Properties of FunctionsA function, f, is defined as a rule which assigns each member of a set ‘A’ uniquely to a member of a set ‘B’.

A function f assigns exactly one value y to each x. We write y = f (x) orf: x y (‘f maps x to y’).

y is referred to as the image of x in f.

Set ‘A’ is referred to as the Domain of the function and the set ‘B’ as the co-domain.

The subset of ‘B’ which is the set of all images of the function is called the range of the function.

It is possible for f (a) = f (b) and yet a ≠ b.

Often restrictions have to be placed on these sets to provide a suitable domain and range.

1. Consider the function ( ) .

What would be a suitable domain and range?

f x x

is an unsuitable domain since negative values have no image in .

{0} is a suitable domain

{0} is a suitable range

12. Consider the function ( ) .

1What would be a suitable domain and range?

f xx

{1} is a suitable domain

{0} is a suitable range

The modulus functionThe modulus or absolute value of x is denoted by | x | and is defined as

when 0

when 0

x xx

x x

Properties of |x| include:

21. x x

2. x y x y

3. x y x y

4. x a a x a

5. or x a a x x a

y

x

y x

Inverse functionsThe graph of and inverse function f -1 (x) is obtained by reflecting f (x) in the line y = x

f -1 (x) is obtained by making x the subject of the formula.

For example find the inverse function of f (x) = 2x +1

2 1y x

2 1x y

12

yx

1 1( )

2x

f x

(2 1) 12

x

22x

x

1 1Note that ( ( )) (2 1)f f x f x

12. Find ( ) when ( ) .1

xf x f x

x

1x

yx

( 1)y x x

yx y x

yx x y

( 1)x y y

1y

xy

1( )1

xf x

x

1Here ( ) ( )f x f x

Placing restrictions on the domain and range can often ensure that a function has an inverse.

2 13. ( ) 4 1. Find ( ) and state a suitable domain and range for the function.f x x f x

24 1y x 24 1x y

2 14

yx

12

yx

1 1( )

2x

f x When x < 1, f -1 (x) is undefined

Hence largest suitable domain is x ≥ 1 giving a range y ≥ 0.

Domain: : 0,x x x

The domain and range of the inverse function give us the range and domain respectively of the function.

Range: : 1,y y y

Inverse Trig Functionsy

x

2

2

–

–

– 2

– 2

1

1

2

2

– 1

– 1

– 2

– 2

siny x

Reflecting the curve in y = x.

We need the largest region over which the curve is always increasing (or always decreasing).

y

x

0.5

0.5

1

1

1.5

1.5

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

2

2

–2

–2

1siny x

y

x

2

2

–

–

– 2

– 2

1

1

2

2

– 1

– 1

– 2

– 2

siny x

Domain: : 1 1,x x x

Range: : ,2 2

y y y

Odd and Even Functions

Even Functions

( ) ( )f x f x

Odd Functions

( ) ( )f x f x

22

11. Consider ( ) and ( )f x x f x

x

2( )f x x2( ) ( )f x x

2x

( ) ( )f x f x

Hence this is an even function.

2

1( )f x

x

2

1( )

( )f x

x

2

1x

( ) ( )f x f x

Hence this is an even function.

32. Consider ( )f x x

3( )f x x3( ) ( )f x x

3x

( ) ( )f x f x

Hence this is an odd function.

4 23. Prove ( ) 8 3 is an even function.f x x x

4 2( ) 8 3f x x x

4 2( ) ( ) 8( ) 3f x x x

4 28 3x x

( ) ( )f x f x

Hence this is an even function.

34. Prove ( ) 2 is an odd function.f x x x 3( ) 2f x x x

3( ) ( ) 2( )f x x x 3 2x x

( ) ( )f x f x

Hence this is an odd function.

Page 100 Exercise 3 Questions 1(a), (e), (g), (h), 2(e), (c), (g), 3Page 102 Exercise 4 Questions 1 and 2Page 108 Exercise 8 Question 3.

Vertical asymptotes

( )( ) is a rational function if it can be expressed in the form

( )g x

f xh x

where ( ) and ( ) are real polynomial functions and ( ) is of

degree 1 or greater.

g x h x h x

When h (a) = 0, the function is not defined at a.

As , ( ) and the function is discontinuous at .x a f x a

A function is said to be continuous if lim ( ) ( )x a

f x f a

11. How does ( ) behave in the neighbourhood of 1?

1f x x

x

As 1, the denominator, 1 0 and ( ) .x x f x

As 1 from the left, 1 0, as the numerator 0, ( ) 0x x f x

(to the left of the line x = 1, the curve f (x) ∞-, negative infinity).

As 1 from the right, 1 0, as the numerator 0, ( ) 0x x f x

(to the right of the line x = 1, the curve f (x) ∞+, positive infinity).

The line x = 1, is called a vertical asymptote of the function. The function is said to approach the line x = 1 asymptotically.

y

x

1

1

2

2

3

3

– 1

– 1

– 2

– 2

– 3

– 3

( )y f x

2 32. Consider ( )

( 4)( 1)x

f xx x

Vertical asymptotes occur at x = -4 and x = -1.

As 4x (-4 from the negative direction)

( )( )( )

y

As 4x

( )( )( )

y

As 1x

( )( )( )

y

As 1x

( )( )( )

y

( )Remember for ( )

( )g x

f xh x

Divide if the deg ( ) deg ( ) (partial fractions)g x h x

Page 109 Exercise 9 Question 1.

TJ Exercise 1.

