5.2 Inverse Functions


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1 78 Further Topics in Functions. Inverse Functions Thinking of a function as a process like we did in Section., in this section we seek another function which might reverse that process. As in real life, we will find that some processes like putting on socks and shoes) are reversible while some like cooking a steak) are not. We start b discussing a ver basic function which is reversible, f) = +. Thinking of f as a process, we start with an input and appl two steps, as we saw in Section.. multipl b. add To reverse this process, we seek a function g which will undo each of these steps and take the output from f, +, and return the input. If we think of the realworld reversible twostep process of first putting on socks then putting on shoes, to reverse the process, we first take off the shoes, and then we take off the socks. In much the same wa, the function g should undo the second step of f first. That is, the function g should. subtract. divide b Following this procedure, we get g) =. Let s check to see if the function g does the job. If =, then f) = ) + = + = 9. Taking the output 9 from f, we substitute it into g to get g9) = 9 = =, which is our original input to f. To check that g does the job for all in the domain of f, we take the generic output from f, f) = +, and substitute that into g. That is, gf)) = g + ) = +) = =, which is our original input to f. If we carefull eamine the arithmetic as we simplif gf)), we actuall see g first undoing the addition of, and then undoing the multiplication b. Not onl does g undo f, but f also undoes g. That is, if we take the output from g, g) =, and put that into f, we get fg)) = f ) = ) + = ) + =. Using the language of function composition developed in Section., the statements gf)) = and fg)) = can be written as g f)) = and f g)) =, respectivel. Abstractl, we can visualize the relationship between f and g in the diagram below. f = gf)) = f) g
2 . Inverse Functions 79 The main idea to get from the diagram is that g takes the outputs from f and returns them to their respective inputs, and conversel, f takes outputs from g and returns them to their respective inputs. We now have enough background to state the central definition of the section. Definition.. Suppose f and g are two functions such that. g f)) = for all in the domain of f and. f g)) = for all in the domain of g then f and g are said to be invertible. inverses of each other. The functions f and g are said to be We now formalize the concept that inverse functions echange inputs and outputs. Theorem.. Properties of Inverse Functions: Suppose f and g are inverse functions. The range a of f is the domain of g and the domain of f is the range of g fa) = b if and onl if gb) = a a, b) is on the graph of f if and onl if b, a) is on the graph of g a Recall this is the set of all outputs of a function. Theorem. is a consequence of Definition. and the Fundamental Graphing Principle for Functions. We note the third propert in Theorem. tells us that the graphs of inverse functions are reflections about the line =. For a proof of this, see Eample..7 in Section. and Eercise 7 in Section.. For eample, we plot the inverse functions f) = + and g) = below. = f) = = g) If we abstract one step further, we can epress the sentiment in Definition. b saing that f and g are inverses if and onl if g f = I and f g = I where I is the identit function restricted to the domain of f and I is the identit function restricted to the domain of g. In other words, I ) = for all in the domain of f and I ) = for all in the domain of g. Using this description of inverses along with the properties of function composition listed in Theorem., we can show that function inverses are unique. Suppose g and h are both inverses of a function The identit function I, which was introduced in Section. and mentioned in Theorem., has a domain of all real numbers. Since the domains of f and g ma not be all real numbers, we need the restrictions listed here. In other words, invertible functions have eactl one inverse.
