be satisfied on a symmetric domain, $x$ is neither even nor odd — by default, on domain grounds. We never needed to compute $s (- x)$ .

Verify: $s (4) = 2$ exists but $s (- 4) = - 4$ is undefined in reals ⇒ pairing broken ⇒ neither. ✓ (Contrast: $∣ x ∣$ is even, because $∣ x ∣$ restores the symmetric domain.)

Worked example Example 8 — Cell H: the piecewise spot-check trap

Statement. Let $f (x) = {x^{2}, x + 5, ∣ x ∣ \leq 1 ∣ x ∣ > 1$ A student checks $f (- 0.5) = f (0.5)$ and declares "even." Are they right?

Forecast: one lucky point proves nothing (Mistake 2). Guess: not even — the outer branch is asymmetric.

Domain check. Defined for all $x$ , domain $R$ , symmetric. ✓
The lucky point. For $∣ x ∣ \leq 1$ the rule is $x^{2}$ , which is even, so $f (- 0.5) = 0.25 = f (0.5)$ . True but local. Why this step? Shows why the spot-check misleads — it landed inside the symmetric branch.
Probe the outer branch. Take $x = 2$ ( $∣ x ∣ > 1$ ): $f (2) = 2 + 5 = 7$ . Now $f (- 2)$ : $∣ - 2 ∣ > 1$ so $f (- 2) = (- 2) + 5 = 3$ . Why this step? Even requires $f (- 2) = f (2)$ for all $x$ ; here $3 \neq = 7$ .
Conclusion. Even fails at $x = 2$ ; also $- f (2) = - 7 \neq = 3$ , so odd fails too. The function is neither.

Verify: $f (2) = 7, f (- 2) = 3$ . $3 \neq = 7$ (not even) and $3 \neq = - 7$ (not odd). ✓ The single interior point was a false positive.

Worked example Example 9 — Cell I: word problem (physics)

Statement. A pendulum's horizontal displacement is $d (t) = A sin (ω t)$ , with $A = 0.10 m$ , $ω = 2 rad/s$ . Time $t = 0$ is the instant it swings through the centre. Is displacement even or odd in time about $t = 0$ , and what does that predict for $t = - 1.5 s$ given $t = + 1.5 s$ ?

Forecast: swinging through centre and reversing — a movie run backwards should mirror-flip the position. Guess: odd.

Translate to symmetry. "Run time backwards" means replace $t$ with $- t$ . We test $d (- t)$ vs $d (t)$ . Why this step? Even/odd in time = time-reversal symmetry, a real physical statement.
Compute $d (- t)$ . $d (- t) = A sin (ω (- t)) = A sin (- ω t) = - A sin (ω t) = - d (t),$ > using that $\sin$ is odd. So $d$ is odd in $t$ .
Apply to the numbers. Compute $d (1.5) = 0.10 sin (2 \times 1.5) = 0.10 sin (3) \approx 0.10 \times 0.14112 = 0.014112 m$ . Oddness predicts $d (- 1.5) = - d (1.5) \approx - 0.014112 m$ — same distance, opposite side. Why this step? Physical meaning: 1.5 s before centre-crossing, the bob is the mirror distance on the far side.

Conclusion: odd — and it saves work: knowing one side's motion gives the other by a sign flip. Units check: $A$ in metres, $sin$ dimensionless ⇒ displacement in metres. ✓

Verify: $d (1.5) \approx 0.014112$ m and $d (- 1.5) \approx - 0.014112$ m; their sum is $0$ . ✓

Worked example Example 10 — Cell J: exam twist (decompose then integrate)

Statement. Compute $\int_{- 2}^{2} (x^{3} + x^{2}) d x$ by splitting the integrand into its even and odd parts (see Symmetry in calculus).

Forecast: $x^{3}$ is odd → integrates to $0$ over the symmetric interval; only the even $x^{2}$ survives. Guess: a positive number from $x^{2}$ alone.

Split by parity. $f (x) = x^{3} + x^{2}$ has odd part $O (x) = x^{3}$ and even part $E (x) = x^{2}$ . Why this step? Over $[- a, a]$ , the odd part cancels itself; splitting isolates the work.
Kill the odd part. $\int_{- 2}^{2} x^{3} d x = 0$ because for every strip at $+ x$ there is an equal-and-opposite strip at $- x$ . Why this step? This is the whole payoff of odd symmetry in integration.
Do only the even part, doubled. $\int_{- 2}^{2} x^{2} d x = 2 \int_{0}^{2} x^{2} d x = 2 \cdot \frac{x ^{3}}{3}_{0}^{2} = 2 \cdot \frac{8}{3} = \frac{16}{3} .$ Why this step? An even function's left half mirrors its right half, so double the $[0, 2]$ area.
Assemble. $\int_{- 2}^{2} (x^{3} + x^{2}) d x = 0 + \frac{16}{3} = \frac{16}{3} \approx 5.333$ .

Verify: direct antiderivative $\frac{x ^{4}}{4} + \frac{x ^{3}}{3}$ at $2$ gives $4 + \frac{8}{3}$ , at $- 2$ gives $4 - \frac{8}{3}$ ; difference $= \frac{16}{3}$ . ✓ Matches.

Recall drill

Recall Why does a lopsided domain (Ex 7) block even/odd before any algebra?

Because the definitions compare $f (x)$ with $f (- x)$ ; if $- x$ isn't in the domain the comparison can't be made ::: the domain must contain $- x$ whenever it contains $x$ , else the function is neither by default.

Recall Which single function is both even and odd, and why?

The zero function ::: because it needs $z (- x) = z (x)$ and $z (- x) = - z (x)$ simultaneously, i.e. $0 = - 0$ , which only $0$ satisfies.

Recall Why is

$cos (x^{2})$ even even though... it involves nothing odd? The inner $x^{2}$ is even, so the sign is destroyed before the outer function acts ::: an even inner layer forces the whole composite to be even.

An odd function integrated over

[- a, a]

equals

0

An even inner function makes any composite

even

The only function that is both even and odd is

the zero function