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Understanding Functions: Determining if a Relation is a Function and Finding Function Values

This text explains the concept of functions and how to determine if a relation is a function. It also provides information on finding function values and using the vertical line test. Nonlinear functions are also briefly discussed. Written by S. Calahan in 2008.

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Understanding Functions: Determining if a Relation is a Function and Finding Function Values

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  1. Functions 4-6 I can determine whether a relation is a function and find function values. S. Calahan 2008

  2. What is a function? • A function is a relation in which each element of the domain is paired with exactly one element of the range.

  3. Identify Functions x y -4 9 This is a function since each element -1 -6 of the domain (x) corresponds to only one 1 11 element in the range (y). 3 It doesn’t matter if two elements of the domain are paired with the same element in the range.

  4. Identify Functions x y -3 6 2 5 3 1 2 4 The table represents a relation that is not a function because there are two of the same elements in the domain (x).

  5. Vertical Line Test • Use the vertical line test to determine if a graph represents a function. • If a vertical line can be drawn so that it intersects the graph no more than once the graph is a function. • If a vertical line can be drawn so that it intersects the graph at two or more points the graph is not a function.

  6. Vertical Line Test • Not a function, because it crosses twice

  7. Vertical line test • This is a function because it crosses only once no matter where you draw the line.

  8. Function Notation • Equations that are functions can be written in function notation form • Example: y = 3x – 8 equation f(x) = 3x – 8 function notation

  9. Function Values • f(x) = 2x + 5 for f(-2) f(-2) = 2(-2) + 5 = 1 Substitute the value inside the ( ) in for the variable inside the ( ).

  10. f(x) = 2x + 5 • f(1) + 4 = [2(1) + 5] + 4 Substitute the 1 in for the x then add the 4 at the end of the expression. = [2 + 5] + 4 simplify = 7 + 4 = 11 add

  11. Nonlinear Functions • A function that is nonlinear can be solved the same way as a linear function. Nonlinear functions can have an exponent on one of the variables. • f(n) = n2 – 3n + 4

  12. f(n) = n2 – 3n + 4 • f(4) = 42 – 3(4) + 4 = 16 – 12 + 4 = 4 + 4 = 8

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