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Inverse Functions Inverse Functions In this chapter an inverse function is, when a function has has a function functioll is, when whatwhat an inverse we’llwe’ll covercover chapter In this an inverse, inverse functions are useful to graph an inverse an inverse for, for, how how to graph functions are useful inverse whatwhat an inverse, function, to find the inverse a function. of a of function. to find the inverse and and howhow function, One-to-one One-to-one Suppose a function. f one-to-one if every distinct distinct if every f : Af :—*A B!isBa isfunction. We We call call Suppose f one-to-one of objects is assigned a distinct of objects B. other In other in B.in In pair pair of objects to atodistinct in AinisAassigned of objects pair pair words, object of the target at most object domain fromfrom the the domain one one object target at most of the has has eacheach object words, assigned to it.to it. assigned There a way of phrasing the previous definition a more mathematical mathematical in a in more definition the previous of phrasillg There is a is way language: language: f is one-to-one if whenever wewe have two objects, have one-to-one if whenever f ais and c, in the of f with a 6= c, C, a, c domain two objects E A with a then are guaranteed that4ff(c). (a) 6= f (c). that f(a) we are we guaranteed picture below illustrates a function is not one-to-one, sincesince as as one-to-one, is not thatthat a function picture below illustrates TheThe in the picture, di↵erent objects in the domain of of in the domain differeilt objects two two are are picture, therethere can can see see in the you you g :—+D T!that T that assigned to the object in the target. Namely, Namely, target. in the samesame object are are assigned by gbytogthe g: D 3 6= 7 but g(3) = g(7). 37butg(3)=g(7). D 1 riot 36 38 Example. 2x2isisnot Example. hh: :RR-+!IRRwhere whereh(x) h(x)==x notone-to-one one-to-onebecause because3 3 6=—3 3 and andyet yeth(3) h(3)==h(—3) h( 3)since sinceh(3) h(3)and andh(—3) h( 3)both bothequal equal9.9. zz Horizontal Horizontalline linetest test IfIfaahorizontal horizontalline lineintersects intersectsthe thegraph graphofoff(x) f (x)ininmore morethan thanone onepoint, point, then thenf(x) f (x)isisnot notone-to-one. one-to-one. The Thereason reasonf(x) f (x)would wouldnot notbebeone-to-one one-to-oneisisthat thatthe thegraph graphwould wouldcontain contain two points that have the coordinate. two points that have thesame samesecond second coordinate.InInthe thepicture picturebelow, below,wewe have havedrawn drawnaafunction functionffthat thatintersects intersectsthe thehorizontal horizontalline lineatatheight heightz zininmore more than thanone onepoint, point,for forexample exampleininthe thepoints points(a, (a,z)z)and and(b,(b,z).z).That Thatmeans meansthat that (a)==zzand andf(b) f (b)==z,z,and andone-to-one one-to-onefunctions functionscan’t can’tassign assigntwo twodifferent di↵erent ff(a) objects in the domain to the same object of the target. objects in the domain to the same object of the target. b IfIfevery everyhorizontal horizontalline lineininthe theplane plane intersects intersectsthe thegraph graphofofa afunction functionatatmost mostonce, once, then thenthe thefunction functionisisone-to-one. one-to-one. 3739 Example. BelowBelow is the isgraph g : Rof!gR: where g(x) = x3 . g(x) Any horizontal Examples. the of graph IR R where = x . Any 3 linehorizontal that couldline be that drawn would the graph of gthe in graph at most intersect could be intersect drawn would of one at most g inpoint, so gone is one-to-one. point, so g is one-to-one. — Range Range Recall that thethe target of aoffunction f : fD: ! T is setset T . T.That means Recall that target a function D —f T the is the That means that for any d 2 D, we have that f (d) 2 T , or in other words, whenever wewe that for any d e D, we have that f(d) T, or in other words, whenever putput an an object from thethe domain intointo thethe function, thethe output is an object in in object from domain function, output is a object thethe target. target. ButBut it might notnot be be that every object in T an a object from DD assigned it might that every object in has T has object from assigned to it by the function f . For example, the function g : R ! R defined by by to it by the function f. For example, the function g : R —+ R defined 2 g(x) = x= x hashas R as its its target. if you square a real number, thethe g(x) 2 R as target.However, However, if you square a real number, 2 2 (_3)2 result is never a negative number (2 (22 = 4, = 9, That is, is, none of of result is never a negative number = ( 4, 3) That none = etc.). 