# 1. Introduction

The concept of a function is essential in mathematics. There are two common notations in use:

 (a) ƒ(x) = x2 + 2, and (b) ƒ : x x2 + 2.

Form (a) is commonly used. Form (b) is interpreted as:

 the function ƒ maps x to x2 + 2

Example 1: If two functions are given as ƒ(x) = 2x + 3 and g(x) = 3 − x2 , then:

(a) ƒ(2) = 2 × 2 + 3 = 7
(d) ƒ(−3) = 2 × (−3) + 3 = −3
(c) g(0) = 3 − (0)2 = 3
(d) g(4) = 3 − (4)2 = 3 − 16 = −13

Example 2: Find the numbers which map to zero under the function:

h : x x2 − 9

Solution: The function can also be written as h(x) = x2 − 9 and if x maps to zero then h(x) = 0, i.e.:

x2 − 9  x2 = 0 = 9

The numbers are 3 and −3, as when squared they both equal 9.

Exercise 1: Two functions are given as: h : x x2 − 4 and g : x 10x + 5 . Find the following:

(a) h(1)

Solution: The function is h(x) = x2 − 4 so:

h(1) = 12 − 4 = 1 − 4 = −3
(b) h(-2)

Solution: The function is h(x) = x2 − 4 so:

h(-2) = -22 − 4 = 4 − 4 = 0
(c) h(0)

Solution:The function is h(x) = x2 − 4 so:

h(0) = (3)2 − 4 = 0 − 4 = −4
(d) g(3)

Solution: The function is g(x) = 10x + 5 so:

g(3) = 10 × (3) + 5 = 30 + 5 = 35
(e) g(−1)

Solution: The function is g(x) = 10x + 5 so:

g(−1) = 10 × (−1) + 5 = −10 + 5 = −5
(f) the values of x such that h(x) = 12

Solution: If h(x) = 12 , then since h(x) = x2 − 4 :

x2 − 4  x2 = 12 = 12 + 4 = 16 = ±4

Click on questions to reveal their solutions

# 2. Flow Diagrams

The function ƒ(x) = 2x + 3 in Example 1 may be represented as a flow diagram:

x multiply by 2 2x add 3 2x + 3

From the diagram it is clear that the order of the operations cannot be confused: first multiply by 2 and then add 3.

Example 3: Draw a ﬂow diagram for each of the following functions:

(a) k : x (3x − 2)2

Solution:

x multiply by 3 3x subtract 2 3x − 2 square (3x − 2)2
(b) g : x 4x − 2 / 3

Solution:

x multiply by 4 4x subtract 2 4x − 2 divide by 3 4x − 2 / 3

The exercise below is designed to give you some practice at addition: and subtraction of fractions.

Exercise 2: Draw flow diagrams for each of the following functions:

(a) h : x 6x + 1

Solution: For h : x 6x + 1 the flow diagram is:

x multiply by 6 6x add 1 6x + 1
(b) h : x 4(3 − 2x)

Solution: For h : x 4(3 − 2x) the flow diagram is:

x multiply by −2 −2x add 3 3 − 2x multiply by 4 4(3 − 2x)
(c) h : x (2x − 5)2

Solution: For h : x (2x − 5)2 the flow diagram is:

x multiply by 2 2x subtract 5 2x − 5 square (2x − 5)2
(d) g : x 3x2 − 4

Solution:

x square x2 multiply by 3 3x2 subtract 4 3x2 − 4
(e) g : x 2x2 /3 + 5

Solution:

x square x2 multiply by 2 2x2 divide by 3 2x2 /3 add 5 2x2 /3 + 5
(f) g : x x2 + 2

Solution:

x square x2 add 2 x2 + 2 take square root x2 + 2

Click on questions to reveal their solutions

Now try this short quiz:

Quiz 1: From the functions listed below choose the one which is given by the following ﬂow diagram:

x cube   divide by 4   subtract 1   square   add 4
(a) ( x3 / 4 − 1 ) 2 + 4 Correct - well done!
(b) ( x3 / 4 − (1)2 ) 2 + 4 Incorrect - please try again!
(c) ( x3 / 4 − 1 ) 2 + 4 Incorrect - please try again!
(d) ( ( x / 4 ) 3 + (1)2 ) + 4 Incorrect - please try again!