Non vertical asymptotes

(i) If the degree of the numerator < degree of the denominator, divide each term of the numerator and denominator by the highest power of x.

( ) ( )If the function ( ) can be expressed as ( ) , where is

( ) ( )

a proper rational function, then,

m x m xf x g x

n x n x

as , ( ) ,x n x ( )

0( )

m xn x

and ( ) ( ).f x g x

If g(x) is a linear function, it is known as either a horizontal asymptote or an oblique asymptote depending on the gradient of the line y = g(x).

The behaviour of the function as x ∞+ and as x ∞- can be considered for each particular case.

(ii) If the degree of the numerator = degree of the denominator or the degree of the numerator > degree of the denominator, divide the numerator by the denominator.

2

2 31. Find the non vertical asymptote for the function ( )

5 4x

f xx x

2

2 3( )

5 4x

f xx x

2 2

2

2 2 2

2 3

5 4

xx x

x xx x x

2

2

2 3

5 41

x x

x x

0As , ( ) 0

1x f x

0As , ( ) 0

1x f x

Hence equation of asymptote is y = 0.

2

2

2 3

( )5 4

1

x xf x

x x

Let us now consider how the curve approaches y = 0.

As , ( ) 01

x f x

As , ( ) 01

x f x

2

2

1. Find the non vertical asymptote and determine the behaviour of the curve

2 1for the function ( )

5 4x x

f xx x

The degree of the numerator = the degree of the denominator so we divide.

2 25 4 2 1x x x x 1

2 5 4x x 3 3x

2

2 2

2 1 3 3( ) 1

5 4 5 4x x x

f xx x x x

2

2

3 3

15 4

1

x x

x x

0As , ( ) 1 1

1x f x Hence y = 1 is a horizontal asymptote.

As , ( ) 1 11

x f x

As , ( ) 1 11

x f x

1y

1y

2

1. Find the non vertical asymptote and determine the behaviour of the curve

4 3for the function ( )

2x x

f xx

The degree of the numerator > the degree of the denominator so we divide.

22 4 3x x x x

2 2x x2 3x

1( ) 2

2f x x

x

1

22

1

xx

x

0As , ( ) 2 2

1x f x x x Hence y = x + 2 is an oblique asymptote.

As , ( ) 2 21

x f x x x

As , ( ) 2 21

x f x x x

( 2)y x

( 2)y x

+2

2 4x -1

Page 110 Exercise 10 Questions 1(a), (b), (g), (f), (k), (l).

TJ Exercise 2

Curve SketchingBringing it all together

When sketching curves, gather as much information as possible form the list below.

1. Intercepts.

(a) Where does the curve cut the y axis? x = 0(b) Where does the curve cut the x axis? y = 0

2. Extras.

(a) Where is the function increasing? f `(x) > 0(b) Where is the function decreasing? f `(x) < 0

3. Stationary Points.

(a) Where is the curve stationary? f `(x) = 0(b) Are there points of inflexion? f ``(x) = 0

4. Asymptotes.

(a) Where are the vertical asymptotes? denominator = 0

(b) Where are the non vertical asymptotes? x ± ∞

22 11. Sketch the graph of

1x x

yx

(2 1)( 1)1

x xy

x

When 0, 1 (0,1)x y

1 1When y 0, and 1 1,0 , ,0

2 2x

Using the quotient rule.

2( ) 2 1 ( ) 1

`( ) 4 1 `( ) 1

f x x x g x x

f x x g x

2

2

(4 1)( 1) (2 1)( 1)

dy x x x xdx x

2

2 ( 2)( 1)x xx

0 for stationary points.dydx

2

2 ( 2)0

( 1)x xx

2 ( 2) 0x x

0 and 2x x

When 0, 1 (0,1)

When 2, 9 (2,9)

x y

x y

Using the quotient rule for the second derivative.

2 2( ) 2 4 ( ) ( 1)

`( ) 4 4 `( ) 2( 1)

f x x x g x x

f x x g x x

2 2 2

2 4

(4 4)( 1) 2( 1)(2 4 )( 1)

d y x x x x xdx x

``(0) 0 Max T.P. @ (0,1)f

``(2) 0 Min T.P. @ (2,9)f

Asymptotes

Vertical asymptote at x = 1.

As 1 , ( ) 0,x f x

(2 1)( 1)( )

1x x

f xx

( )f x

As 1 , ( ) 0,x f x

( )f x

Non vertical asymptotes

The degree of the numerator > the degree of the denominator so we divide.

21 2 1x x x 2x

22 2x x3 1x

+3

3 3x 2

2( ) 2 3

1f x x

x

2

2 31

1

xx

x

Non vertical asymptote at y = 2x + 3

2

2 31

1

xy x

x

As , ( ) 2 3 2 31

x f x x x

As , ( ) 2 3 2 31

x f x x x

(2 3) (above the line)y x

(2 3) (below the line)y x

Now lets put this information on a set of axis.

y

x

5

5

10

10

– 5

– 5

– 10

– 10

5

5

10

10

15

15

20

20

25

25

30

30

– 5

– 5

– 10

– 10

-1 ½ 1

3

(2,9)

(0,1)

You should also be able to do the following from higher.

From f(x), you should be able to sketch

f(x - a), f(x + a), a f(x), f(x) ± a, f `(x), f -1(x) etc.

Page 112 Exercise 11 Questions 1(a), (c), (e), (g), (i), (k), (m)Page 114 Exercise 12A

TJ Exercise 3, 4 and 5

Do the review on Page 117