Unit 5: Inverse Functions

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Haberman MTH Section I: Functions and Their Graphs Unit 5: Inverse Functions EXAMPLE: Consider the unction n( ), which is graphed in Figure Figure : Graph o n( ) Since n is a unction, each input corresponds to eactl one output In this case it is also true that each output corresponds to eactl one input For eample, i n ( ) 8 then must be, ie, there are no other values that correspond with the output 8 Similarl, i n ( ) 7 then must be, ie, the onl value that corresponds with the output 7 is We sa that unctions like n( ) are one-to-one DEFINITION: A unction is one-to-one i each output corresponds to eactl one input, ie, i ( a) ( b) then a b

Haberman MTH Section I: Unit 5 EXAMPLE: Now consider the unction p( ), which is graphed in Figure Figure : Graph o p( ) Here, it is not the case that each output corresponds to eactl one input, so p is not oneto-one (I anthing, p is two-to-one ) For eample, i p ( ) then ou cannot be sure what is since it could be or Similarl, i p ( ) 6 then could be or We can determine i a unction is one-to-one b appling the horizontal line test to the graph o the unction The Horizontal Line Test I a unction is one-to-one then no horizontal line intersects the graph o the unction more than once EXAMPLE: Which o the ollowing graphs represent one-to-one unctions? Figure Figure Figure 5 The onl graph that represents a one-to-one unction is the one in Figure The graphs in Figures and 5 do not represent one-to-one unctions because the do not pass the horizontal line test

Haberman MTH Section I: Unit 5 I a unction ( ) is one-to-one, then, in addition to having the outputs a unction o the inputs, it is also true that the inputs are a unction o the outputs, ie, is a unction o, ie, can pla the role o the independent variable and can pla the role o the dependent variable (This is equivalent to saing that i ou turn our head 90 when looking at the graph o, the graph ou see passes the vertical line test and is a unction Thus, we see that the horizontal line test is just the vertical line test ater reversing the roles o and ) Inverse Functions I a unction is one-to-one then its inverse is a unction The unction that reverses the roles o and is the inverse unction o and is denoted CAUTION: The notation eponent The notation represents the inverse o the unction The is NOT an does NOT mean EXAMPLE: I t and t are inverses o one another, then t() implies that t () Since the inverse o a unction reverses the role o the inputs and the outputs, to ind the inverse o a unction, all that we have to do is reverse the roles o and EXAMPLE: Find the inverse o the unction n( ) First we write n in equation orm: solving the equation or : We can then reverse the roles o and b

Haberman MTH Section I: Unit 5 The equation represents the inverse, but we would like to use standard unction notation to represent the inverse Since usuall is used to represent the independent variable, we write n ( ) In Figure 6, both n and graph o n are graphed, along with the line n is the graph o n relected about the line As ou can see, the Figure 6: n( ) is graphed in blue and n ( ) is graphed in green and the line is graphed in red KEY POINT: The graph o the inverse o a unction is the same as the graph o the original unction but relected about the line EXAMPLE: Find the inverse o the unction ( ) To ind the inverse o solve or ( ) we irst write the unction as and then

Haberman MTH Section I: Unit 5 5 ( ) ( ) Since we usuall use as our independent variable, we switch our and to obtain the inverse unction ( ) Recall that the inverse o a unction reverses the roles o and Thus, i ou compose with its inverse ou take to and then to, so is sent to itsel, ie, the composition o a unction with its inverse is the identit unction I g ( ) and ( ) inverses o one another Inverse Function Test g, then the unctions and g are

Haberman MTH Section I: Unit 5 6 EXAMPLE: Check that ( ) and the unction that we ound in the eample above are in act inverses o one another b perorming the inverse unction test In the eample above we determined that the inverse o ( ) is ( ) To check this conclusion we can perorm the inverse unction test To perorm the test, we need to veri that ( ) and ( ) ( ) ( ) AND ( ) ( ) 6 ( ) Since ( ) ( ) and ( ), ( ) is in act the inverse o

Haberman MTH Section I: Unit 5 7 EXAMPLE: Find the inverse o the unction g( ) 7 Then veri that the unction ou obtain is in act the inverse o g( ) 7 To ind the inverse o solve or g( ) 7 we irst write the unction as 7 and then 7 7 7 7 Since we usuall use as our independent variable, we switch our and to obtain the inverse unction g ( ) 7 To veri that g g g ( ) ( ) 7 and g g ( ) g g ( ) is the inverse o g g ( ) : 7 g 7 7 7 7 7 7 AND g( ) 7, we need to check i g g ( ) g g( ) g 7 7 7 Since g g( ) g( ) 7 and g g ( ), g ( ) 7 is in act the inverse o