Odd and Even Functions: Check the Domain, Then f(−x) — One Picture Explains It

1Warm-up

Start with a question, on instinct:

MockSource: Math Mock Exam 1

Among the following functions, which one is both an odd function and an increasing function?

2The idea: parity is symmetry of the graph

Parity asks two things: is the domain symmetric? and what is the graph symmetric about? About the y-axis (a mirror) means even; about the origin (a half-turn lands on itself) means odd.

Pick a function and guess odd, even or neither, then read the verdict. Drag a: the filled dots are the graph at a and −a; the two rings are P's two reflections. See which one the graph's own point lands on.

Filled: P and the graph's point at −a. Dashed ring: P reflected in the y-axis. Solid ring: P turned 180° about the origin.

a = 1.5

$f(1.5)=3.38$ , $f(-1.5)=-3.37$

Guess first: it is…

Symmetry → equation → name: two lines

graph symmetric about the y-axis (mirror)

$f(-x)=f(x)$

even function

graph symmetric about the origin (half-turn)

$f(-x)=-f(x)$

odd function

Prerequisite: the domain is symmetric about the origin. If it is not, the function is neither.

Common functions at a glance

Odd	Even	Neither
$x^3$ , $x$	$x^2$ , $x^4$ , $\|x\|$	$2^x$ , $\ln x$
$\dfrac1x$ , $x+\dfrac1x$	$x^4+3$ , $\ln x^2$	$\sqrt{x}$ (domain not symmetric)
$\sin x$ , $\tan x$	$\cos x$ , $x\sin x$	$x^3+1$ (the constant spoils it)
$2^x-2^{-x}$	$\cos(\sin x)$	$\dfrac{1}{x-1}$ (domain not symmetric)

Combining functions

Combination	Result	Example
odd ± odd	odd	$x^3+x$
even ± even	even	$x^2+\|x\|$
odd × odd	even	$x\sin x$
odd × even	odd	$x\cos x$
even(odd( $x$ ))	even	$\cos(\sin x)$
odd ± even	usually neither	$x+x^2$

Colour key: blue = the result is odd, amber = the result is even.

Read it as text

In symbols: even $f(-x)=f(x)$ ; odd $f(-x)=-f(x)$ . Both need one thing first: a domain symmetric about the origin — if $x$ is in it, so is $-x$ . The domain of $\sqrt{x}$ is $[0,+\infty)$ : $2$ is in, $-2$ is not, so it is neither, and you never need $f(-x)$ .

When unsure, do not go by looks: replace every $x$ with $-x$ and compare with $f(x)$ and $-f(x)$ . $x^3+1$ looks odd, but $f(-x)=-x^3+1$ equals neither $f(x)$ nor $-f(x)=-x^3-1$ .

Combining: odd + odd = odd, even + even = even, odd × odd = even, odd × even = odd; odd + even is usually neither. For compositions look at the outside: even(odd(x)) = even, odd(odd(x)) = odd.

34 steps

Walking through $f(x)=x+\dfrac1x$

1
Check the domain first
$x\neq0$ , i.e. $(-\infty,0)\cup(0,+\infty)$
- symmetric → carry on
- not symmetric → neither, stop
e.g. $\sqrt{x}$ has domain $[0,+\infty)$ : $2$ is in, $-2$ is not.
2
Compute $f(-x)$
$f(-x)=(-x)+\dfrac{1}{-x}=-x-\dfrac1x$
Replace every $x$ with $-x$ , the one in the denominator too.
3
Compare with $f(x)$ and $-f(x)$
- $f(-x)=f(x)$ → even
- $f(-x)=-f(x)$ → odd
- neither → neither
Here $-x-\frac1x=-\left(x+\frac1x\right)=-f(x)$ : odd.
4
Read the stem before you choose
- asked for the "wrong / incorrect" one → pick the false statement
- asked for parity and monotonicity → check both
The data in "Common mistakes" below comes from exactly this step.

