8.4 Inverse Functions

Transcription

1 Section 8. Inverse Functions Inverse Functions As we saw in the last section, in order to solve application problems involving eponential unctions, we will need to be able to solve eponential equations such as 100 = 1000e 0.06t or 300 =. However, we currentl don t have an mathematical tools at our disposal to solve or a variable that appears as an eponent, as in these equations. In this section, we will develop the concept o an inverse unction, which will in turn be used to deine the tool that we need, the logarithm, in Section 8.. One-to-One Functions Deinition 1. A given unction is said to be one-to-one i or each value in the range o, there is just one value in the domain o such that = (). In other words, is one-to-one i each output o corresponds to precisel one input. It s easiest to understand this deinition b looking at mapping diagrams and graphs o some eample unctions. Eample. Consider the two unctions h and k deined according to the mapping diagrams in Figure 1. In Figure 1(a), there are two values in the domain that are both mapped onto 3 in the range. Hence, the unction h is not one-to-one. On the other hand, in Figure 1(b), or each output in the range o k, there is onl one input in the domain that gets mapped onto it. Thereore, k is a one-to-one unction. 1 h 3 1 k 3 (a) (b) Figure 1. Mapping diagrams help to determine i a unction is one-to-one. Eample 3. The graph o a unction is shown in Figure (a). For this unction, the -value is the output corresponding to two input values, = 1 and = 3 (see the corresponding mapping diagram in Figure (b)). Thereore, is not one-to-one. Graphicall, this is apparent b drawing horizontal segments rom the point (0, ) on the -ais over to the corresponding points on the graph, and then drawing vertical segments to the -ais. These segments meet the -ais at 1 and 3. 1 Coprighted material. See:

2 80 Chapter 8 Eponential and Logarithmic Functions (a) Figure. (b) A unction which is not one-to-one. Eample. In Figure 3, each -value in the range o corresponds to just one input value. Thereore, this unction is one-to-one. Graphicall, this can be seen b mentall drawing a horizontal segment rom each point on the -ais over to the corresponding point on the graph, and then drawing a vertical segment to the -ais. Several eamples are shown in Figure 3. It s apparent that this procedure will alwas result in just one corresponding point on the -ais, because each -value onl corresponds to one point on the graph. In act, it s easiest to just note that since each horizontal line onl intersects the graph once, then there can be onl one corresponding input to each output. Figure 3. A one-to-one unction The graphical process described in the previous eample, known as the horizontal line test, provides a simple visual means o determining whether a unction is one-toone.

3 Section 8. Inverse Functions 80 Horizontal Line Test I each horizontal line intersects the graph o at most once, then is one-to-one. On the other hand, i some horizontal line intersects the graph o more than once, then is not one-to-one. Remark. It ollows rom the horizontal line test that i is a strictl increasing unction, then is one-to-one. Likewise, ever strictl decreasing unction is also oneto-one. Inverse Functions I is one-to-one, then we can deine an associated unction g, called the inverse unction o. We will give a ormal deinition below, but the basic idea is that the inverse unction g simpl sends the outputs o back to their corresponding inputs. In other words, the mapping diagram or g is obtained b reversing the arrows in the mapping diagram or. Eample 6. The unction in Figure (a) maps 1 to and to 3. Thereore, the inverse unction g in Figure (b) maps the outputs o back to their corresponding inputs: to 1 and 3 to. Note that reversing the arrows on the mapping diagram or ields the mapping diagram or g. 1 g (a) (b) Figure. Reversing the arrows on the mapping diagram or ields the mapping diagram or g. Since the inverse unction g sends the outputs o back to their corresponding inputs, it ollows that the inputs o g are the outputs o, and vice versa. Thus, the unctions g and are related b simpl interchanging their inputs and outputs. The original unction must be one-to-one in order to have an inverse. For eample, consider the unction h in Eample. h is not one-to-one. I we reverse the arrows in the mapping diagram or h (see Figure 1(a)), then the resulting relation will not be a unction, because 3 would map to both 1 and. Beore giving the ormal deinition o an inverse unction, it s helpul to review the description o a unction given in Section.1. While unctions are oten deined b means o a ormula, remember that in general a unction is just a rule that dictates how to associate a unique output value to each input value.

