Even and odd functions — graphical and algebraic tests: Exercises

This page is a self-testing ladder. Each problem has a fully worked solution hidden inside a collapsible callout — try the problem first, then reveal. Levels climb from recognising symmetry to creating with it.

Everything here rests on two equations you must have burned into memory from the parent note:

Recall The two definitions (reveal to confirm you know them)

A function is even when $f (- x) = f (x)$ — its graph is a mirror across the $y$ -axis. A function is odd when $f (- x) = - f (x)$ — its graph has 180° rotational symmetry about the origin. "Neither" means at least one $x$ breaks both rules.

Before any test, always check the domain is symmetric (if $x$ is allowed, $- x$ must be allowed too) — this connects to Domain and range.

The picture below is the mental model for every problem on this page.

Figure — Even and odd functions — graphical and algebraic tests

Level 1 — Recognition

Goal: given a rule, decide even / odd / neither by direct substitution.

Problem 1.1. Classify $f (x) = x^{4} - 3 x^{2} + 7$ .

Recall Solution 1.1

Step 1 — domain. Polynomial, so domain is all real numbers $R$ ; symmetric. ✓ What we did: checked $- x$ is always allowed. Why: the definitions require it.

Step 2 — substitute $- x$ for every $x$ . $f (- x) = (- x)^{4} - 3 (- x)^{2} + 7 = x^{4} - 3 x^{2} + 7$ Every power is even, so the minus sign vanishes (an even power of a negative is positive).

Step 3 — compare. $f (- x) = x^{4} - 3 x^{2} + 7 = f (x)$ . Conclusion: even.

Problem 1.2. Classify $g (x) = 2 x^{5} + x^{3} - 4 x$ .

Recall Solution 1.2

Domain: $R$ , symmetric. ✓

Substitute: $g (- x) = 2 (- x)^{5} + (- x)^{3} - 4 (- x) = - 2 x^{5} - x^{3} + 4 x$ Factor out the shared minus sign: $= - (2 x^{5} + x^{3} - 4 x) = - g (x)$ Conclusion: odd.

Problem 1.3. Classify $h (x) = x^{2} + 2 x + 1$ .

Recall Solution 1.3

Domain: $R$ , symmetric. ✓

Substitute: $h (- x) = (- x)^{2} + 2 (- x) + 1 = x^{2} - 2 x + 1$ Compare against the two targets:

$h (x) = x^{2} + 2 x + 1$ → not equal, so not even.
$- h (x) = - x^{2} - 2 x - 1$ → not equal, so not odd.

Conclusion: neither. The $x^{2}$ and $1$ terms want mirror symmetry; the $2 x$ term wants rotational symmetry. They cannot both win.

Level 2 — Application

Goal: handle trig, roots, quotients, and the domain check.

Problem 2.1. Classify $f (x) = x^{2} sin (x)$ .

Recall Solution 2.1

Domain: $R$ , symmetric. ✓

Test each factor first:

$x^{2}$ is even: $(- x)^{2} = x^{2}$ .
$sin x$ is odd: $sin (- x) = - sin x$ (see Trigonometric functions).

By the product rule from the parent note, even × odd = odd. Confirm directly: $f (- x) = (- x)^{2} sin (- x) = x^{2} (- sin x) = - x^{2} sin x = - f (x)$ Conclusion: odd.

Problem 2.2. Classify $g (x) = \frac{x}{x ^{2} + 1}$ .

Recall Solution 2.2

Domain: $x^{2} + 1$ is never zero, so domain is $R$ , symmetric. ✓

Substitute: $g (- x) = \frac{- x}{( - x ) ^{2} + 1} = \frac{- x}{x ^{2} + 1} = - \frac{x}{x ^{2} + 1} = - g (x)$ The denominator is even (unchanged), the numerator is odd (flips sign). Odd ÷ even = odd. Conclusion: odd.

Problem 2.3. Classify $p (x) = x$ .

Recall Solution 2.3

Domain: $x$ needs $x \geq 0$ , so domain is $[0, \infty)$ . Is this symmetric? Take $x = 4$ : it is in the domain, but $- 4$ is not.

