WEEK 5

1. WEEK 5 FUNCTIONS AND GRAPHS BASIC OF FUNCTIONS

2. 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.

3. INDEX • Definitions • Relations • Domain and Range • Functions • Difference between Functions and Relations • Determining whether a Relation is a Function • Function as equation • Determining whether an Equation is a Function • Function Notation • Evaluating a Function • Functions and Difference Quotients • Piecewise Functions • Domain of a Function • Summary

4. 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)

5. 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 Relationbetween the name and age. • So, we can say that a Relation is a relationship between sets of information. • Thus we can define Relation as anyset of ordered pairs.

6. 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} Relation: {(x1,y1), (x2,y2), (x3,y3)} y1 x1 y2 x2 y3 x3 Range Domain

7. 4 2 9 3 25 5 36 6 Domain Range 4. FUNCTIONS Relation: {(2,4), (3, 9), (5,25), (6, 36)} • Consider 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.

8. Relation A Relation may or may not be a function. 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)} Function All functions are relations. Functions are sub-classification of relations 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 4 2 4 6 3 6 4 9 8 4 8 12 10 5 5. DIFFERENCE BETWEEN FUNCTION AND RELATION

9. 1 2 4 3 6 5 8 Domain Range 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 isnot a function Relation: {(1,2),(3,4),(5,6),(5,8)}

10. 1 2 4 2 6 3 Domain Range DIFFERENCE BETWEEN FUNCTIONS AND RELATIONS (Cont….) Relation: {(1, 2), (2, 4), (3, 6)} 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.

11. 1 2 4 3 6 5 8 Domain Range 5. DIFFERENCE BETWEEN FUNCTIONS AND RELATIONS (Cont….) Relation: {(1,2), (3,4), (6,5), (8,5)} 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.

12. 6. FUNCTIONS AS EQUATION Do 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. (a)

13. 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 = 

14. 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 notdefine y as a function of x.

15. 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 thatf 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”.

16. 8. FUNCTION NOTATION (Cont…) Now lets understand function notation with the help of an example: • We have already seenthat 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.

17. 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) • Givenf(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)

18. 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.

19. 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

20. 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

21. 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.

22. 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)

23. 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}.

24. 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}

25. 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.