We would now like to turn our attention to a specific family of functions, the one to one functions.

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1 9.6 Inverse Functions We would now like to turn our attention to a speciic amily o unctions, the one to one unctions. Deinition: One to One unction ( a) (b A unction is called - i, or any a and b in the domain o, a b. ) implies that The one to one idea is similar to that o the basic unction idea. Whereas, the basic unction idea is that or every one domain value, you only have one range value, or a one to one unction, the idea is every one range value only comes rom one domain value. That is, the unction cannot repeat any y-value to be -. Eample : Show that the unction + We need to start with ( b) b + we get So + is -. is -. a b and show that b ( a) ( b) a + a a b b + b a. So, since ( a) a + Subtract rom both sides Multiply on both sides and Since the - idea is so similar to the unction idea, we have a test similar to the vertical line test to test i a unction is -. The Horizontal Line Test The graph o a unction represents the graph o a - unction i any horizontal line intersects the graph at no more than one point. So to test a graph o a unction to see i it is -, we simply need to see i any horizontal line intersects more than one point. Eample : Determine which o the ollowing unctions are - a. b. c.

2 Clearly parts a. and c. pass the horizontal line test and part b. does not. Thereore, a. and c. are one to one unctions and b. is not. Although one to one unctions are important, there main application comes in the way o the inverse unction, which is our net deinition. Deinition: Inverse unction Let and g be two unctions such that ( g ) or all in the domain o g and g or all in the domain o, then the unction g is called the inverse unction o ( ), written. For the sake o simpliication, we will not worry about the domains involved. Readers curious about this can read about inverse unctions in a precalculus tetbook. The idea o the inverse unction is that two unctions are inverses i the composition in both directions always gives you. Eample : Veriy that the unctions and ( + ) g are inverse unctions. In order to veriy that the unctions are inverses we simply need to show that g ( ). We will start with ( g ). So ( g ). Now we show So since ( g ) and +. ( g ) ( ( + ) ) ( ( + )) ( + ) ( ) g( ) (( ) + ) g. g ( g ) g, the unctions are inverse unctions. We write g as As a consequence o the above deinition we clearly get the ollowing property. Property o Inverse Functions ( ) and ( ) and

3 We don t want to simply be able to show that two unctions are inverse unctions. We would actually like to start with a given unction and ind what the inverse unction is, that is, i it eists. We have the ollowing steps to inding the inverse unction. Finding an Inverse Function. In the equation or, replace with y.. Interchange and y.. Solve the resulting equation or y.. Replace y by. Eample : 6 + Let. Find. First we replace with y to get y 6 +. Now we interchange and y to get 6 y +. Net we solve the equation or y. 6y + Finally, we replace y by Eample 5:. Find. Let 6y y 6 to get. 6 Again we will ollow the steps outlined above.

4 ( y ) y y Replace y y + + with y Interchange and y Solve or y Multiply by the LCD y Divide by Replace y by Eample 6: +. Find. Let y y + y y + y + 8 Replace with y Interchange and y Solve or y Cube both sides to eliminate the radical Replace y by We can see rom the last eample that powers and roots are inverses o one another. So, when trying to ind an inverse o an equation that contains eponents we take a radical o both sides and vice versa. Notice by the way we ind an inverse unction, that since we interchange the roles o and y, it would make sense that the domains and ranges would interchange as well, since they are associated with the and y This is eactly the case. We state this as the ollowing act. Fact: The domain o is the range o and the range o is the domain o. Or alternately, the domain o is the range o and the range o is the domain o. Finally we would like to talk about the geometry o a unction and its inverse unction. We start with the ollowing condition.

5 Condition or an Inverse Function A unction has an inverse unction i and only i it is - What that means or us is that a unction has an inverse unction only when it passes the horizontal line test. More importantly though is the relationship between the graphs o a unction and its inverse unction. Since in order to ind an inverse unction we interchange the roles o and y, then it would make sense that i the point (a, b) is on the graph o a unction then (b, a) would have to be on the inverse unction. This relationship corresponds to a relection across the line y. So i the graph o is Then the graph o would be I we graph these on the same ais along with the line y y we can really see the relection. Eample 7: Given the graph o, graph. a. b. We simply need to relect each graph over the line y.

6 a. y b. y 9.6 Eercises Show that the ollowing unctions are g 6. g 7. g 8. g 9. h 0. h + Determine i the graph represents a - unction

7 Veriy that the two unctions are inverse unctions , 7 g 7., g , g ,.. ( u), g u u u , 6. ( n), n + n n , , g ( ) 5 +, g t t g t t + 6 g + g + g + g + + g,., g 5., g 7., g 9., Find the inverse o the ollowing g 9 7. g 8. g 9. g 0. g +. g +. g g + h 5. h h 7. h 8. h h g g 56. g 57. h g g h g 65. h +

8 Given the graph o, graph In Eercises -65 we ound the inverse o a unction using the techniques we learned in this section. However, sometimes these techniques ail us. For eample, i then we can calculate. But, is not - and thereore cannot have an inverse unction. However, i we ch ange the domain to make it a - unction, then it can have an inverse unction. We simply change the domain o to be 0. Similarly, i we started with to get we change the domain o For the ollowing eercises, ind State the domain and range o each eby making it -. to ther. Be sure to consider all domain stipulations. and ( ) 8. +, 0 8. ( ), 8. ( + ), 8. +, 0

9 Let. Find the ollowing ( 89. ( + h ) 90. ) 88. ( a) ( + h) + h