Read it as text

1. Check the domain first
Is the domain symmetric about the origin? If not, it is neither and you are done.
2. Compute $f(-x)$
Replace every $x$ with $-x$ and simplify.
3. Compare with $f(x)$ and $-f(x)$
Equal to $f(x)$ : even. Equal to $-f(x)$ : odd. Neither: neither.
4. Read the stem before you choose
If it asks for the "wrong / incorrect" statement pick the false one; if it asks about parity and monotonicity, check both.

Choose first, then check: the test on 5 functions

Commit to each step first; a wrong pick tells you why.

Step 1Step 1, the domain: $x\neq0$ — symmetric about the origin?

Read all 5 worked examples in full

$f(x)=x+\frac{1}{x}$

Step 1Step 1, the domain: $x\neq0$ — symmetric about the origin?
Symmetric, so carry on.
✗ No, because $0$ is missing:The point removed is the origin itself; what is left, $(-\infty,0)\cup(0,+\infty)$ , is still symmetric.
Step 2Step 2, compute $f(-x)$ :
$f(-x)=-x-\frac1x$ .
✗ $-x+\dfrac1x$ :The $x$ in the denominator changes too: $\frac{1}{-x}=-\frac1x$ .
✗ $x+\dfrac1x$ :That is $f(x)$ itself. Replace every $x$ by $-x$ .
Step 3Step 3, compare:
Odd.
✗ $f(-x)=f(x)$ : even: $-x-\frac1x=-\left(x+\frac1x\right)$ is the opposite of $f(x)$ , not equal to it.
✗ neither:It satisfies $f(-x)=-f(x)$ ; one of the two is enough.

Result: $f(-x)=-f(x)$

$f(x)=\sqrt{x}$

Step 1Step 1, the domain $[0,+\infty)$ : symmetric?
The domain is not symmetric.
✗ Yes, because it contains $0$ :Symmetry means "if $x$ is in, so is $-x$ ", not whether $0$ is in. For $x=2$ , $-2$ is missing.
Step 2Step 2, the verdict:
An asymmetric domain means neither, straight away.
✗ even, because $\sqrt{(-x)^2}=\sqrt{x^2}$ :That is $\sqrt{x^2}=|x|$ , a different function. $\sqrt{x}$ is undefined for $x<0$ , so $f(-x)$ cannot even be compared with $f(x)$ .

Result: $2\in[0,+\infty),\ -2\notin[0,+\infty)$

$f(x)=x^3+1$

Step 1Step 1, the domain:
The domain is $\mathbb{R}$ .
✗ must exclude $x=-1$ : $x^3+1$ is defined for every real number. $x=-1$ is just a zero.
Step 2Step 2, compute $f(-x)$ :
$f(-x)=-x^3+1$ .
✗ $-x^3-1$ :The sign only hits $x^3$ : $(-x)^3=-x^3$ ; the constant $1$ does not change sign.
Step 3Step 3, compare: $f(x)=x^3+1$ , $-f(x)=-x^3-1$
Neither — unlike $x^4+3$ (even) or $x^3$ (odd).
✗ odd: $-x^3+1\ne-x^3-1$ . The constant $1$ spoils oddness.
✗ even: $-x^3+1\ne x^3+1$ .

Result: $f(-x)\neq f(x),\ f(-x)\neq -f(x)$

$f(x)=\ln(x^{2})$

Step 1Step 1, the domain:
$x^2>0$ means $x\neq0$ : symmetric.
✗ $x>0$ , not symmetric:The condition is $x^2>0$ , not $x>0$ : for negative $x$ , $x^2$ is still positive.
Step 2Step 2, compute $f(-x)$ :
$f(-x)=\ln((-x)^2)=\ln(x^2)=f(x)$ .
✗ $-\ln(x^2)$ : $(-x)^2=x^2$ : the square swallows the sign, nothing flips.
Step 3Step 3, the verdict:
Even.
✗ parity cannot be decided:Every function has a definite answer: $f(-x)=f(x)$ makes it even.

