WEEK 5

FUNCTIONS AND GRAPHSBASIC OF FUNCTIONS

OBJECTIVES:

At the end of this session, you will be able to: Find the domain and range of a relation. Determine whether a relation is a function. Determine whether an equation represents a function. Evaluate a function. Find and simplify a function’s difference quotient. Understand and use piecewise functions. Find the domain of a function.

INDEX

1. Definitions

2. Relations

3. Domain and Range

4. Functions

5. Difference between Functions and Relations

6. Determining whether a Relation is a Function

7. Function as equation

8. Determining whether an Equation is a Function

9. Function Notation

10. Evaluating a Function

11. Functions and Difference Quotients

12. Piecewise Functions

13. Domain of a Function

14. Summary

1. DEFINITIONS

Let us recall some definitions:

Set: A set is a collection of objects whose contents can be clearly determined.The objects which belong to the set, are called elements or members of the set.

For example: Set of natural numbers less than 5 i.e

Set = {1, 2, 3, 4} Elements = 1, 2, 3, 4

Set of vowels in English alphabet i.e Set = {a, e, i, o, u}

Set of letters of the word MATHEMATICS i.e Set = {M, A, T, H, E, I, C, S}.

Ordered pairs: By an ordered pair we mean any two elements a and b, listed in a specific order. Its denoted by (a, b).

For example: (4, 2), (-5, -3)

2. RELATION

Let us understand what is a relation?

Consider set A which gives the names of 4 people.

i.e. A = { John, Paul, Danny, Mario}

Consider set B which gives their age.

i.e. B = { 20, 25, 30, 35}

From the sets A and B if we want to relate the name of the person with his age we get a set {(John, 20), (Paul, 25), (Danny, 30), (Mario, 35)}. This set shows the relationship between the person and his age. We can call this set as a Relation between the name and age.

So, we can say that a Relation is a relationship between sets of information.

Thus we can define Relation as any set of ordered pairs.

3. DOMAIN AND RANGE

Domain is the set of all first components of the ordered pairs of the relation.

Range is the set of all second components of the ordered pairs of the relation.

Example:

Consider the relation {(x1,y1), (x2,y2), (x3,y3)}.

Domain is the set of all first components. The domain of the above relation is {x1, x2, x3}

Range is the set of all second components. The range of the above relation is {y1, y2, y3}

x1

x2

x3

y1

y2

y3

Domain Range

Relation:{(x1,y1), (x2,y2), (x3,y3)}

4. FUNCTIONSConsider set A = {2, 3, 5, 6}

Consider set B as the set of squares of elements of set A.i.e Set B = {4, 9, 25, 36}

Now we have a relation {(2,4), (3, 9), (5,25), (6, 36)}.

Domain of the above relation = { 2, 3, 5, 6}

Range of the above relation = { 4, 9, 25, 36}

A relation in which each member of the domain corresponds to exactly one member of the range is called a function.

We observe from the figure that each element of the domain corresponds to exactly one member of the range, so this relation is a function.

Thus, the concept of function is special case of relation.

Domain Range

2

3

5

6

4

9

25

36

Relation:{(2,4), (3, 9), (5,25), (6, 36)}

5. DIFFERENCE BETWEEN FUNCTION AND RELATION

Relation1. A Relation may or may not

be a function.

2. Relation may relate one

member of the domain to

more than one member of

the range.

Example: Consider an example of relation which is not a function {(2, 4), (2, 6), (4, 9), (8, 12)}

Function1. All functions are relations.

Functions are sub-classification of relations

2. Function relates each member of the domain to one and only one member of the range i.e. for given x there is one and only one y.

Example: Consider an example of relation which is a function {(2, 4), (3, 6), (4, 8), (5, 10)}

2

3

4

5

4

68

10

2

4

8

46

9

12

5. DIFFERENCE BETWEEN FUNCTIONS AND RELATIONS (Cont….)

Now, lets find out if the following relation is a function:

{(1,2),(3,4),(5,6),(5,8)} 

A relation is a function only if each element of the domain corresponds to one and only one element of the range.

For a relation to be a function, two ordered pairs should have different first components but can have the same second component.

From the figure, we observe that the element 5 of the domain corresponds to two elements 6 and 8 of the range i.e the first component is same thus this relation is not a function

Domain Range

1

3

5 6

8

2

4

Relation: {(1,2),(3,4),(5,6),(5,8)} 

DIFFERENCE BETWEEN FUNCTIONS AND RELATIONS (Cont….)

Consider another example:

{(1, 2), (2, 4), (3, 6)}

Now we know that a relation is a function only if each element of the domain corresponds to one and only one element of the range.

From the figure, we observe that each element of the domain corresponds to only one member of the range i,e first component is different. So, this relation is a function.

Domain Range

1

2

3 6

2

4

Relation:{(1, 2), (2, 4), (3, 6)}

5. DIFFERENCE BETWEEN FUNCTIONS AND RELATIONS (Cont….)

Let us take another example:

{(1,2), (3,4), (6,5), (8,5)}  The figure shows that every

element in the domain corresponds to exactly one element in the range.