Explanation: The flow diagram:

x cube   divide by 4   subtract 1   square   add 4

is completed as:

x cube x3 divide by 4 x3/4 subtract 1 x3/4 − 1 square ( x3/4 − 1)2 add 4 ( x3/4 − 1)2 + 4

- which is the function:

k : ( x3 / 4 − 1 ) 2 + 4

# Composite Functions

The function ƒ(x) = 2x + 3 , from Example 1, is composed of two simpler functions, i.e. multiply by 2 and add 3. If these two functions are written as h : 2x and g : x + 3 then the composition of these two functions is written gh (sometimes as gοh or g (h (x))) .

Example 4: If h : 2x2 and g : x + 5 , find the composite function gh

Solution: Applying first h and then g results in the composite function:

gh = 2x2 + 5

This can best be seen by using a flow diagram:

 x square x2 multiply by 2 2x2 add 5 2x2 + 5 square root √2x2 + 5 h g
Note: The composition of two functions, fg , is NOT the same as their product.

Exercise 3: Evaluate each of the following in their lowest terms.

(a) ƒ(x) = 2x + 1, g(x) = x − 3

Solution: For the functions: ƒ(x) = 2x + 1 and g(x) = x − 3 the function ƒg is:

ƒg : x 2(x − 3) + 1 = 2x − 6 + 1 = 2x − 5

The flow diagram is:

 x subtract 3 x − 3 multiply by 2 2(x − 3) add 1 2(x − 3) + 1 g f

The function fg can also be determined as follows. The two functions can be written as ƒ(z) = 2z + 1 and g(x) = x − 3. Then, by substituting z = g(x) into ƒ(z) = 2z + 1:

ƒg(x) = ƒ(g(x)) = 2g(x) + 1 = 2(x − 3) + 1 = 2x − 5
(b) ƒ : x 2x − 1, g : x x2

Solution: For the functions: ƒ :  x 2x − 1 and g : x x2 the function ƒg is:

ƒg : x 2x2 − 1

The flow diagram is:

 x square x2 multiply by 2 2x2 subtract 1 2x2 − 1 g f

The composition may also be determined by writingƒ(z) = 2z − 1 and g(x) = x2 and by substituting z = g(x) obtaining:

ƒg(x) = ƒ(g(x)) = 2g(x) + 1 = 2(x2) − 1 = 2x2 − 5
(c) ƒ(x) = x2, g(x) = 2x −1

Solution: For the functions: ƒ : x  x2 and g : x 2x − 1 the function ƒg is:

ƒg : x (2x− 1)2

The flow diagram is:

 x multiply by 2 2x subtract 1 2x − 1 square (2x − 1)2 g f

Alternatively, writingƒ(z) = z2 and g(x) = 2x − 1, then substituting for z = g(x):

ƒg(x) = ƒ(g(x)) = (g(x))2 = (2x − 1)2
(d) ƒ : x x + 3, g : x x − 3

Solution: For the functions: ƒ : x x + 3 and g : x x − 3 the function ƒg is:

ƒg : x x

The flow diagram is:

 x subtract 3 x − 3 add 3 (x − 3) + 3 g f

The function which maps x to x is called the identity function. The identity function ƒ : x x does not change the value of x.

(e) ƒ : x x/3 − 2, g : x 3x2

Solution: For the functions: ƒ : x x/3 − 2 andg : x 3x2 the function ƒg is:

ƒg : x x2 − 1

The flow diagram is:

 x square x2 multiply by 3 3x2 divide by 3 x2 subtract 2 x2 − 2 g f

Writingƒ(x) = z/3 and g(x) = 3x2, and substituting z = g(x):

ƒg(x) = ƒ(g(x)) = g(x)/3 − 2 = 3x2/3 − 2 = x2 − 2
(f) ƒ : x 3x2, g : x x/3 − 2

Solution: For the functions: ƒ : x 3x2 and g : x x/3 − 2 the function ƒg is:

ƒg : x 3(x/3 − 2 )2

The flow diagram is:

 x divide by 3 x/3 subtract 2 x/3 − 2 square (x/3 − 2)2 multiply by 3 3(x/3 − 2)2 g f

Alternatively, the functions are ƒ(z) = 3z2 and g(x) = x/3 − 2. Thus:

ƒg(x) = ƒ(g(x)) = 3(g(x))2 = 3(x/3 − 2)2

Click on questions to reveal their solutions

The functions in question (b) above are reversed in question (c). The results show that, in general, reversing the order of two functions changes the composite function. In simpler terms, if h and k are two functions then, in general, hkkh.

Quiz 2: Two functions are given as: ƒ : x 2x2 − 1 and g : x x − 3. Which of the following is a solution to ƒ(g(x)) = 7?

(a) x = 5 Correct - well done!
(b) x = −5 Incorrect - please try again!
(c) x = 2 Incorrect - please try again!
(d) x = −2 Incorrect - please try again!

Explanation: For the functions ƒ : x 2x2 − 1 and g : x x − 3, the composite function is ƒg(x) = 2(x − 3)2 − 1:

 2(x − 3)2 − 1 = 7 2(x − 3)2 = 7 + 1 = 8 (x − 3)2 = 4 (x − 3) = ±2 x = 3 ± 2

- so that x = 5 and x = 1 are both solutions.

# 3. Inverse Functions

If a function ƒ maps m to n, then the inverse function, written as ƒ−1, maps n to m.

Example 5: Find the inverse of the function: h : x 4x − 3 / 2

Solution: First draw a flow diagram for the function:

x multiply by 4 4x subtract 3 4x − 3 divide by 2 4x − 3/2

Now draw a flow diagram, starting from the right, with each operation replaced by its inverse:

2x + 3/4 divide by 42x + 3 add 3 2xmultiply by 2x

The inverse of h : x 4x − 3 / 2 is thus h−1 : x 2x + 3 / 4

If a function ƒ has an inverse ƒ−1 then the composite function ƒƒ−1 is the identity function which was mentioned in Exercise 3(d), i.e. ƒƒ−1 : x x. It is also true that ƒ−1ƒ : x x.

Example 6: For the function h in Example 5, show that the composite function hh−1 is the identity function.

Solution: First note that h−1 : x 2x + 3 / 4. For the composition hh−1, therefore, this must be operated on by the function h, i.e. in the first flow diagram of Example 5 the input on the right hand side must be 2x + 3 / 4.

 x h−1(x) 2x + 3/4 multiply by 4 2x + 3 subtract 3 2x divide by 2 x h(x) = 4x − 3 / 2

Quiz 3: Two functions are given as: ƒ : x 1 / 2 x and g : x 3x + 16. If h = ƒg, which of the following is h−1?

(a) h−1 : x 1 / 3 (2x + 16) Incorrect - please try again!
(b) h−1 : 3(2x + 16) Incorrect - please try again!
(c) h−1 : x 1 / 3 (2x − 16) Correct - well done!
(d) h−1 : 2( x / 3 − 16) Incorrect - please try again!

Explanation: The function h = ƒg is h : x 1 / 2(3x + 16). This is the function obtained by simplifying the function of Exercise 4(e). The inverse of the function is thus:

h−1 : x 1 / 3(2x − 16)

# 4. Quiz on Functions (need to check answers)

If three functions are given as: ƒ : x x2, g : x 4x, h : x x + 5, choose the correct options for the following

1. ƒg : x ...
(a) 8x2
(b) 4x2
(c) 16x
(d) 16x2
2. gƒ : x ...
(a) 8x2
(b) 4x2
(c) 16x
(d) 16x2
3. gh : x ...
(a) 5 + 4x
(b) 4 + 5x
(c) 4x + 5
(d) 4(x + 5)
4. (hg)−1 : x ...
(a) 1/4(x − 5)
(b) 1/4(x + 5)
(c) 1/4(x + 4)
(d) 1/4(x − 4)