3 80 Further Topics in Functions f. B Theorem., the domain of g is equal to the domain of h, since both are the range of f. This means the identit function I applies both to the domain of h and the domain of g. Thus h = h I = h f g) = h f) g = I g = g, as required. We summarize the discussion of the last two paragraphs in the following theorem. Theorem.. Uniqueness of Inverse Functions and Their Graphs : Suppose f is an invertible function. There is eactl one inverse function for f, denoted f read finverse) The graph of = f ) is the reflection of the graph of = f) across the line =. The notation f is an unfortunate choice since ou ve been programmed since Elementar Algebra to think of this as f. This is most definitel not the case since, for instance, f) = + has as its inverse f ) =, which is certainl different than f) = +. Wh does this confusing notation persist? As we mentioned in Section., the identit function I is to function composition what the real number is to real number multiplication. The choice of notation f alludes to the propert that f f = I and f f = I, in much the same wa as = and =. Let s turn our attention to the function f) =. Is f invertible? A likel candidate for the inverse is the function g) =. Checking the composition ields g f)) = gf)) = =, which is not equal to for all in the domain, ). For eample, when =, f ) = ) =, but g) = =, which means g failed to return the input from its output. What g did, however, is match the output to a different input, namel, which satisfies f) =. This issue is presented schematicall in the picture below. f = = g We see from the diagram that since both f ) and f) are, it is impossible to construct a function which takes back to both = and =. B definition, a function matches a real number with eactl one other real number.) From a graphical standpoint, we know that if It is an ecellent eercise to eplain each step in this string of equalities. In the interests of full disclosure, the authors would like to admit that much of the discussion in the previous paragraphs could have easil been avoided had we appealed to the description of a function as a set of ordered pairs. We make no apolog for our discussion from a function composition standpoint, however, since it eposes the reader to more abstract was of thinking of functions and inverses. We will revisit this concept again in Chapter 8.
4 . Inverse Functions 8 = f ) eists, its graph can be obtained b reflecting = about the line =, in accordance with Theorem.. Doing so produces 7 6, ), ), ) 6 7 = f) = reflect across = switch and coordinates, ) = f )? We see that the line = intersects the graph of the supposed inverse twice  meaning the graph fails the Vertical Line Test, Theorem., and as such, does not represent as a function of. The vertical line = on the graph on the right corresponds to the horizontal line = on the graph of = f). The fact that the horizontal line = intersects the graph of f twice means two different inputs, namel = and =, are matched with the same output,, which is the cause of all of the trouble. In general, for a function to have an inverse, different inputs must go to different outputs, or else we will run into the same problem we did with f) =. We give this propert a name. Definition.. A function f is said to be onetoone if f matches different inputs to different outputs. Equivalentl, f is onetoone if and onl if whenever fc) = fd), then c = d. Graphicall, we detect onetoone functions using the test below. Theorem.. The Horizontal Line Test: A function f is onetoone if and onl if no horizontal line intersects the graph of f more than once. We sa that the graph of a function passes the Horizontal Line Test if no horizontal line intersects the graph more than once; otherwise, we sa the graph of the function fails the Horizontal Line Test. We have argued that if f is invertible, then f must be onetoone, otherwise the graph given b reflecting the graph of = f) about the line = will fail the Vertical Line Test. It turns out that being onetoone is also enough to guarantee invertibilit. To see this, we think of f as the set of ordered pairs which constitute its graph. If switching the  and coordinates of the points results in a function, then f is invertible and we have found f. This is precisel what the Horizontal Line Test does for us: it checks to see whether or not a set of points describes as a function of. We summarize these results below.
5 8 Further Topics in Functions Theorem.. Equivalent Conditions for Invertibilit: Suppose f is a function. following statements are equivalent. f is invertible The f is onetoone The graph of f passes the Horizontal Line Test We put this result to work in the net eample. Eample... Determine if the following functions are onetoone in two was: a) analticall using Definition. and b) graphicall using the Horizontal Line Test.. f) =. g) =. h) = +. F = {, ), 0, ),, )} Solution.. a) To determine if f is onetoone analticall, we assume fc) = fd) and attempt to deduce that c = d. Hence, f is onetoone. fc) = fd) c = d c = d c = d c = d b) To check if f is onetoone graphicall, we look to see if the graph of = f) passes the Horizontal Line Test. We have that f is a nonconstant linear function, which means its graph is a nonhorizontal line. Thus the graph of f passes the Horizontal Line Test.. a) We begin with the assumption that gc) = gd) and tr to show c = d. We have shown that g is onetoone. gc) = gd) c d = c d c d) = d c) c cd = d dc c = d c = d
6 . Inverse Functions 8 b) We can graph g using the si step procedure outlined in Section.. We get the sole intercept at 0, 0), a vertical asmptote = and a horizontal asmptote which the graph never crosses) =. We see from that the graph of g passes the Horizontal Line Test. 6 = f) = g). a) We begin with hc) = hd). As we work our wa through the problem, we encounter a nonlinear equation. We move the nonzero terms to the left, leave a 0 on the right and factor accordingl. hc) = hd) c c + = d d + c c = d d c d c + d = 0 c + d)c d) c d) = 0 c d)c + d) ) = 0 factor b grouping c d = 0 or c + d = 0 c = d or c = d We get c = d as one possibilit, but we also get the possibilit that c = d. This suggests that f ma not be onetoone. Taking d = 0, we get c = 0 or c =. With f0) = and f) =, we have produced two different inputs with the same output meaning f is not onetoone. b) We note that h is a quadratic function and we graph = h) using the techniques presented in Section.. The verte is, ) and the parabola opens upwards. We see immediatel from the graph that h is not onetoone, since there are several horizontal lines which cross the graph more than once.. a) The function F is given to us as a set of ordered pairs. The condition F c) = F d) means the outputs from the function the coordinates of the ordered pairs) are the same. We see that the points, ) and, ) are both elements of F with F ) = and F ) =. Since, we have established that F is not onetoone. b) Graphicall, we see the horizontal line = crosses the graph more than once. Hence, the graph of F fails the Horizontal Line Test.
