The Language of Relation and Functions Mathematical Language RELATION Relation A relation is a set of ordered pairs. If x and y are elements of these sets and if a relation exists between x and y, then we say that x corresponds to y or that y depends on x and is represented as the ordered pair of (x, y). A relation from set A to set B is defined to be any subset of Aο΄B. If R is a relation from A to B and (a, b) ο R, then we say that “a is related to b” and it is denoted as a R b. Relation Let A {0, 1, 2} and B = {1, 2, 3} Then Aο΄B gives all possible pairings of the elements of A and B. Aο΄B = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)} Let us say the an element π₯ in A is related to an element π¦ in B if and only if, π₯ is less than π¦. R= {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)} Relation Example: Let us say the an element π₯ in π΄ is related to an element π¦ in π΅ if and only if, π₯ is less than π¦. R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)} βͺ βͺ βͺ βͺ βͺ βͺ 0 R 1 since 0<1 0 R 2 since 0<2 0 R 3 since 0<3 1 R 2 since 1<2 1 R 3 since 1<3 2 R 3 since 2<3 R means “not related to” π₯ π π¦ means the (π₯, π¦) ∈ π π₯ π π¦ means the (π₯, π¦) ∉ π Relation Example: Let A = {1, 2} and B = {1, 2, 3} and define a relation R from A to B as follows: Given any π₯, π¦ ∈ π΄ × π΅, π₯, π¦ ∈ π means that x−y 2 is an integer. π΄ × π΅ = { 1, 1 , 1, 2 , 1, 3 , 2, 1 , 2, 2 , 2, 3 } πΉ = { π, π , π, π , π, π } True or False: Is 1 π 3? Is 2 π 3? Is 2 π 2? True False True Relation Ways How to Write or Represent Relations Example: R = {(-2, 1), (-2, 3), (0, -3), (1, 4), (3, 1)} Tabular: Graphical: Mapping Diagram (Arrow Diagram): Relation Let π΄ and π΅ be sets. A relation πΉ from π¨ to π© is a subset of π΄ × π΅. Given an ordered pair (π₯, π¦) in π΄ × π΅, π is related to π by πΉ, written π₯ π π¦, if, and only if, (π₯, π¦) is in π . The set π΄ is called the domain of π and the set π΅ is called the co-domain (image). Example: A = {4, 7}, π΄ο΄π΄ = {(4, 4), (4, 7), (7, 4),(7, 7)}. Let R on A be the description: x R y ο« x + y is even. Then (4, 4) ο R, and (7, 7) ο R. R = { (4, 4), (7, 7)} dom R = {4, 7}, im R = {4, 7} Relation Example: R = {(-2, 1), (-2, 3), (0, -3), (1, 4), (3, 1)} dom R = {-2,0,1,3}, im R = {1, 3, -3, 4} Example: R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)} dom R = {0, 1, 2}, im R = {1, 2, 3} FUNCTIONS FUNCTIONS A function π from a set π¨ to a set π© is a relation with domain π΄ and co-domain π΅ that satisfies the following properties: β For every element π₯ in π΄, there is an element π¦ in π΅ such that (π₯, π¦) ∈ πΉ. β For all elements π₯ in π΄ and π¦ and π§ in π΅, if (π₯, π¦) ∈ πΉ and (π₯, π§) ∈ πΉ, then π¦ = π§ FUNCTIONS A function is a relation that for each input, there is only one output. Example 1: Example2: FUNCTION FUNCTION FUNCTIONS Example 3: Example 4: FUNCTION FUNCTION FUNCTIONS Example 5: Example 6: NOT a function NOT a function FUNCTIONS Example 7: NOT a function Example 8: NOT a function What Makes a Relation a Function? A function is actually a “special” kind of relation because it follows an extra rule. Just like a relation, a function is also a set of ordered pairs; however, every x-value must be associated to only one y-value. π¦ = π(π₯) Dependent variable OUTPUT Name of the function Independent variable INPUT Graph of a Function Given: Graph: Note: you can determine if a relation is a function by using the vertical line test FUNCTIONS Constant Function π π₯ =π The c-value can be any number, so the graph of a constant function is a