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Function And Relations Review

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Presentation on theme: "Function And Relations Review"— Presentation transcript:

1 Function And Relations Review
f(x) Vertical Line Test

2 Relations and Functions
Relation: a set of ordered pairs. {(3, 2), (–1, 5), (0, 2)} Domain: all of the x values or inputs {3, –1, 0} Range: all of the y values or outputs {2, 5, 2}

3 Function: a relation in which each element of the domain is paired with exactly one element of the range. Are the following relations also functions? Ex. {(2, 3), (–1, 4), (4, 5), (–2, 3)} Yes, each element of the domain is paired with 1 element of the range. Ex. {(4, 4), (–2, 3), (4, 2), (3, 4)} No, 4 in the domain is paired with 2 different outputs.

4 Example: Which mapping represents a function? a. b.
–2 Yes No, –1 is paired with two elements of the range

5 This graph represents a function
Vertical Line Test: a relation is a function if a vertical line drawn through its graph, passes through only one point. Example: Do the following graphs represent functions? hits twice hits only once This graph represents a function This graph is not a function

6 4.2 Perform Function Operations and Composition
Pg. 112 f(x) Vertical Line Test

7 Function Operations You can perform operations, such as addition, subtraction, multiplication, and division, with functions… For example: If f(x) = 3x and g(x) = x – 5 f(x) + g(x) = 3x + (x – 5) = f(x) – g(x) = 3x – (x – 5) f(x) • g(x) = (3x)(x – 5) What about f(x) ÷ g(x) ???????

8 Set the denominator equal to zero. x – 5 = 0
Fraction can not be simplified (they are factors) BUT we can tell where the domain is restricted !!!!! Set the denominator equal to zero. x – 5 = 0 Therefore, D: All Real Numbers except where x = 5 or The denominator tells us the domain of the function

9 Find the following. where

10 Composition of Functions
Functions can be put together in a composite function that looks like this… f(g(x)) or (f ° g)(x)= “f of g of x” 2 + 4 2x 2

11 Evaluating functions: Example: If f(x) = 3x – 1, find f(5)
“f of x” is a function. Find “f of 5”. f(5) = 3(5) – substitute 5 for x = 15 – 1 therefore, f(5) = 14

12 Example: If g(x) = 3x2 – 2x find g(–2)
Substitute –2 for each x g(–2) = 3(–2)2 – 2(–2) = 3(4) = g(–2) = 16

13 Examples: If f(x) = x2 – 5 and g(x) = 3x2 + 1 find f(g(2)) and g(f(2))

14 Example: If f(x) = x2 + 6 and g(x) = 3x – 4 find f(g(x))
F(g(x)) = Put g(x) in for the (x) in f(x)

15 Homework Pg. 114, 1 – 24 all


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