Result: $f(-x)=f(x)$

$f(x)=\cos(\sin x)$

Step 1Step 1, the domain:
The domain is $\mathbb{R}$ .
✗ must restrict $\sin x$ :Whatever $\sin x$ is, $\cos$ is defined: the domain is $\mathbb{R}$ .
Step 2Step 2, $f(-x)=\cos(\sin(-x))=\cos(-\sin x)=$ ?
$\cos(-\sin x)=\cos(\sin x)$ .
✗ $-\cos(\sin x)$ : $\cos$ is even: $\cos(-t)=\cos t$ , and the sign does not come out.
Step 3Step 3, the verdict:
Even.
✗ odd, because the inner $\sin x$ is odd:An odd inside does not make the whole odd: look at the outside. The outer $\cos$ is even and swallows the sign, so the whole is even: even(odd( $x$ )) = even.

Result: $\cos(\sin(-x))=\cos(\sin x)$

4Worked examples

Try it yourself, then open the steps

Real examSource: CSCA Math March 2026 Past Paper

On its domain, the function

f ( x ) = \ln \left( x ^ { 2 } \right)

is

Step 1Domain first: $x^2>0$ means $x\neq0$ , which is symmetric about the origin.
Step 2Compute $f(-x)=\ln\left((-x)^2\right)=\ln(x^2)=f(x)$ .
Step 3 $f(-x)=f(x)$ : even, answer D.

Try it yourself, then open the steps

MockSource: Math Mock Exam 3

Given

f ( x ) = \frac { \sqrt { 1 - x ^ { 2 } } } { | x + 2 | - 2 }

, then

f ( x ) \quad ( \quad )

Step 1Find the domain first (before simplifying): $1-x^2\ge0$ gives $-1\le x\le1$ , and $|x+2|-2\ne0$ gives $x\ne0$ , so the domain is $[-1,0)\cup(0,1]$ , symmetric.
Step 2On this range $x+2>0$ , so $|x+2|-2=x$ and $f(x)=\dfrac{\sqrt{1-x^2}}{x}$ .
Step 3 $f(-x)=\dfrac{\sqrt{1-x^2}}{-x}=-f(x)$ : odd, answer B.

Try it yourself, then open the steps

MockSource: Math Mock Exam 4

If the function

f ( x ) = 1 + \frac { m } { \mathrm { e } ^ { x } + 1 }

is an odd function, then the value of

m

is ( )

Step 1An odd function whose domain is $\mathbb{R}$ ( $e^x+1\neq0$ ) must have $f(0)=0$ .
Step 2 $f(0)=1+\dfrac m2=0$ gives $m=-2$ .
Step 3Check it: $f(x)=\dfrac{e^x-1}{e^x+1}$ and $f(-x)=\dfrac{e^{-x}-1}{e^{-x}+1}=\dfrac{1-e^x}{1+e^x}=-f(x)$ ✓. Answer B.

5Common mistakes

Answer data: CSCA.app students' answers over the 90 days before 2026-10-23.

1. Asked for the wrong statement, choosing a true one

What the data shows:On the June 2026 real exam, asked which statement about an odd function is wrong, only about four in ten students get it right. About a third choose D, $f(-x)=-f(x)$ , which is exactly the definition of an odd function — a true statement.

How to avoid it:When you see "wrong / incorrect", circle it in the stem, judge each of the four statements true or false, then pick the false one. Two true facts about odd functions to keep: the domain is symmetric about the origin, and $f(-x)=-f(x)$ (the graph is symmetric about the origin); $f(-x)=f(x)$ belongs to even functions.

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

Which of the following statements about the odd function

y = f(x)

is incorrect?

2. Judging parity by feel instead of computing $f(-x)$

What the data shows:On the mock question $f(x)=2^x-2^{-x}+x^3$ , about three in ten students call it even (options C and D); only a little over half get it right. About one in seven get the parity right but call it decreasing.

How to avoid it:Do not go by looks: compute $f(-x)$ , replacing every $x$ by $-x$ . Here $2^{-x}$ becomes $2^{x}$ and $x^3$ becomes $-x^3$ , so the whole becomes $-f(x)$ : odd. Once parity is settled, check monotonicity separately.