Because $p (- x)$ is undefined for positive $x$ , the equations $p (- x) = p (x)$ and $p (- x) = - p (x)$ can never even be asked. Conclusion: neither (the symmetry test does not apply). See Domain and range.

Problem 2.4. Classify $q (x) = cos (x) + x^{2}$ .

Recall Solution 2.4

Domain: $R$ . ✓ $cos x$ is even, $x^{2}$ is even; even + even = even. Confirm: $q (- x) = cos (- x) + (- x)^{2} = cos x + x^{2} = q (x)$ Conclusion: even.

Level 3 — Analysis

Goal: prove general facts, use decomposition, reason about pieces.

Problem 3.1. Find the even part and odd part of $f (x) = e^{2 x}$ .

Recall Solution 3.1

Use the decomposition formulas: $f_{even} (x) = \frac{f ( x ) + f ( - x )}{2} = \frac{e ^{2 x} + e ^{- 2 x}}{2} = cosh (2 x)$ $f_{odd} (x) = \frac{f ( x ) - f ( - x )}{2} = \frac{e ^{2 x} - e ^{- 2 x}}{2} = sinh (2 x)$ Check: $cosh (2 x) + sinh (2 x) = \frac{e ^{2 x} + e ^{- 2 x} + e ^{2 x} - e ^{- 2 x}}{2} = \frac{2 e ^{2 x}}{2} = e^{2 x}$ . ✓ This is exactly why Hyperbolic functions exist — even/odd parts of exponentials.

Problem 3.2. Prove that the derivative of an odd function is even. (Uses Symmetry in calculus.)

Recall Solution 3.2

Let $o (x)$ be odd, so $o (- x) = - o (x)$ for all $x$ . Differentiate both sides with respect to $x$ . The left side needs the chain rule because of the inner $- x$ : $\frac{d}{d x} o (- x) = o^{'} (- x) \cdot (- 1) = - o^{'} (- x)$ The right side: $\frac{d}{d x} (- o (x)) = - o^{'} (x)$ Set them equal: $- o^{'} (- x) = - o^{'} (x) ⟹ o^{'} (- x) = o^{'} (x)$ That is precisely the even condition. The derivative of an odd function is even. ∎

Problem 3.3. If $f$ is even and odd simultaneously (both $f (- x) = f (x)$ and $f (- x) = - f (x)$ hold), what is $f$ ?

Recall Solution 3.3

Both hold, so set the right sides equal: $f (x) = - f (x) ⟹ 2 f (x) = 0 ⟹ f (x) = 0$ for every $x$ in the domain. Conclusion: the only function that is both even and odd is the zero function $f (x) = 0$ . Geometrically: the only graph that is both mirror-symmetric across the $y$ -axis and 180°-rotation-symmetric is the flat line along the $x$ -axis.

Level 4 — Synthesis

Goal: combine tools — decomposition, integration, and structure.

Problem 4.1. Evaluate $\int_{- 2}^{2} (x^{3} + x^{2}) d x$ using symmetry, not brute antiderivatives where avoidable.

Recall Solution 4.1

Split by symmetry (linearity of the integral): $\int_{- 2}^{2} x^{3} d x + \int_{- 2}^{2} x^{2} d x$

$x^{3}$ is odd, and any odd function integrated over a symmetric interval $[- a, a]$ gives $0$ — positive and negative halves cancel. So the first integral is $0$ .
$x^{2}$ is even, so $\int_{- 2}^{2} x^{2} d x = 2 \int_{0}^{2} x^{2} d x = 2 \cdot \frac{x ^{3}}{3}_{0}^{2} = 2 \cdot \frac{8}{3} = \frac{16}{3}$ .

Answer: $\frac{16}{3}$ . The odd term vanishing is the payoff of symmetry — see Symmetry in calculus.

Problem 4.2. A function is built as $f (x) = (even) \cdot (odd) + (odd) \cdot (odd)$ . Classify $f$ , then verify with $f (x) = x^{2} \cdot x + x \cdot x^{3}$ .

Recall Solution 4.2

Use the product rules:

even × odd = odd
odd × odd = even

So $f = (odd) + (even)$ = neither (a sum of one odd and one even piece has no single symmetry — unless one piece is zero).