4 806 Chapter 8 Eponential and Logarithmic Functions Deinition 7. Suppose that is a given one-to-one unction. The inverse unction g is deined as ollows: or each in the range o, deine g() to be the unique value such that = (). To understand this deinition, it s helpul to look at a diagram: g() Figure. The input or g is an -value in the range o. Thus, the input in the above diagram is a value on the -ais. The output o g is the corresponding value on the -ais which satisies the condition = (). Note in particular that the -value is unique because is one-to-one. The relationship between the original unction and its inverse unction g can be described b: Propert 8. I g is the inverse unction o, then = g() = (). In act, this is reall the deining relationship or the inverse unction. An eas wa to understand this relationship (and the entire concept o an inverse unction) is to realize that it states that inputs and outputs are interchanged. The inputs o g are the outputs o, and vice versa. It ollows that the Domain and Range o and g are interchanged: Propert 9. I g is the inverse unction o, then Domain(g) = Range() and Range(g) = Domain(). The smbol means that these two statements are equivalent: i one is true, then so is the other.

5 Section 8. Inverse Functions 807 The deining relationship in Propert 8 is also equivalent to the ollowing two identities, so these provide an alternative characterization o inverse unctions: Propert 10. I g is the inverse unction o, then g(()) = or ever in Domain() and (g()) = or ever in Domain(g). Note that the irst statement in Propert 10 sas that g maps the output () back to the input. The second statement sas the same with the roles o and g reversed. Thereore, and g must be inverses. Propert 10 can also be interpreted to sa that the unctions g and undo each other. I we irst appl to an input, and then appl g, we get back again. Likewise, i we appl g to an input, and then appl, we get back again. So whatever action perorms, g reverses it, and vice versa. Eample 11. Suppose () = 3. Thus, is the cubing unction. What operation will reverse the cubing process? Taking a cube root. Thus, the inverse o should be the unction g() = 3. Let s veri Propert 10: and g(()) = g( 3 ) = 3 3 = (g()) = ( 3 ) = ( 3 ) 3 =. Eample 1. Suppose () = 1. acts on an input b irst multipling b, and then subtracting 1. The inverse unction must reverse the process: irst add 1, and then divide b. Thus, the inverse unction should be g() = ( + 1)/. Again, let s veri Propert 10: and g(()) = g( 1) = ( 1) + 1 = = ( ) ( ) (g()) = = 1 = ( + 1) 1 =.

6 808 Chapter 8 Eponential and Logarithmic Functions Remarks The computation g(()), in which the output o one unction is used as the input o another, is called the composition o g with. Thus, inverse unctions undo each other in the sense o composition. Composition o unctions is an important concept in man areas o mathematics, so more practice with composition o unctions is provided in the eercises.. I g is the inverse unction o, then is also the inverse o g. This ollows rom either Propert 8 or Propert 10. (Note that the labels and or the variables are unimportant. The ke idea is that two unctions are inverses i their inputs and outputs are interchanged.) Notation: In order to indicate that two unctions and g are inverses, we usuall use the notation 1 or g. The smbol 1 is read inverse. In addition, to avoid conusion with the tpical roles o and, it s oten useul to use dierent labels or the variables. Rewriting Propert 8 with the 1 notation, and using new labels or the variables, we have the deining relationship: Propert 1. v = 1 (u) u = (v) Likewise, rewriting Propert 10, we have the composition relationships: Propert 1. 1 ((z)) = z or ever z in Domain() and ( 1 (z)) = z or ever z in Domain( 1 ) However, the new notation comes with an important warning: Warning does not mean 1 The 1 eponent is just notation in this contet. When applied to a unction, it stands or the inverse o the unction, not the reciprocal o the unction. The Graph o an Inverse Function How are the graphs o and 1 related? Suppose that the point (a, b) is on the graph o. That means that b = (a). Since inputs and outputs are interchanged or the inverse unction, it ollows that a = 1 (b), so (b, a) is on the graph o 1. Now (a, b)

7 Section 8. Inverse Functions 809 and (b, a) are just relections o each other across the line = (see the discussion below or a detailed eplanation), so it ollows that the same is true o the graphs o and 1 i we graph both unctions on the same coordinate sstem (i.e., as unctions o ). For eample, consider the unctions rom Eample 11. The unctions () = 3 and 1 () = 3 are graphed in Figure 6 along with the line =. Several relected pairs o points are also shown on the graph. = 1 Figure 6. Graphs o () = 3 and 1 () = 3 are relections across the line =. To see wh the points (a, b) and (b, a) are just relections o each other across the line =, consider the segment S between these two points (see Figure 7). It will be enough to show: (1) that S is perpendicular to the line =, and () that the intersection point P o the segment S and the line = is equidistant rom each o (a, b) and (b, a). = (a,b) P S (b,a) Figure 7. Switching the abscissa and ordinate relects the point across the line =. 1. The slope o S is a b b a = 1,