Check yourself

MockSource: Math Mock Exam 1

Given the function

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

, which of the following conclusions is correct?

3. Parity right, range or monotonicity wrong

What the data shows:On the real-exam question about $y=x^4$ , only about one in ten calls it odd, so parity itself is fairly solid. But about a quarter choose "the range is all real numbers", and about one in ten choose "decreasing on $(0,+\infty)$ ".

How to avoid it:Parity, range and monotonicity are three separate questions; check each: $x^4\ge0$ , so the range is $[0,+\infty)$ ; on $(0,+\infty)$ a larger $x$ gives a larger $x^4$ , so it increases. $x^4$ is symmetric about the $y$ -axis, so it decreases on $(-\infty,0)$ and increases on $(0,+\infty)$ ; that is $x^4$ 's own property, not something to extend to every even function ( $\cos x$ is even too, and its monotonicity is different).

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

Which of the following statements about the function

y = x^4

is correct?

4. An odd inner function taken to make the whole odd

What the data shows:On the mock question $f(x)=\cos(\sin x)$ , only about a quarter of students get it right; about one in five choose "it is odd" — perhaps because the inner $\sin x$ is odd — and more than a third choose "the range is $[-1,1]$ ".

How to avoid it:An odd inside does not make the whole odd: look at the outside. The outer $\cos$ is even, $\cos(-t)=\cos t$ , so the sign is swallowed and the whole is even. Rule: even(odd( $x$ )) = even, odd(odd( $x$ )) = odd. The range is not $[-1,1]$ either: $\sin x\in[-1,1]$ and $\cos t$ on that interval is at least $\cos1$ , so the range is $[\cos1,1]$ .

Check yourself

MockSource: Math Mock Exam 1

Given

f(x)=\cos(\sin x)

, which of the following statements is correct?

6Practice

Question 1 of 5

Real examSource: CSCA Math December 2025 Past Paper

Which of the following functions is odd?

0 of 5 right in this set

7Mastery check

5 questions; 4 right means you have it. No solutions until the end.

Odd and Even Functions

1Warm-up

2The idea: parity is symmetry of the graph

Symmetry → equation → name: two lines

Common functions at a glance

Combining functions

34 steps

Walking through $f(x)=x+\dfrac1x$

Choose first, then check: the test on 5 functions

$f(x)=x+\frac{1}{x}$

$f(x)=\sqrt{x}$

$f(x)=x^3+1$

$f(x)=\ln(x^{2})$

$f(x)=\cos(\sin x)$

4Worked examples

5Common mistakes

1. Asked for the wrong statement, choosing a true one

2. Judging parity by feel instead of computing $f(-x)$

3. Parity right, range or monotonicity wrong

4. An odd inner function taken to make the whole odd

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

1Warm-up

2The idea: parity is symmetry of the graph

Symmetry → equation → name: two lines

Common functions at a glance

Combining functions

34 steps

Walking through f(x)=x+1xf(x)=x+\dfrac1xf(x)=x+x1​

Choose first, then check: the test on 5 functions

f(x)=x+1xf(x)=x+\frac{1}{x}f(x)=x+x1​

f(x)=xf(x)=\sqrt{x}f(x)=x​

f(x)=x3+1f(x)=x^3+1f(x)=x3+1

f(x)=ln⁡(x2)f(x)=\ln(x^{2})f(x)=ln(x2)

f(x)=cos⁡(sin⁡x)f(x)=\cos(\sin x)f(x)=cos(sinx)

4Worked examples

5Common mistakes

1. Asked for the wrong statement, choosing a true one

2. Judging parity by feel instead of computing f(−x)f(-x)f(−x)

3. Parity right, range or monotonicity wrong

4. An odd inner function taken to make the whole odd

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

Walking through $f(x)=x+\dfrac1x$

$f(x)=x+\frac{1}{x}$

$f(x)=\sqrt{x}$

$f(x)=x^3+1$

$f(x)=\ln(x^{2})$

$f(x)=\cos(\sin x)$

2. Judging parity by feel instead of computing $f(-x)$