# Lesson 3.8 Functions.

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Lesson 3.8 Functions

No input value is repeated so the relation is a function
Definition A function is a special relation between values: The input value (domain) gives back exactly one unique Output value (range). Class Grade Function Student Grade 80 85 96 Bilal Peter Selma Ahmad George {(B, 85), (P, 96), (S, 80), (A, 85), (G, 96)} No input value is repeated so the relation is a function

Example of a relation that is not a function
Has visited Person Country Paris London Dubai New York Ahmed Rami Sally Nancy {(A, D), (A, P), (R, N), (S, P), (S, L), (S, D), (N, D)} The inputs A and S are repeated so the relation is not a function

x y F(x) = x2 – 1 { } { } Function Notation (x2 – 1) function Domain
{ } { } Domain 2, 5, 7, 10 Range 3, 24, 48, 99 F(x) = x2 – 1 Output input Function name

F: x 3 – 2x; D = {0, 1, 2, 3} f(x) = 3 – 2x R = {3, 1, -1, -3} f(0) =
Example The Domain D and the rule of some function are given. Find the range F: x – 2x; D = {0, 1, 2, 3} f(x) = 3 – 2x f(0) = 3 – 2(0) = 3 f(1) = 3 – 2(1) = 1 f(2) = 3 – 2(2) = -1 f(3) = 3 – 2(3) = -3 R = {3, 1, -1, -3}

1) F(x) = x3 – 3 2) F(x) = Domain of the function
The domain of a function is all the values that an input is allowed to take on. Give the domain of each function 1) F(x) = x3 – 3 There are no values that I can't plug in for x. when I have a polynomial, the answer is always that the domain is ”all real numbers” 2) F(x) = The only values that x can not take on are those which would cause division by zero. Here x can not take (1), so my domain will be “All real numbers except 1”

3) F(x) = x – 2 ≥ 0 x ≥ 2 Domain of the function
The only problem I have with this function is that I cannot have a negative inside the square root. So I'll set the insides ≥ 0, and solve. The result will be my domain: x – 2 ≥ 0 x ≥ 2 Then the domain is “all x ≥ 2”

f(g(1)) f(g(1)) Composite functions
Consider the functions: f(x) = 3x2 + 2 and g(x) = x + 1 is a composite function, where g is performed first and then f is performed on the result of g. The function fg may be found using a flow diagram Examples: find the indicated value: f(g(1)) f(g(1)) g(x) x + 1 f(x) 3x2 + 2 Method 1 1 input 2 input fg = 14 f(g(1)) Method 2 Step1: find g(1) = (1) + 1 = 2 Step2: find f(2) = 3(2)2 + 2 = = 14

H.W (4, 8, 12) + (30 – 40) Even page 144, 145