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 hand-side of the y-axis, then to complete the sketch, draw the mirror image on the left-side of the y-axis.

Y-axis

Right-hand side of f(x)

y = f(x)

Y-axis

Y-axis is an axis of symmetry

y = f(x)

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);

f(x) = x2 - 3

f(3) = (3)2 - 3

f(3) = 6

and f(x) = x2 - 3

f(-3) = (-3)2 - 3

f(-3) = 6

Now, f(-3) = f(3)Hence, f(x) = x2 - 3 is an even function.

But, is this 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

f(a) = (a)2 - 3

f(a) = a2 - 3

and f(x) = x2 - 3

f(-a) = (-a)2 - 3

f(-a) = a2 - 3

Now, f(-a) = f(a)Hence, f(x) = x2 - 3 is an even function.

Evaluate f(-3) in f(x);

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) = a2[a2 - 4]

g(-a) = g(a)

Hence, g(x) = x2(x2 – 4) is an even function.

g(-a) = (-a)2[(-a)2 - 4]

but, g(a) = a2(a2 - 4)

2) Show that is an even function

A show statement requires you to evaluate LHS and the RHS and show they are equal.

At x = a,

At x = -a,

Hence, f(-a) = f(a)

Therefore is an even function

f(x) is even if f(-x) = f(x) by definition

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 top-side 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

Top-side of y = f(x)

y = f(x)

Y-axisy = f(x)

Bottom-side of y = f(x)

180° rotation

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);

f(x) = x3 – 4x

f(4) = (4)3 – 4(4)

f(4) = 48

and f(x) = x3 – 4x

f(-4) = (-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

f(a) = (a)3 – 4(a)

f(a) = a3 – 4a

and f(x) = x3 - 4x

f(-a) = (-a)3 - 4(-a)

f(-a) = -a3 + 4a

Now, f(-a) = –f(a)Hence, f(x) = x3 - 4x is an odd function.

Evaluate f(-4) in f(x);

But, –f(a) = –(a3 – 4a)

f(-a) = –(a3 – 4a)

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.

2) Show that is an odd function.

f(x) is an odd function if f(-x) = – f(x)

At x = a

At x = -a

= – f(a)

Hence, f(-a) = –f(a),

Therefore is an odd function.

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 NEITHER.

(1)

(2)

(3)

(4)

(5)

Answer: neither

Answer: even

Answer: odd

Answer: even

Answer: odd