horizontal line. Here is the graph of f(x) = 4 FUNCTIONS Identity Function π π₯ =π₯ For the identity function, the xvalue is the same as the y-value. The graph is a diagonal line going through the origin. FUNCTIONS Linear Function π π₯ = ππ₯ + π An equation written in the slopeintercept form is the equation of a linear function, and the graph of the function is a straight line. Here is the graph of π π₯ = 3π₯ + 4 FUNCTIONS Absolute Value Function π π = π The absolute value function is easy to recognize with its Vshaped graph. The graph is in two pieces and is one of the piecewise functions. Inverse Functions An inverse function reverses the inputs with its outputs. Given the function π π₯ = 3π₯ − 4 π π₯ =π¦ The inverse of a function is denoted by the symbol π−π π π−1 π+4 π = 3 How do we solve for the Inverse of a Function? Given : π π₯ = 3π₯ − 4 re-write: π¦ = 3π₯ − 4 Interchange the variables: π₯ = 3π¦ − 4 Solve for y in terms of x: 3π¦ = π₯ + 4 π₯+4 π¦= 3 Change π¦ to π −1 (π₯): π+4 −1 π π = 3 Function Operations Addition of functions: π π₯ +π π₯ = π+π π₯ Subtraction of functions: π π₯ −π π₯ = π−π π₯ Multiplication of functions: π π₯ β π π₯ = Division of functions: π π₯ π π₯ = π π π₯ πβπ π₯ Function Operations Example1 : Let π π₯ = 2π₯ + 1 and π π₯ = π₯ 2 − 4 (a) Find π + π π₯ π+π π =π π +π π (b) Find π − π π₯ π−π π =π π −π π = 2π₯ + 1 + π₯ 2 − 4 = 2π₯ + 1 − π₯ 2 − 4 = ππ + ππ − π = −ππ + ππ + π Function Operations 1 1 πππ π π₯ = , π‘βππ π€βππ‘ ππ π − π π₯ ? Example 2 : πΌπ π π₯ = π₯−2 π₯+2 π − π π₯ = π π₯ − π(π₯) 1 1 = − π₯−2 π₯+2 π₯+2−π₯+2 = π₯−2 π₯+2 π₯+2 − π₯−2 = π₯−2 π₯+2 π = π π −π Function Operations Example 3 : If π π₯ = π₯ 2 + π₯ − 6 and π₯ = 1 , then what π₯+3 is ππ π₯ ? ππ π₯ = π π₯ π(π₯) = π₯2 +π₯−6 π₯2 + π₯ − 6 = π₯+3 1 π₯+3 π₯−2 π₯+3 = π₯+3 = π−π Function Operations Example 4 : π π₯ = π₯−3 2 πππ π π₯ = 5 − π₯ 2 , what is π − π π₯ ? π − π π₯ = π π₯ − π(π₯) = π₯−3 2 − 5−π₯ 2 = π₯ 2 − 6π₯ + 9 − 25 − 10π₯ + π₯ 2 = π₯ 2 − 6π₯ + 9 − 25 + 10π₯ − π₯ 2 = ππ − ππ Function Operations Composition of Functions: π β π = π π(π₯) read as “π of π of π₯" It means that whenever there is an x in the function f, it is replaced with the function g(x). The domain of π°π is the set of all x in the domain of g such that g(x) is in the domain of f. Example 1: Let π π₯ = π₯ 2 and π π₯ = π₯ − 3. Find π π π₯ π π(π₯) = π π₯ − 3 = π₯−3 2 = ππ − ππ + π Composition of Functions: Example 2: π β π = π π(π₯) Let π π₯ = 2π₯ − 1 and π π₯ = π₯ + 2. Find π π π₯ π π π₯ = π 2π₯ − 1 =2 π₯+2 −1 = 2π₯ + 4 − 1 = ππ + π Composition of Functions: π β π = π π(π₯) Example 3: Let π π₯ = 3π₯ + 1 and π π₯ = 2π₯ − 3. Find π π π₯ π π π₯ = π 2π₯ − 3 π π π₯ and π π π₯ = π 3π₯ + 1 = 3 2π₯ − 3 + 1 = 2 3π₯ + 1 − 3 = 6π₯ − 9 + 1 = 6π₯ + 2 − 3 = ππ − π = ππ − π HOME WORK #5 A. Give the domain and range of each relation Domain Range 1. 1,2 , 5,0 , 0, −2 , −3,4 _________________ __________________ 2. −4,6 , 4,0 , −3, −1 , 5,2 _________________ __________________ 3. 0,4 , −5,1 , −3, −2 , 3, −5 _________________ __________________ 4. 1,2 , 2,3 , 3,4 , 4,5 _________________ __________________ 5. 1,3 , 0,3 , 2,1 , 4,2 _________________ __________________ HOME WORK #5 B. Determine if the given mapping is a Functions or Not a Function 6. 7. HOME WORK #5 B. Determine if the given mapping is a Functions or Not a Function 8. 9. HOME WORK #5 B. Determine if the given mapping is a Functions or Not a Function 10. C. Given: π π₯ = 2π₯ + 7 and π π₯ = 3π₯ + 1 Perform the indicated operations 11. π π₯ + π π₯ 12. π π₯ − π π₯ 13. π π π₯ 14. π π(−2) 15. π 5 + π −3