8 810 Chapter 8 Eponential and Logarithmic Functions and the slope o the line = is 1, so the are perpendicular.. The line containing S has equation b = ( a), or equivalentl, = +(a+b). To ind the intersection o S and the line =, set = + (a + b) and solve or to get = a + b. Since =, it ollows that the intersection point is ( a + b P =, a + b ). Finall, we can use the distance ormula presented in section 9.6 to compute the distance rom P to (a, b) and the distance rom P to (b, a). In both cases, the computed distance turns out to be a b. Computing the Formula o an Inverse Function How does one ind the ormula o an inverse unction? In Eample 11, it was eas to see that the inverse o the cubing unction must be the cube root unction. But how was the ormula or the inverse in Eample 1 obtained? Actuall, there is a simple procedure or inding the ormula or the inverse unction (provided that such a ormula eists; remember that not all unctions can be described b a simple ormula, so the procedure will not work or such unctions). The ollowing procedure works because the inputs and outputs (the and variables) are switched in step 3. Computing the Formula o an Inverse Function 1. Check the graph o the original unction () to see i it passes the horizontal line test. I so, then is one-to-one and ou can proceed.. Write the ormula in -equation orm, as = (). 3. Interchange the and variables.. Solve the new equation or, i possible. The result will be the ormula or 1 ().

9 Section 8. Inverse Functions 811 Eample 17. Eample 1. Let s start b inding the inverse o the unction () = 1 rom Step 1: A check o the graph shows that is one-to-one (see Figure 8). Figure 8. The graph o () = 1 passes the horizontal line test. Step : Write the ormula in -equation orm: = 1 Step 3: Interchange and : = 1 Step : Solve or : Thus, 1 () = + 1. = 1 = + 1 = = + 1 Figure 9 demonstrates that the graph o 1 () = ( + 1)/ is a relection o the graph o () = 1 across the line =. In this igure, the ZSquare command in the ZOOM menu has been used to better illustrate the relection (the ZSquare command equalizes the scales on both aes). = (a) Figure 9. (b) Smmetr across the line =

10 81 Chapter 8 Eponential and Logarithmic Functions Eample 18. This time we ll ind the inverse o () = + 3. Step 1: veri). A check o the graph shows that is one-to-one (this is let or the reader to Step : Write the ormula in -equation orm: = + 3 Step 3: Interchange and : = + 3 Step : Solve or : 3 Thus, 1 () =. = + 3 = 3 = = 3 = 3 = Again, note that the graph o 1 () = ( 3)/ is a relection o the graph o () = + 3 across the line = (see Figure 10). = (a) Figure 10. (b) Smmetr across the line = Eample 19. Find the inverse o () = /(7 + ). Step 1: veri). A check o the graph shows that is one-to-one (this is let or the reader to Step : Write the ormula in -equation orm: = 7 + Step 3: Interchange and : = 7 +

11 Section 8. Inverse Functions 813 Step : Solve or : = 7 + = (7 + ) = Thus, 1 () = 7. = 7 + = = = 7 = 7 Eample 0. This eample is a bit more complicated: ind the inverse o the unction () = ( + )/( 3). Step 1: veri). A check o the graph shows that is one-to-one (this is let or the reader to Step : Write the ormula in -equation orm: = + 3 Step 3: Interchange and : = + 3 Step : Solve or : = + 3 = ( 3) = + = 3 = + This equation is linear in. Isolate the terms containing the variable on one side o the equation, actor, then divide b the coeicient o. Thus, 1 () = = + = = 3 + = ( ) = 3 + = = 3 + Eample 1. According to the horizontal line test, the unction h() = is certainl not one-to-one. However, i we onl consider the right hal or let hal o the unction (i.e., restrict the domain to either the interval [0, ) or (, 0]), then

12 81 Chapter 8 Eponential and Logarithmic Functions the unction would be one-to-one, and thereore would have an inverse (Figure 11(a) shows the let hal). For eample, suppose is the unction () =, 0. In this case, the procedure still works, provided that we carr along the domain condition in all o the steps, as ollows: Step 1: Step : The graph in Figure 11(a) passes the horizontal line test, so is one-to-one. Write the ormula in -equation orm: =, 0 Step 3: Interchange and : =, 0 Note how and must also be interchanged in the domain condition. Step : Solve or : = ±, 0 Now there are two choices or, one positive and one negative, but the condition 0 tells us that the negative choice is the correct one. Thus, the last statement is equivalent to =. Thus, 1 () =. The graph o 1 is shown in Figure 11(b), and the graphs o both and 1 are shown in Figure 11(c) as relections across the line =. = 1 1 (a) (b) (c) Figure 11.