Functions 1
Odd and Even Functions
By Mr Porter
Even Functions
A function, f(x), is an even function if f(x) = f(-x) , for all x
in its domain.
An even function has the important property line symmetry
with the y-axis as it axis. This allows us to plot the right handside of the y-axis, then to complete the sketch, draw the mirror
image on the left-side of the y-axis.
Y-axis
Y-axis
y = f(x)
y = f(x)
Right-hand
side of f(x)
Y-axis is
an axis of
symmetry
Testing for an Even Function
To test whether a function is even or odd or neither, work with f(-x) and compare it to f(x).
Usually, it is also better to use x = a and x = -a as the two values for comparison.
Examples
Show that f(x) = x2 - 3 is an even function.
Lets use numerical values x = 3 and x = -3.
Evaluate f(3) in f(x); Evaluate f(-3) in f(x);
f(x) =
x2
f(3) =
(3)2
f(3) = 6
-3
-3
and
f(x) =
f(-3) =
x2
-3
(-3)2
-3
f(-3) = 6
Now, f(-3) = f(3)
Hence, f(x) = x2 - 3 is an even function.
But, is the same true for x = ±4, ±12, ..
To be certain, you should test every value of x!
This is not practical.
A better method, is to use algebraic
values of x, say x = ±a
f(x) = x2 - 3
and
f(x) = x2 - 3
f(a) = (a)2 - 3
f(-a) = (-a)2 - 3
f(a) = a2 - 3
f(-a) = a2 - 3
Now, f(-a) = f(a)
Hence, f(x) = x2 - 3 is an even function.
Examples
1) Prove that g(x) = x2(x2 - 4) is an
even function.
A prove statement requires you to start with the
LHS and arrive at the RHS.
g(x) is even if g(-x) = g(x) by definition
At x = a, g(a) = a2(a2 - 4)
At x = -a
g(-a) =
(-a)2[(-a)2
g(-a) = a2[a2 - 4]
- 4]
but, g(a) = a2(a2 - 4)
g(-a) = g(a)
Hence, g(x) = x2(x2 – 4) is an
even function.
2) Show that f (x)  x 2  4 is an
even function
A show statement requires you to evaluate LHS and
the RHS and show they are equal.

f(x) is even if f(-x) = f(x) by definition
At x = a,
f (a)  a 2  4
At x = -a, f (a)  [a]2  4

f (a)  a 2  4

Hence,
f(-a) = f(a)

Therefore f (x)  x 2  4 is an even
function

Odd Functions
A function, f(x), is an odd function if f(-x) = –f(x) , for all x
in its domain.
An odd function has the important property point symmetry,
180° rotation about the origin. This allows us to plot the topside of the x-axis, then to complete the sketch, draw the 180°
rotated mirror image on the bottom-side of the x-axis.
Y-axis
Y-axis
y = f(x)
y = f(x)
Top-side of
y = f(x)
180° rotation
Bottom-side of y = f(x)
Testing for an Odd Function
To test whether a function is even or odd or neither, work with f(-x) and compare it to f(x).
Usually, it is also better to use x = a and x = -a as the two values for comparison.
Examples
Show that f(x) = x3 - 4x is an odd function.
Lets use numerical values x = 4 and x = -4.
Evaluate f(4) in f(x); Evaluate f(-4) in f(x);
– 4x and
f(x) =
x3
f(4) =
(4)3
f(4) = 48
– 4(4)
f(x) =
f(-4) =
x3
– 4x
(-4)3
– 4(-4)
f(-2) = -48
Now, f(-4) = – f(4)
Hence, f(x) = x3 - 4x is an odd function.
But, is the same true for x = ±4, ±12, ..
To be certain, you should test every value of x!
This is not practical.
A better method, is to use algebraic
values of x, say x = ±a
f(x) = x3 – 4x and
f(x) = x4 - 4x
f(a) = (a)4 – 4(a)
f(-a) = (-a)3 - 4(-a)
f(a) = a4 – 4a
f(-a) = -a3 + 4a
f(-a) = –(a3 – 4a)
But, –f(a) = –(a3 – 4a)
Now, f(-a) = –f(a)
Hence, f(x) = x3 - 4x is an odd function.
Examples
1) Prove that f(x) = 5x - x3 is an odd
function.
f(x) is an odd function if f(-x) = – f(x)
At x = a
f(x) = 5x - x3
f(a) = 5a - a3
At x = -a
f(x) = 5x - x3
f(-a) = 5(-a) - (-a)3
= -5a - -a3
= -5a + a3
= -(5a – a3)
= – f(a)
Hence, f(-a) = –f(a),
Therefore f(x) = 5x - x3 is an odd function.
x
2) Show that f (x)  2
is an odd
x 1
function.
f(x) is an odd function if f(-x) = – f(x)
x
f
(x)

At x= a
x 2 1
a
f (a)  2
a 1
x
f (x)  2
At 
x = -a
x 1
a
f (a) 

(a)2 1
a

f (a)  2
a 1
 a 

f (a)    2 
a 1
= – f(a)

Hence, f(-a) = –f(a),

Therefore
function.
f (x) 
x
x 2 1
is an odd
Example: Show that f(x) = x2 + x is neither an even or odd function.
To show or prove that a function is neither odd or even, you need to show both
the following two conditions.
I
f(x) ≠ f(-x)
II
f(-x) ≠ – f(x)
At x = a,
f(x) = x2 + x
f(a) = a2 + a
At x = -a,
f(x) = x2 + x
f(-a) = (-a)2 + (-a)
f(-a) = a2 – a
But,
– f(a) = –a2 – a
Now, f(a) ≠ f(-a) and f(-a) ≠ – f(a)
Hence, f(x) = x2 + x is neither an even or odd function.
Exercise
Show whether the following functions are EVEN, ODD or NEITHRER.
(1)
f (x)  4x 1
Answer: neither
(2)
h(x)  x 2  5
Answer: even
(3)
x2  5
g(x) 
x
Answer: odd
 (4)
f (x)  3x
Answer: even
 (5)
h(x)  x 3  2x
Answer: odd