7 8 Further Topics in Functions 6 = F ) = h) We have shown that the functions f and g in Eample.. are onetoone. This means the are invertible, so it is natural to wonder what f ) and g ) would be. For f) =, we can think our wa through the inverse since there is onl one occurrence of. We can track stepbstep what is done to and reverse those steps as we did at the beginning of the chapter. The function g) = is a bit trickier since occurs in two places. When one evaluates g) for a specific value of, which is first, the or the? We can imagine functions more complicated than these so we need to develop a general methodolog to attack this problem. Theorem. tells us equation = f ) is equivalent to f) = and this is the basis of our algorithm.. Write = f). Interchange and Steps for finding the Inverse of a Onetoone Function. Solve = f) for to obtain = f ) Note that we could have simpl written Solve = f) for and be done with it. The act of interchanging the and is there to remind us that we are finding the inverse function b switching the inputs and outputs. Eample... Find the inverse of the following onetoone functions. analticall using function composition and graphicall. Check our answers. f) =. g) = Solution.. As we mentioned earlier, it is possible to think our wa through the inverse of f b recording the steps we appl to and the order in which we appl them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algorithm. We write = f) and proceed to switch and
8 . Inverse Functions 8 = f) = = = = = switch and = + We have f ) = +. To check this answer analticall, we first check that f f ) ) = for all in the domain of f, which is all real numbers. f f ) ) = f f)) = f) + = ) + = ) + = + + = We now check that f f ) ) = for all in the range of f which is also all real numbers. Recall that the domain of f ) is the range of f.) f f ) ) = ff )) = f ) = + = + = = To check our answer graphicall, we graph = f) and = f ) on the same set of aes. The appear to be reflections across the line =. Note that if ou perform our check on a calculator for more sophisticated functions, ou ll need to take advantage of the ZoomSquare feature to get the correct geometric perspective. )
9 86 Further Topics in Functions = = f) = f ). To find g ), we start with = g). We note that the domain of g is, ), ). = g) = = ) = = = + switch and = + ) factor = + We obtain g ) = +. To check this analticall, we first check g g ) ) = for all in the domain of g, that is, for all. g g ) ) = g g)) ) = g ) = = ) + ) ) ) ) + clear denominators
10 . Inverse Functions 87 = = + ) + = = Net, we check g g ) ) = for all in the range of g. From the graph of g in Eample.., we have that the range of g is, ), ). This matches the domain we get from the formula g ) =, as it should. + g g ) ) = g g ) ) ) = g + ) + = ) = = + ) ) ) + ) + ) clear denominators = = Graphing = g) and = g ) on the same set of aes is bus, but we can see the smmetric relationship if we thicken the curve for = g ). Note that the vertical asmptote = of the graph of g corresponds to the horizontal asmptote = of the graph of g, as it should since and are switched. Similarl, the horizontal asmptote = of the graph of g corresponds to the vertical asmptote = of the graph of g.
