Unit 9: Symmetric Functions

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1 Haberman MTH 111 Section I: Functions and Their Graphs Unit 9: Symmetric Functions Some functions have graphs with special types of symmetries, and we can use the reflections we just studied to analyze these symmetries EXAMPLE: The graph of y u( )? symmetry about the y-ais u( ) is given below How does it compare with the graph of The graph of u( ) Notice that if we reflect the graph of u( ) about the y-ais, we obtain the eact same graph! This can be represented algebraically with the statement u( ) u( ) We say that graphs like u( ) have symmetry about the y-ais and call functions with this type of symmetry even functions (It helps me to remember that u( ) is an eample of an even function since the power on (ie, ) is an even number) A function y f ( ) is even if its graph is symmetric about the y-ais, (which means that the graph of the function isn t changed if it is reflected about the y-ais) An algebraic test to determine if a function is even is given below: A function f is even if, for all in its domain, f ( ) f ( ) Notice that the test is an algebraic representation of the statement a function is even if a reflection about the y-ais does not change the graph

2 Haberman MTH 11 Section I: Unit 9 symmetry about the origin EXAMPLE: The graph of v( ) is given below How does the graph of y v( ) compare with the graph of y v( )? The graph of v( ) Notice that if you anchor the graph of v( ) at the origin (0, 0) and rotate it 180 in either direction, the graph ends up in the same place it started We say that graphs with this sort of symmetry have symmetry about the origin We can also study this symmetry by considering reflections Let s reflect the graph of v( ) about both the - and y-aes, ie, let s graph both y v( ) (reflection about the y-ais) and y v( ) (reflection about the -ais): The graph of y v( ) (blue) (reflection of y v( ) (grey) about y-ais) The graph of y v( ) (blue) (a reflection of y v( ) (grey) about -ais) These graphs show us that, when we reflect v( ) about the y-ais, we obtain the same graph as we do when we reflect the graph about the -ais We can summarize this fact with the following algebraic statement: v( ) v( )

3 Haberman MTH 11 Section I: Unit 9 Functions with symmetry about the origin are called odd functions (It helps me to remember that v( ) is an eample of an odd function since the power on (ie, ) is an odd number) A function y f ( ) is odd if its graph is symmetric about the origin, which means that if you rotate its graph 180 about the origin, you obtain the original graph (Equivalently, a function is odd if reflection about the y-ais gives you the same graph as reflection about the -ais) An algebraic test to determine if a graph is odd is given below: For all in its domain, f ( ) f ( ) Notice that the algebraic test is an algebraic representation of the statement a function is odd if a reflection about the y-ais produces the same graph as a reflection about the -ais EXAMPLE: Determine if the functions graphs below appear to be even, odd, or neither The graph of y a( ) The graph of y b( ) The graph of y c( ) The graph of y d( )

4 Haberman MTH 11 Section I: Unit 9 4 SOLUTION: The functions y a( ) and y d( ) appear to be ODD since they are symmetric about the origin (ie, if we rotate the graphs 180 about the origin, we ll obtain the original graphs) The function y c( ) appears to be EVEN since it is symmetric about the y-ais (ie, if we reflect the graph about the y-ais we ll obtain the original graph) The function y b( ) appears to neither even nor odd since it s not symmetric about the origin or the y-ais EXAMPLE: Perform the appropriate algebraic test to determine if the following functions are even, odd, or neither a a ( ) b b( ) c c ( ) d 6 4 d( ) 8 SOLUTIONS: It is important to notice that the algebraic tests for both even and odd symmetry start with making the input the opposite sign, so when we test for symmetry, we need to start with this step, simplify, and observe the result We only need to perform one test and observe the result, rather than performing a separate test for both even and odd a ( ) a( ) ( ) a ( ) Since a( ) a( ), we can conclude that a is an odd function

5 Haberman MTH 11 Section I: Unit 9 b b( ) ( ) ( ) Since this is neither the original function nor the opposite of the original function, ie, b( ) b( ) and b( ) b( ), we see that b is neither even nor odd c c( ) ( ) ( ) ( ) ( ) c ( ) Since c( ) c( ), we can conclude that c is an even function d 6 4 d( ) ( ) ( ) 8( ) d( ) Since d( ) d( ), we can conclude that d is an even function