The element 1 in the domain corresponds to exactly one element 2 in the range. Similarly, element 3 of the domain corresponds to 4 of the range, 6 corresponds to 5 and 8 corresponds to 5.

Here it is important to note that first component is different, however the second component can be same as stated in previous slide thus, this relation is a function.

Domain Range

1

3

6 5

2

4

8

Relation: {(1,2), (3,4), (6,5), (8,5)} 

6. FUNCTIONS AS EQUATIONDo you know that equations can also be functions?

Well, functions are usually given in terms of equations rather than as sets of ordered pairs.

For example: Consider an equation y = 4 – x2

From the given equation we can see that for each value of x, there is one and only one value of y, so we say that the variable y is a function of the variable x.

Assuming different values of x, we compute the value of y and we get the table (a).

Thus, we can see that for each value of x there is one and only one value of y. So, the given equation is a function.

The variable x is called the independent variable because it can be assigned any value from the domain.

The variable y is the dependent variable because it’s value depends on x.

x y

0 4

1 3

2 0

(a)

7. DETERMINING WHETHER AN EQUATION REPRESENTS A FUNCTION

As we have already seen that not all relations are functions. Similarly, not all equations with variables x and y define a function.

An equation represents a function only if for each value of x, there is one and only one value of y.

We solve the given equation for y in terms of x. If two or more values of y can be obtained for a given x, then the given equation is not a function.

Lets take another example:

x2 + y2 = 4 (Given equation describes a circle)

x2 + y2 - x2 = 4 - x2 (Isolate y2 by subtracting x2 from both

sides)

y2 = 4 - x2 (Simplify)

y =

24 x

7. DETERMINING WHETHER AN EQUATION REPRESENTS A FUNCTION (Cont…)

The in this last equation tells us that for certain values of x,

there are two values of y.

For instance, if we assume x = 1,

then y = (Substituting x = 0 in the given

equation)

y = (Simplify)

Thus the given equation does not define y as a function of x.

24 1

3

8. FUNCTION NOTATION

When an equation represents a function, the function is often named by a letter such as f, g, h, F, G, or H. Any letter can be used to name a function.

Suppose that f names a function. Think of domain as the set of the function’s input and the range as the set of function’s output.

The special notation f(x), read “ f of x” or “ f at x”, represents the value of the function at the number x.

Example: f(x) = 4 – x2 . This equation is read as “f of x equals 4 – x2”.

IMPORTANT: Parentheses until now, always indicated multiplication, but in the function notation parentheses do not indicate multiplication. The notation f(x) does not mean “f times x”. f(x) simply means “ plug in a value for x”, it does not mean “multiply f and x”.

8. FUNCTION NOTATION (Cont…)

Now lets understand function notation with the help of an example:

We have already seen that the equation y = 6 – 2x defines y as a function of x.

If a function is named f and x represents the independent variable, then the notation f(x) gives the y-value for a given x.

Thus, f(x) = 6 – 2x and y = 6 – 2x define the same function. This function can be written as y = f(x) = 6 – 2x.

To find the value of the function at 1, we simply substitute 1 for x in the given equation.When we do this, we are evaluating the function at 1.

f(x) = 6 – 2x (This is the given function)

f(1) = 6 – 2(1) (Replace x by 1 in the given equation)

f(1) = 6 – 2 (Simplify)

f(1) = 4 (Subtract)

The statement f(1) = 4, read as “ f of 1 equals 4”, tells us that the value of the function at 1 is 4.

When the function’s input is 1, its output is 4.

9. EVALUATING A FUNCTION

We will understand how to evaluate a function with the help of an example.Given f(x) = x2 - 2x +7, and we have to find f(-5).

In order to evaluate f(-5) for the given definition of f, we simply replace x by -5 in the given equation and simplify.

For instance, in the given equation, we have f(x) = x2 - 2x + 7

f (-5) = (-5)2 - 2(-5) + 7 ( Replace x by -5) f (-5) = 25 + 10 + 7 (-5 multiplied by –5 = +25) f (-5) = 42 (add)

Given f(x) = x2 - 2x +7, and now we have to find f(x + 4). In order to evaluate f(x + 4) for the given equation, we replace x by (x + 4)

in the equation. In the given equation we have, f(x) = x2 - 2x + 7 f (x + 4) = (x + 4)2 - 2(x + 4) + 7 ( Replace x by (x + 4))

f (x + 4) = x2 + 8x + 16 – 2x – 8 + 7 (Square x + 4 using (A + B)2 = A2 + 2AB + B2 and distribute 2 throughout the parentheses)

f (x + 4) = x2 + 6x + 15 (Combining like terms)

10. FUNCTIONS AND DIFFERENCE QUOTIENTS

Now we will understand what is a difference quotient of a function:The ratio, called difference quotient helps us in understanding the rate at which functions change.The expression for h 0,

is called the difference quotient.