Verify concretely: $f (x) = x^{2} \cdot x + x \cdot x^{3} = x^{3} + x^{4}$ . $f (- x) = (- x)^{3} + (- x)^{4} = - x^{3} + x^{4}$

vs $f (x) = x^{3} + x^{4}$ : not equal → not even.
vs $- f (x) = - x^{3} - x^{4}$ : not equal → not odd.

Conclusion: neither, matching the structural prediction.

Problem 4.3. Show that in a Fourier series of an even periodic function, all sine coefficients are zero.

Recall Solution 4.3

A Fourier sine coefficient is $b_{n} = \frac{1}{L} \int_{- L}^{L} f (x) sin (\frac{nπ x}{L}) d x .$

$f (x)$ is even (given).
$sin (\frac{nπ x}{L})$ is odd (sine is odd).

Their product is even × odd = odd. Integrating an odd function over the symmetric interval $[- L, L]$ gives $0$ . Therefore every $b_{n} = 0$ : an even function is built from cosines alone. ∎

Level 5 — Mastery

Goal: create and reason at the highest level.

Problem 5.1. Prove the decomposition is unique: if $f = E_{1} + O_{1} = E_{2} + O_{2}$ with $E_{i}$ even and $O_{i}$ odd, then $E_{1} = E_{2}$ and $O_{1} = O_{2}$ .

Recall Solution 5.1

From $E_{1} + O_{1} = E_{2} + O_{2}$ , rearrange: $E_{1} - E_{2} = O_{2} - O_{1} .$ The left side is even (difference of evens); the right side is odd (difference of odds). Call this common function $g$ . Then $g$ is both even and odd. By Problem 3.3, the only such function is $g (x) = 0$ . Hence $E_{1} - E_{2} = 0 and O_{2} - O_{1} = 0,$ so $E_{1} = E_{2}$ and $O_{1} = O_{2}$ . The split is unique. ∎

Problem 5.2. Construct a function that is neither even nor odd, whose even part is $cos x$ and whose odd part is $x$ . Then confirm the parts.

Recall Solution 5.2

By the decomposition theorem, $f = f_{even} + f_{odd}$ , so simply add them: $f (x) = cos x + x .$ Confirm even part: $\frac{f ( x ) + f ( - x )}{2} = \frac{( cos x + x ) + ( cos x - x )}{2} = \frac{2 cos x}{2} = cos x .$ ✓ Confirm odd part: $\frac{f ( x ) - f ( - x )}{2} = \frac{( cos x + x ) - ( cos x - x )}{2} = \frac{2 x}{2} = x .$ ✓ And $f$ itself is neither (odd $x$ term + even $cos x$ term), as required.

Problem 5.3. Let $f$ be any function with symmetric domain. Prove $\int_{- a}^{a} f (x) d x = \int_{- a}^{a} f_{even} (x) d x = 2 \int_{0}^{a} f_{even} (x) d x .$

Recall Solution 5.3

Write $f = f_{even} + f_{odd}$ and integrate over $[- a, a]$ : $\int_{- a}^{a} f = \int_{- a}^{a} f_{even} + \int_{- a}^{a} f_{odd} .$ The odd integral over a symmetric interval is $0$ , killing the second term: $\int_{- a}^{a} f = \int_{- a}^{a} f_{even} .$ Now $f_{even}$ is even, so its two halves $[- a, 0]$ and $[0, a]$ contribute equally: $\int_{- a}^{a} f_{even} = 2 \int_{0}^{a} f_{even} .$ Chaining gives the required identity. ∎ Meaning: only the even part of a function survives a symmetric integral — the deepest single sentence about why even/odd analysis saves so much work.

Recall Rapid self-check (cover the right side)

$x^{2} sin x$ is ::: odd (even × odd) $\frac{x}{x ^{2} + 1}$ is ::: odd $x$ is ::: neither (domain not symmetric) The only even-and-odd function is ::: $f (x) = 0$ $\int_{- 2}^{2} (x^{3} + x^{2}) d x$ = ::: $\frac{16}{3}$ The even part of $e^{2 x}$ is ::: $cosh (2 x)$

Related deep threads: Polynomial functions, Function transformations, Fourier series, Hyperbolic functions, Symmetry in calculus.