Week 1 · Day 4 Lesson 1.10 About 10 minutes Free

Odd and even functions

By the end: Decide whether a function is even, odd or neither, use the symmetry to find unknown values and parameters, and spot the domain trap.

Understand it

Parity describes a symmetry of the graph, and it is tested with one substitution: replace $x$ by $-x$ .

Step 1: check the domain. A function can be even or odd only if its domain is symmetric about $0$ : whenever $x$ is in the domain, so is $-x$ . The domain $[-2,2]$ is symmetric. The domain $[-1,2]$ or $[-2,3)$ is not, and then the function is neither even nor odd, whatever the formula looks like.

Step 2: compare $f(-x)$ with $f(x)$ .

If $f(-x)=f(x)$ for every $x$ , the function is even. Its graph is symmetric about the $y$ -axis.
If $f(-x)=-f(x)$ for every $x$ , the function is odd. Its graph is symmetric about the origin: turning it half a turn about the origin leaves it unchanged.
If neither holds, it is neither. Most functions are neither.

Quick recognition. $x^2$ , $x^4$ , $|x|$ , $\cos x$ and constants are even. $x$ , $x^3$ , $\dfrac1x$ and $\sin x$ are odd. For a polynomial: all exponents even gives an even function, all exponents odd gives an odd function, and a mix (such as $x^2+x$ ) gives neither. A constant $5$ is even, and the function $0$ is both even and odd.

Combining. Even $\pm$ even is even. Odd $\pm$ odd is odd. Even $\times$ even and odd $\times$ odd are even. Even $\times$ odd is odd.

Using oddness. If $f$ is odd, then $f(-a)=-f(a)$ , so one value gives another. If $f$ is odd and $0$ is in its domain, then $f(0)=0$ (but $\frac1x$ is odd and has no value at $0$ ). The classic exam trick is a function of the form “odd part plus a constant”, for example $f(x)=x^5+ax^3+bx-8$ : the part $x^5+ax^3+bx$ is odd, so you only need to deal with the constant.

Essential formulas

\text{even: } f(-x)=f(x)\qquad \text{odd: } f(-x)=-f(x)

These must hold for every

x

in the domain, and the domain must be symmetric about

0

: if

x

is in it, so is

-x

f\ \text{odd and defined at }0\ \Rightarrow\ f(0)=0

Put

x=0

f(-x)=-f(x)

to get

f(0)=-f(0)

, so

f(0)=0

. This does not apply if

0

is not in the domain.

\text{even}\times\text{even}=\text{even},\quad \text{odd}\times\text{odd}=\text{even},\quad \text{even}\times\text{odd}=\text{odd}

Sums follow the same type only when both functions have the same type: even + even is even, odd + odd is odd.

See it

y = x² is even: f(−1) = f(1) = 1. y = x³ is odd: f(−1) = −1 = −f(1).

Worked examples

Worked example 1

Decide whether each function is even, odd or neither: (a) $f(x)=x^3-x$ ; (b) $g(x)=x^2+x$ ; (c) $h(x)=x^2$ on $[-2,3)$ .

(a) $f(-x)=(-x)^3-(-x)=-x^3+x=-(x^3-x)=-f(x)$ . All exponents are odd, so $f$ is odd.
(b) $g(-x)=x^2-x$ . This is neither $g(x)=x^2+x$ nor $-g(x)=-x^2-x$ (check: $g(1)=2$ , $g(-1)=0$ ). So $g$ is neither.
(c) The formula $x^2$ looks even, but the domain $[-2,3)$ is not symmetric: $x=2.5$ is allowed while $-2.5$ is not. A function on a non-symmetric domain is neither even nor odd.

Answer: (a) odd, (b) neither, (c) neither.

Worked example 2

Let $f(x)=x^5+ax^3+bx-8$ with $f(-2)=10$ . Find $f(2)$ .

Let $g(x)=x^5+ax^3+bx$ . All its exponents are odd, so $g$ is odd, and $f(x)=g(x)-8$ .
$f(-2)=g(-2)-8=10$ , so $g(-2)=18$ .
Because $g$ is odd, $g(2)=-g(-2)=-18$ .
So $f(2)=g(2)-8=-18-8=-26$ .

Answer: $-26$

Worked example 3

$f$ is an odd function on $\mathbb{R}$ , and $f(x)=x^2+2x$ for $x>0$ . Find $f(-3)$ , $f(0)$ and $f(x)$ for $x<0$ .

$f(0)=0$ , because $f$ is odd and defined at $0$ .
$f(-3)=-f(3)=-(9+6)=-15$ .
For $x<0$ we have $-x>0$ , so $f(-x)=(-x)^2+2(-x)=x^2-2x$ . By oddness, $f(x)=-f(-x)=-(x^2-2x)=-x^2+2x$ .
Check: the formula for $x<0$ gives $f(-3)=-9-6=-15$ , the same as before.

Answer: $f(-3)=-15$ , $f(0)=0$ , and $f(x)=-x^2+2x$ for $x<0$ .

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

The function $f(x)=ax^2+bx+3a+b$ is even, and its domain is $[a-1,2a]$ . Find $a+b$ .

Hint 1 of 2

The domain of an even function must be symmetric about $0$ . What does that say about the two endpoints?

Hint 2 of 2

Then use $f(-x)=f(x)$ and compare the $x$ -terms.

Show the step-by-step solution

A symmetric domain $[a-1,2a]$ needs its endpoints to be opposites: $(a-1)+2a=0$ , so $a=\tfrac13$ .
$f(-x)=ax^2-bx+3a+b$ and $f(x)=ax^2+bx+3a+b$ must be equal for every $x$ , so $-bx=bx$ , which forces $b=0$ .
Then $a+b=\tfrac13+0=\tfrac13$ .

Answer: $\tfrac13$

Common mistakes

Checking $f(-x)=f(x)$ at a single value. It must hold for every $x$ in the domain.
Skipping the domain. $x^2$ on $[-1,2]$ is not even.
Assuming $f(0)=0$ for every odd function. $\dfrac1x$ is odd and $f(0)$ does not exist.
Forgetting that most functions, such as $x^2+x$ or $x+1$ , are neither even nor odd.
Treating “the graph is symmetric about the $y$ -axis” and “about the origin” as the same thing. The first is even, the second is odd.

Exam tips

First look at the domain. If it is not symmetric about $0$ , the answer is “neither” immediately.
For polynomials, look at the exponents. A mix of even and odd exponents means neither.
For “odd part plus a constant” questions, define the odd part as $g(x)$ , use $g(-a)=-g(a)$ , then put the constant back.
For parameter questions, write $f(-x)$ , set it equal to $f(x)$ (or $-f(x)$ ), and compare the terms with the same power of $x$ .

In short

Domain symmetric about $0$ first; then $f(-x)=f(x)$ is even and $f(-x)=-f(x)$ is odd.
Even is symmetric about the $y$ -axis; odd is symmetric about the origin.
Odd means $f(-a)=-f(a)$ , and $f(0)=0$ if $0$ is in the domain.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

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