11 88 Further Topics in Functions 6 = = g) and = g ) We now return to f) =. We know that f is not onetoone, and thus, is not invertible. However, if we restrict the domain of f, we can produce a new function g which is onetoone. If we define g) =, 0, then we have = f) = restrict domain to 0 = g) =, 0 The graph of g passes the Horizontal Line Test. To find an inverse of g, we proceed as usual = g) =, 0 =, 0 switch and = ± = since 0
12 . Inverse Functions 89 We get g ) =. At first it looks like we ll run into the same trouble as before, but when we check the composition, the domain restriction on g saves the da. We get g g ) ) = g g)) = g ) = = =, since 0. Checking g g ) ) = g g ) ) = g ) = ) =. Graphing 6 g and g on the same set of aes shows that the are reflections about the line = = g) = = g ) Our net eample continues the theme of domain restriction. Eample... Graph the following functions to show the are onetoone and find their inverses. Check our answers analticall using function composition and graphicall.. j) = +,.. k) = + Solution.. The function j is a restriction of the function h from Eample... Since the domain of j is restricted to, we are selecting onl the left half of the parabola. We see that the graph of j passes the Horizontal Line Test and thus j is invertible. 6 = j) 6 We graphed = in Section.7.
13 90 Further Topics in Functions We now use our algorithm 7 to find j ). = j) = +, = +, switch and 0 = + = ± ) ) ) quadratic formula, c = ) = ± = ± ) = ± = ± ) = ± = since. We have j ) =. When we simplif j j ) ), we need to remember that the domain of j is. Checking j j, we get j j ) ) = j j)) = j + ), = + ) = + = ) = = )) since = j j ) ) = j j ) ) = j ) = ) ) + = + ) + + = + = 7 Here, we use the Quadratic Formula to solve for. For completeness, we note ou can and should!) also consider solving for b completing the square.
14 . Inverse Functions 9 Using what we know from Section.7, we graph = j ) and = j) below. 6 = j) = 6 = j ). We graph = k) = + using what we learned in Section.7 and see k is onetoone. = k) We now tr to find k. = k) = + = + + = + + ) = + ) switch and + + = + = + We have k ) = +. Based on our eperience, we know something isn t quite right. We determined k is a quadratic function, and we have seen several times in this section that these are not onetoone unless their domains are suitabl restricted. Theorem. tells us that the domain of k is the range of k. From the graph of k, we see that the range is [, ), which means we restrict the domain of k to. We now check that this works in our compositions.
15 9 Further Topics in Functions and k k ) ) = k k)) = k + ), = + ) + + ) = + ) = + = k k ) ) = k + ) = + ) + = + + = + ) = + = + since = Graphicall, everthing checks out as well, provided that we remember the domain restriction on k means we take the right half of the parabola. = k) = k ) Our last eample of the section gives an application of inverse functions. Eample... Recall from Section. that the pricedemand equation for the PortaBo game sstem is p) =. + 0 for 0 66, where represents the number of sstems sold weekl and p is the price per sstem in dollars.
16 . Inverse Functions 9. Eplain wh p is onetoone and find a formula for p ). State the restricted domain.. Find and interpret p 0).. Recall from Section. that the weekl profit P, in dollars, as a result of selling sstems is given b P ) = Find and interpret P p ) ).. Use our answer to part to determine the price per PortaBo which would ield the maimum profit. Compare with Eample... Solution.. We leave to the reader to show the graph of p) =. + 0, 0 66, is a line segment from 0, 0) to 66, ), and as such passes the Horizontal Line Test. Hence, p is onetoone. We find the epression for p ) as usual and get p ) = 00. The domain of p should match the range of p, which is [, 0], and as such, we restrict the domain of p to 0.. We find p 0) = 00 0) = 0. Since the function p took as inputs the weekl sales and furnished the price per sstem as the output, p takes the price per sstem and returns the weekl sales as its output. Hence, p 0) = 0 means 0 sstems will be sold in a week if the price is set at $0 per sstem.. We compute P p ) ) = P p ) ) = P ) 00 =. 00 ) ) 0. After a heft amount of Elementar Algebra, 8 we obtain P p ) ) = To understand what this means, recall that the original profit function P gave us the weekl profit as a function of the weekl sales. The function p gives us the weekl sales as a function of the price. Hence, P p takes as its input a price. The function p returns the weekl sales, which in turn is fed into P to return the weekl profit. Hence, P p ) ) gives us the weekl profit in dollars) as a function of the price per sstem,, using the weekl sales p ) as the middle man.. We know from Section. that the graph of = P p ) ) is a parabola opening downwards. The maimum profit is realized at the verte. Since we are concerned onl with the price per sstem, we need onl find the coordinate of the verte. Identifing a = and b = 0, we get, b the Verte Formula, Equation., = b a = 6. Hence, weekl profit is maimized if we set the price at $6 per sstem. Comparing this with our answer from Eample.., there is a slight discrepanc to the tune of $0.0. We leave it to the reader to balance the books appropriatel. 8 It is good review to actuall do this!
