Even and odd functions — graphical and algebraic tests: Question bank

This is a question bank for the parent topic Even and odd functions. No heavy computation here — every item hunts for a misconception or a boundary case. Read the prompt, answer out loud, then reveal.

True or false — justify

True or false: a function whose graph passes through the origin must be odd.

False. Passing through

(0, 0)

is necessary for odd (if

0

is in the domain), but not sufficient —

f (x) = x^{2}

passes through the origin yet is even.

True or false: if

f (- x) = f (x)

holds at

x = 3

only, then

f

is even.

False. Even requires

f (- x) = f (x)

for all

x

in the domain; a single agreeing point proves nothing (see Mistake 2).

True or false: the sum of two odd functions is always odd.

True.

(o_{1} + o_{2}) (- x) = o_{1} (- x) + o_{2} (- x) = - o_{1} (x) - o_{2} (x) = - (o_{1} + o_{2}) (x)

, so the sum flips sign as a whole.

True or false: the product of two odd functions is odd.

False — it is even. Two sign-flips multiply to a plus:

(- o_{1}) (- o_{2}) = o_{1} o_{2}

, like negative × negative = positive.

True or false: a constant function

f (x) = 5

is even.

True.

f (- x) = 5 = f (x)

for every

x

, so it is symmetric about the y-axis (a flat horizontal line).

True or false: the only function that is both even and odd is

f (x) = 0

True. Both conditions force

f (x) = - f (x)

, so

2 f (x) = 0

, hence

f (x) = 0

everywhere on its symmetric domain.

True or false: if a polynomial contains only odd powers of

x

, it is an odd function.

True. Each term

x^{2 k + 1}

flips sign under

x \to - x

, and a sum of odd functions stays odd — see Polynomial functions.

True or false:

f (x) = x^{3} + x^{2}

is neither even nor odd.

True. The odd term

x^{3}

and even term

x^{2}

have different symmetries and do not cancel into a single pattern.

True or false: every function on a symmetric domain can be split into an even part plus an odd part.

True. The decomposition theorem gives

f_{even} = \frac{f ( x ) + f ( - x )}{2}

and

f_{odd} = \frac{f ( x ) - f ( - x )}{2}

, and they add back to

f

True or false:

cosh (x)

and

sinh (x)

are just made-up names with no link to symmetry.

False. They are the even and odd parts of

e^{x}

— see Hyperbolic functions and Example 6 in the parent.

True or false: if

f

is odd and

0

is in its domain, then

f (0) = 0

True. Set

x = 0

f (- 0) = - f (0)

gives

f (0) = - f (0)

, so

2 f (0) = 0

, meaning

f (0) = 0

Spot the error

f (x) = x

is neither even nor odd because

- x \neq = \pm x

." — what's wrong?

The reasoning skips the real issue: the domain

[0, \infty)

isn't symmetric, so

f (- x)

is undefined for

x > 0

. The classification simply doesn't apply — check the domain first (Domain and range).

g (x) = x^{3} + x

mixes powers, so it must be neither." — what's wrong?

Wrong test. Both

x^{3}

and

x

are odd, so

g (- x) = - g (x)

and

g

is odd — see Mistake 1. Judge by algebra, not by appearance.

"The graph looks symmetric about the y-axis in my viewing window, so

f

is even." — what's wrong?

A finite window can hide asymmetry far out or at small scales; only an algebraic check

f (- x) = f (x)

for all

x

proves it.

f (- x) = (- x)^{2} + (- x) = x^{2} - x = f (x)

, so

x^{2} + x

is even." — what's wrong?

The final equality is false:

x^{2} - x \neq = x^{2} + x

(they differ by

2 x

). The function is actually neither.

cos (- x) = - cos (x)

, so cosine is odd." — what's wrong?

The identity is backwards:

cos (- x) = + cos (x)

, so cosine is even. It is sine that satisfies

sin (- x) = - sin (x)

— see Trigonometric functions.

"An odd function times an even function is odd, therefore it must pass through the origin." — what's wrong?

The product being odd is correct, but "passes through origin" only follows if

0

is in the domain; on a domain excluding

0

(like

1/ x

style pieces) the origin conclusion is meaningless.

Why questions

Why does the average

\frac{f ( x ) + f ( - x )}{2}

always give an even function?

Replacing

x

with

- x

swaps the two terms

f (x)

and

f (- x)

, leaving the sum unchanged — so it satisfies

E (- x) = E (x)

by construction.

Why does even × odd give odd rather than even?

The even factor keeps its sign under

x \to - x

while the odd factor flips, so the product picks up exactly one minus sign:

e (x) \cdot (- o (x)) = - (e o) (x)

Why is

\int_{- a}^{a} f_{odd} (x) d x = 0

The area on the left of the y-axis is the exact negative of the area on the right (point-symmetry through the origin), so they cancel — this is the payoff studied in Symmetry in calculus.

Why do even functions in a Fourier series have only cosine terms?

Cosine is the even basis function; an even target has no odd content, so every sine (odd) coefficient must vanish, halving the work.

Why must the domain be symmetric about

0

before we even ask "even or odd?"

The definitions compare

f (x)

with

f (- x)

, so

- x

must exist in the domain whenever

x

does — otherwise the comparison is undefined.

Why does adding an even and an odd function usually give "neither"?

The two pieces obey incompatible sign rules under

x \to - x

; unless one piece is zero, no single symmetry can describe the sum.

Edge cases

f (x) = 0

(the zero function) even, odd, or neither?

Both. It satisfies

f (- x) = f (x)

and

f (- x) = - f (x)

simultaneously, and it is the only such function.

Can a function defined only on

(0, \infty)

ever be classified as odd?

No. Its domain is not symmetric about

0

, so

f (- x)

is undefined and the odd test cannot be applied — a Domain and range gate, not a symmetry failure.

A piecewise function equals

x^{2}

for

∣ x ∣ \leq 1

and

x + 5

for

∣ x ∣ > 1

— is it even?

No. Spot-checks inside

[- 1, 1]

suggest evenness, but

f (- 2) = 3 \neq = 7 = f (2)

; symmetry must hold on the whole domain.

f

is even and differentiable, what symmetry does

f^{'}

have, and why?

f^{'}

is odd. Differentiating

f (- x) = f (x)

(chain rule) gives

- f^{'} (- x) = f^{'} (x)

, i.e.

f^{'} (- x) = - f^{'} (x)

— a Symmetry in calculus fact.

Does shifting an even function horizontally keep it even?

Generally no.

f (x - c)

with

c \neq = 0

moves the axis of symmetry to

x = c

, so it is no longer symmetric about the y-axis — see Function transformations.

Recall Quick self-test

The one-sentence traps ::: domain symmetry first, "one point" proves nothing, cos is even not odd, odd·odd = even, and $f (0) = 0$ only for odd functions with $0$ in the domain.