9, etc.). thethe negative numbers in the target have a number from thethe domain assigned negative numbers in the target have a number from domain assigned to it. to it. TheThe range of aoffunction f isf the subset of the target consisting of objects range is the subset of the target a function consisting of objects that actually “come outout of”of’ f . f. that actually “come • Let h : h: {1, {1, 2, 3,2,3, 4} 4} ! —+ {3,{3, 4, 7,4,7,8, 8, 9} 9} be be thethe function given by by • Let function given h(1) =9 h(1)=9 h(2) =4 h(2)=4 h(3) =4 h(3)=4 h(4) =8 h(4)=8 If we putput thethe numbers from thethe domain “in “in to”to” h, the only numbers that If we numbers domain from k, the only numbers that “come out” areare 9, 4, 8. That means that thethe range of hofish {9, 4, 8}. “come out” 4, and means that range is {9, 4, 8}. 9, and 8. That • If• gIfg : R: ! is defined by byg(x) g(x) = x=x—2, 2, then g(3) = 3= 3—2 2 = 1. We RR —÷R is defined theng(3) = 1. We putput 3 in3 to gotgot 1 out, so 150is1 an object in the range. in g, to and 1 out, is an object in the range. g, and 40 38 Using our notation for sets, we could write the range of f D —+ T as the set Using our notation for sets, we could write the range of f : D ! T as the Using our notation for sets, we could write the range of f D —+ T as the {y Tj f(x) = y for some x D} set set Meaning that the range is| fthe ofsome the target consisting of objects {{y yof2 fTTj (x)subset ==yyfor xx2 DD} } f(x) for some that are assignedthe an objectoffrom domain. the subset Meaning Meaningthat the subsetofofthe range offfisisthe that therange thetarget targetconsisting consistingofofobjects objects Another what way to say the is to say that it is the smallest set range is, that are assigned an object from the domain. that are assigned an object from the domain. thatAnother can serve as to of therange function. the target Another way what the way to say the rangeis,is,isistotosay saywhat saythat thatititisisthe thesmallest smallestset set that can serve as the target of the function. that can serve as the target of the function. Examples. • The range of f : R —+ R where f(x) = x 2 is the set [0, cc). That’s Examples. Examples. 2 square a real if you of number, because the=result will set never be negative, •• The (x) x2 isisthe —+RRwhere range offf: :RR! The range whereff(x) = x the set[0,[0,1). cc). That’s That’s be 0,square and ita could but it could be any positive number. because if you real number, the result will never be negative, because if you square a real number, the result will never be negative, but be0,0, and andititcould but itit could could be couldbe anypositive beany positivenumber. number. • The range of g : IR —+ R where g(x) = eX is the set (0, cc). If x R that ex could then we know of positive be anyg(x) neverIf0,x and •• The R where = ex number. is the setIt’s (0, 1). 2 RR The range range ofgg : : RIRx! —+ R where g(x) = eX is the set (0, cc). If x negative. it’sthen never that eex could then we we know know that couldbe positivenumber. beany anypositive number. It’s It’snever never0,0,and and it’s never negative. it’s never negative. I’ Rane I’ Rane 39 41 39 I e I e loge: is R.real Anynumber real number The range is an output • The •range of logeof: (0, 1)(0, !cc) R is—*R.R Any is an output of logeof . loge. (I OntoOnto Suppose f : D f! T We call ontof ifonto the if range of f equals is a function. Suppose D is—+aTfunction. Wef call the range of f equals T . T. In other words,words, f is onto every in the in target has at has leastatone object if object every object In other the target least one object f isifonto from the assigned to it by fromdomain assigned to fit. by f. the domain Examples. Examples. • The•graph of f : of R f! R f (x) =f(x) x2 is= drawn on theonprevious JR where The graph —+ JR where 2 is drawn x the previous page. page. The range of f isof[0,f 1). of f isofR,f and 1) [0, 6= cc) R, so fR,isso f is is [0,The cc).target The target The range is R,[0,and not onto. not onto. • The•graph of g : of R g: !R g(x) =g(x) ex is= drawn on theonprevious JR where The graph —+ JR where is drawn the previous page. page. The function g hasg the for its its range. range while its target is R. of g, has set the(0, set1) The function JR for This equals the target Because 6= R, g is not onto. so g(0,is1) onto. • The •target of logeof: (0, 1)(0, !cc) R equals the range logeof: (0, 1)(0, !cc) R. —+ JR. loge: The target —* JR equals the of range 1og: They They are both SoR. logSo !cc) R is onto. are R. both loge1) : (0, R is onto. e : (0, • Let’s• compare the first above above — the example f : R f!: R the example first example Let’s compare the example JR —+ JR 2 2 wherewhere f (x) = x — to the function h : R ! [0, 1) where h(x) = x . The to the function h : R - [0, cc) where h(x) = x f(x) = . The 2 two functions are similar, and the between the two two functions are similar, andonly the di↵erence only difference between thefunctions two functions is what is. is their what target their target is. For the reasonreason that the of f isof[0,f 1), of h isofalso Forsame the same thatrange the range is [0,the cc),range the range h is also [0, 1).