Steps for evaluating and simplifying a difference quotient: For a given function f(x), we replace x by x + h. Find f(x + h) – f(x). Open parentheses and apply (A +B)2 = A2 + 2AB + B2

Combine like terms and simplify. Divide both sides of the expression by h. Factoring h from the numerator. Cancel identical factors of h in the numerator and denominator.

( ) ( )f x h f x

h

11. EVALUATING DIFFERENCE QUOTIENT

Lets understand this with help of an example:

Find and simplify difference quotient

for f(x) = x2 – 7x + 3. …………(1)

Find f(x + h) by replacing x by x + h each time that x appears in the equation.

f(x+h) = (x + h)2 – 7(x + h) + 3 …………..(2)

Find f(x + h) – f(x) by subtracting equations (1) and (2)

f(x + h) – f(x) = (x + h)2 – 7(x + h) + 3 – (x2 – 7x + 3).

Open the parentheses and apply (A +B)2 = A2 + 2AB + B2.

f(x + h) – f(x) = (x2 + 2xh + h2) – 7x –7h + 3 - x2 + 7x - 3

Combine like terms and simplify

f(x + h) – f(x) = (x2 - x2) + (-7x + 7x) + (3 - 3) + 2xh + h2 –7h

f(x + h) – f(x) = 2xh + h2 – 7h

( ) ( )f x h f x

h

11. EVALUATING DIFFERENCE QUOTIENT(Cont…)

Dividing by h on both sides

f(x + h) – f(x) = 2xh + h2 – 7h

h h

Factor h from numerator

f(x + h) – f(x) = h (2x + h – 7)

h h

Cancel identical factors of h in the numerator and denominator

f(x + h) – f(x) = (2x + h – 7)

h

So, the value of difference quotient f(x + h) – f(x) for x2 – 7x + 3 is (2x + h – 7)

h

12. PIECEWISE FUNCTIONS

A function that is defines by two (or more) equations over a specified domain is called a piecewise function.

Usual Notation for piecewise functions is f(x) = f1(x) if x a

f2(x) if x > a

indicating a function whose value is f1(x) for x meeting the first condition(x a) and f2(x) for meeting the second condition (x > a). The conditions of the equation can vary but should not overlap.

Example: f(x) = 6x – 1 if x < 0

7x +3 if x 0

In this example if x < 0, then the value of the function f(x) = 6x – 1 and if

x 0, then the value of the function is 7x + 3

Thus we can say that piecewise functions are functions which are defined over a sequence of intervals.

12. PIECEWISE FUNCTIONS(Cont…)Example: f(x) = 2 x 2 - 1, x < 1

x + 4, x 1 As you can see the function is split into two halves, which half of the equation

we use depends on the value of x.

For instance, if we evaluate f(-1) then we substitute the values in the first half of the equation as –1 < 1.

f(x) = 2x2 – 1

f(-1) = 2 (-1) 2 – 1 (Replace x by –1 in first half of the equation)

f(-1) = 2 . 1 – 1 ( (-1)2 = 1)

f(-1) = 2 – 1 = 1 (Simplify)

Next, if we evaluate f(2), then we substitute the values of x in the second half of the equation as 2 > 1.

f(x) = x + 4

f(2) = 2 + 4 (Replace x by 2 in the second half of the equation)

f(2) = 6 (add)

13. DOMAIN OF A FUNCTION

Domain of a function is the largest set of real numbers for which the value

of f(x) is a real number.

We take care of following points while finding domain of a function:

Exclude from a function’s domain real numbers that cause division by

zero.

Exclude from a functions domain real numbers that result in an even

root of a negative number.

Example: Consider a function g(x) =

The equation tells us to take square root of x.

The expression under the radical sign, x, must be greater than or

equal to 0, because the square root of negative numbers is an

imaginary number.

Hence, the domain of g is {x| x 0}.

x

13. DOMAIN OF A FUNCTION ( Cont…)

Let us find the domain of the function f(x) = x2 + 3x – 17

The function f(x) = x2 + 3x – 17 contains neither division nor an even root. So the domain of the given function is a set of all real numbers.

Let us take another example, g(x) =

The given function g(x) = contains division.

As division by zero is undefined so we must exclude such values of x for which the expression x2 – 49 = 0 from the domain of the given function.

On solving the quadratic equation x2 – 49 = 0 for x we get the solution set as x = { -7, 7}

So, The denominator equals zero when x = -7 or x = 7, so we must exclude these values from the domain of the given function.

Thus, domain of g is {x| x -7, x 7}

2

5( )

( ) 49

x

x

2

5( )

( ) 49

x

x

14. SUMMARY

Let us recall what we have learnt so far:

A relation is any set of ordered pairs. The set of first components is the domain and the set of second components is the range.

A function is a correspondence from a first set called domain to a second set called range such that each element in the domain corresponds to exactly one element in the range.

A relation is not a function if any element of the domain corresponds to more than one element of the range.

The special notation f(x), read “ f of x” or “ f at x”, represents the value of the function at the number x.

The difference quotient is , h 0.

The domain of a function is the largest set of real numbers for which value of f(x) is a real number.

( ) ( )f x h f x

h