17 9 Further Topics in Functions.. Eercises In Eercises  0, show that the given function is onetoone and find its inverse. Check our answers algebraicall and graphicall. Verif that the range of f is the domain of f and viceversa.. f) = 6. f) =. f) = +. f) = +. f) = + 6. f) = 7. f) = 8. f) = + 9. f) = 0. f) =. f) = 0,. f) = + ),. f) = 6 +,. f) = + +, <. f) = 7. f) = + 9. f) = + 6. f) = 8. f) = f) = With help from our classmates, find the inverses of the functions in Eercises .. f) = a + b, a 0. f) = a h + k, a 0, h. f) = a +b+c where a 0, b a. a + b. f) =, See Eercise below.) c + d. In Eample.., the price of a dopi media plaer, in dollars per dopi, is given as a function of the weekl sales according to the formula p) = 0 for 0 0. a) Find p ) and state its domain. b) Find and interpret p 0). c) In Eample.., we determined that the profit in dollars) made from producing and selling dopis per week is P ) = , for 0 0. Find P p ) ) and determine what price per dopi would ield the maimum profit. What is the maimum profit? How man dopis need to be produced and sold to achieve the maimum profit?
18 . Inverse Functions 9 6. Show that the Fahrenheit to Celsius conversion function found in Eercise in Section. is invertible and that its inverse is the Celsius to Fahrenheit conversion function. 7. Analticall show that the function f) = + + is onetoone. Since finding a formula for its inverse is beond the scope of this tetbook, use Theorem. to help ou compute f ), f ), and f ). 8. Let f) =. Using the techniques in Section., graph = f). Verif that f is onetoone on the interval, ). Use the procedure outlined on Page 8 and our graphing calculator to find the formula for f ). Note that since f0) = 0, it should be the case that f 0) = 0. What goes wrong when ou attempt to substitute = 0 into f )? Discuss with our classmates how this problem arose and possible remedies. 9. With the help of our classmates, eplain wh a function which is either strictl increasing or strictl decreasing on its entire domain would have to be onetoone, hence invertible. 0. If f is odd and invertible, prove that f is also odd.. Let f and g be invertible functions. With the help of our classmates show that f g) is onetoone, hence invertible, and that f g) ) = g f )).. What graphical feature must a function f possess for it to be its own inverse?. What conditions must ou place on the values of a, b, c and d in Eercise in order to guarantee that the function is invertible?
19 96 Further Topics in Functions.. Answers. f ) = + 6. f ) =. f ) = 0. f ) = +. f ) = ) +, 6. f ) = ) +, 7. f ) = 9 + ) +, 8. f ) = 8 ), 9. f ) = + 0. f ) = ) +. f ) = + +. f + ) =. f ) = +. f ) = +, >. f ) = 7. f ) = + 9. f ) = + 6. f ) = + 8. f ) = f ) =. a) p ) = 0. The domain of p is the range of p which is [0, 0] b) p 0) =. This means that if the price is set to $0 then dopis will be sold. c) P p ) ) = , 0 0. The graph of = P p ) ) is a parabola opening downwards with verte 7, ) 7,.67). This means that the maimum profit is a whopping $.67 when the price per dopi is set to $7. At this price, we can produce and sell p 7) =.6 dopis. Since we cannot sell part of a sstem, we need to adjust the price to sell either dopis or dopis. We find p) = 8 and p) = 70, which means we set the price per dopi at either $8 or $70, respectivel. The profits at these prices are P p ) 8) = and P p ) 70) = 0, so it looks as if the maimum profit is $0 and it is made b producing and selling dopis a week at a price of $70 per dopi. 7. Given that f0) =, we have f ) = 0. Similarl f ) = and f ) =
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