[0, Because the target of h is also [0, 1), the function h is onto, even cc). Because the target of h is also [0, cc), the function h is onto, even though the function f is not What What makesmakes a function onto onto thoughvery the similar very similar function not onto. a function f isonto. is justiswhether its target is chosen to be to theberange. just whether its target is chosen the range. * ** ** ** ** ** ** 42 ** 40 ** ** ** ** ** * Inverse functions The defining characteristic of the cube root function is that for any x 2 R we have that p p 3 x3 = x and ( 3 x)3 = x This is just an example of the definition of inverse functions. Suppose f : A ! B is a function. A function g : B ! A is called the inverse function of f if f g = id and g f = id The line above says that f and g are inverse functions if f g and g f are each the identity function, meaning that for any x we have f g(x) = id(x) and g f (x) = id(x) Remember that id(x) = x, so the line above can be written more simply as f g(x) = x and g f (x) = x If g is the inverse function of f , then we often rename g as f 1 . Then we can rewrite the equations above replacing g with f 1 and we have f f 1 = id and f 1 f = id 1 f (x) = x or equivalently, that for any x f f 1 (x) = x and f Examples. • Let f : R ! R be the function defined by f (x) = 2x + 2. Then f has an inverse function f 1 : R ! R where f 1 (x) = 12 x 1. Notice that ⇣1 ⌘ ⇣1 ⌘ 1 1 f f (x) = f (f (x)) = f x 1 = 2 x 1 + 2 = x 2 2 43 Similarly f 1 f (x) = f 1 (f (x)) = f 1 ⌘ 1⇣ (2x + 2) = 2x + 2 2 1=x • The inverse of addition is subtraction. If g(x) = x + 5, then g 1 (x) = x 5. • The inverse of multiplication is division. If h(x) = 3x, then h 1 (x) = x3 . • The inverse ofpan odd power is an odd root. If f (x) = x3 , then f 1 (x) = 3 x. • The inverse of exponential with base a > 0 (as long as a 6= 1) is logarithm base a. If g(x) = 2x then g 1 (x) = log2 (x). • The inverse of x1 is itself. If h(x) = x1 , then h 1 (x) = x1 . • Even powers do not have inverses exactly. They do have partial inverses though, and those are the even roots. To repeat, the function f : R ! R where f (x) = x2 does not have an inverse function. However, we can restrict the domain and target of f to obtain the function f : [0, 1) ! [0, 1) where f (x) = x2 . This function does have an inverse, the square-root function p 2 : [0, 1) ! [0, 1). Because the domain of the square-root function is [0, 1), it wouldn’t make any sense to try to take the square-root of a negative number. The Inverse of an inverse is the original If f 1 is the inverse of f , then f 1 f = id and f f 1 = id. We can see from the definition of inverse functions above, that f is the inverse of f 1 . That is (f 1 ) 1 = f . For example, as the inverse of addition is subtraction, the inverse of subtraction is addition. As the inverse of an exponential function is a logarithm, the inverse of a logarithm is an exponential function. Inverse functions “reverse the assignment” The definition of an inverse function is given above, but the essence of an inverse function is that it reverses the assignments of the original function. The first chart below shows inputs and their respective outputs for the function x2 . The second chart shows inputs and their respective outputs for the p 2 inverse function x.pIf you reverse the two columns of numbers for the x2 chart, you’ll get the 2 x chart. 44 ______________ x2 x I 1 x2 2 1 1 4 3 1 1 2 2 4 4 2 9 3 3 9 9 4 16 16 16 16 5 1 4 9 4 x I 1 1 p 2 x 3 4 4 25 25 25 5 25 5 5 p The example above isn’t particular to x2 and 2 x. It applies to any pair of The functions example above to x2 and It applies to any pair of inverse as the isn’t claimparticular below demonstrates. inverse functions as the claim below demonstrates. Claim: For inverse functions f and f 1 , if f (a) = b, then f 1 (b) = a. Claim: For inverse functions f and if f(a) b, then f’(b) a. Proof: If f (a) = b, then = Proof: If f(a) = b, then f I 1 = 1 (b) = f 1 (f (a)) f(b)= f f’(f(a)) 1 f (a) 2 = ==fof(a) a =a ⌅ 3 . 4 x. x÷7 ______________ 5 ._- .XZ3X I 45 -F_i ÷5 x2 The example abov inverse functions as I Claim: For invers Proof: If f(a) 43 1 1 = 1 Examples. The claim above states that if f (a) = b, then f 1 (b) = a. The examples below are written in the style of the equivalent statement that if f (a) = b, then a = f 1 (b). • If f (3) = 4, then 3 = f • If f ( 2) = 16, then • If f (x + 7) = • If f 1 (0) = • If f 1 (x2 1 (4). 2=f 1 (16). 1, then x + 7 = f 1 ( 1). 4, then 0 = f ( 4). 3x + 5) = 9, then x2 3x + 5 = f (9). p • Because 52 = 25, we know that 5 = 2 25. • Because 34 = 81, we know that 4 = log3 (81). In the 7 examples above, we “erased” a function from the left side of the equation by applying its inverse function to the right side of the equation. When a function has an inverse A function has an inverse exactly when it is both one-to-one and onto * * * * * * * * * * * * * The Graph of an inverse If f is an invertible function (that means if f is a function that has an inverse function), and if you know what the graph of f looks like, then you can draw the graph of f 1 . The claim below will tell us how. Claim: If (a, b) is a point in the graph of f (x), then (b, a) is a point in the graph of f 1 (x). 46 Proof: If (a, b) is a point in the graph of f (x), then f (a) = b, and our claim on the previous page tells us that f 1 (b) = a. That means f 1 assigns b to a, or in other words, that (b, a) is a point in the graph of f 1 (x). ⌅ New How points in graph of f(x) visual effect New How points in graph of ff(x) visual effect function become new graph New How points inus graph of if (x) visual of effect What the above claim tellspoints is of that we know the graph an invertible function become points of new graph functionwe canbecome pointsallofofnew function, just switch thegraph first and second coordinates of our graph and we’ll be left with a graph of the inverse function. Geometrically, (a, and b) second (b, a) coordinates flipofover thein“xthe = y line” when 1f’(x) you switch all the first points plane, ff 1(x) (a, b) ⇥ ⇤ (b, a) flip over the “x = y line” (x) is to flip the plane (a, b) over ⇤⇥ (b,the a) “x = y line”. flip over the “x = y line” the result Examples. Example. Example. Example. (3, s), c(z) (s, 3’) (a,-a) (-3~.-5) ** ** ** ** ** ** 47 ** 71 76 ** ** ** ** ** ** ci cJ LI > ii c) ci + 0 “I M cJ C I 0 1’ -\> -\> en 48 46 Exercises It’s easy to solve equations that are formed from inverse functions. For example, if f is a function with an inverse function f 1 , and if we are given the equation f (x + 3) = 4, then we can solve for the variable x by erasing the function f and applying its inverse to the other side of the equation: x + 3 = f 1 (4). Next, we can erase addition by 3 by applying its inverse, subtraction by 3, to the other side of the equation: x = f 1 (4) 3. We have solved for x. Just remember, If f (stu↵) = (other stu↵) then (stu↵) = f 1 (other stu↵) For #1-18, solve for x. 1.) x + 3 = 7 7.) p 3 x=2 13.) loge (2x) = 4 2.) 3x = 60 8.) x 4 = 20 14.) (x + 2)3 = 3.) x3 = 7 9.) 1 x = 15.) 1 x+1 10.) 2x + 5 = 3 16.) p 3 5.) ex = 5 11.) 10x 17.) log2 (3x 6.) x 12.) 4.) 1 x = 10 2=4 x 1 3 1 3 1 = 17 =7 18.) 7 1 2x+3 1 8 =2 x= 4 1) = 4 =4 If you know that f is an invertible function, and you have an equation for f (x), then you can find the equation for f 1 in three steps. Step 1 is to replace f (x) with the letter y. Step 2 is to use algebra to solve for x. Step 3 is to replace x with f 1 (y). After using these three steps, you’ll have an equation for the function f 1 (y). 49 Example: To find the inverse of f (x) = loge (x 5), first replace f (x) with y to get y = loge (x 5). Second, solve for x as follows: Erase loge from the right side of the equation by applying its inverse, exponential base e to the left side of the equation and we have ey = x 5. Next, we can erase subtraction by 5 on the right by applying its inverse, addition by 5, on the left. That gives us ey + 5 = x. The third step is to simply replace x with f 1 (y), and that gives us an equation for the inverse function, f 1 (y) = ey + 5. Some people prefer to use x as the variable for their functions whenever they can. If you do, that’s fine, just change the y’s on the last line above to x’s. That is, f 1 (y) = ey + 5 is exactly the same function as f 1 (x) = ex + 5, or for that matter, it’s also exactly the same function as f 1 (z) = ez + 5. For exercises #19-30, find the inverse function of the given function. 23.) g(x) = x 10 24.) h(x) = p 3 21.) h(x) = 2x 25.) f (x) = 3x x 2 22.) f (x) = e4x 26.) g(x) = log10 (x + 2) 19.) f (x) = x + 5 20.) g(x) = 2 x 1 27.) h(x) = (2x + 1)3 + 4 x 4 28.) f (x) = 52x+3 + 1 29.) g(x) = x 4 30.) h(x) = loge (3x For exercises #31-32, find the roots of the quadratic polynomials. 31.) 2x2 3x 32.) 4x2 x 1 3 50 1)