Inverse functions - mathcentre.ac.uk

1 Inverse functionsmc-TY- Inverse -2009-1An Inverse function is a second function which undoes the work of the first one. In this unitwe describe two methods for finding Inverse functions , and wealso explain that the domain of afunction may need to be restricted before an Inverse function can order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: understand the difference between Inverse functions and reciprocal functions , find an Inverse function by reversing the operations appliedtoxin the original function, find an Inverse function by algebraic manipulation, understand how to restrict the domain of a function so that itcan have an Inverse function, sketch the graph of an Inverse function using the graph of theoriginal outf 1by reversing the operations algebraic manipulation to work out Inverse graph off mathcentre 20091.

2 IntroductionSuppose we have a functionfthat takesxtoy, so thatf(x) = Inverse function, which we callf 1, is another function that takesyback tox. Sof 1(y) = 1to be an Inverse off, this needs to work for everyxthatfacts PointThe Inverse of the functionfis the function that sends eachf(x)back tox. We denote theinverse offbyf Working outf 1by reversing the operations offOne way to work out an Inverse function is to reverse the operations thatfcarries out on anumber. Here is a simple example. We shall setf(x) = 4x, so thatftakes a numberxandmultiplies it by 4:f(x) = 4x(multiply by 4).We want to define a function that will take 4 timesx, and send it back tox. This is the sameas saying thatf 1(x)dividesxby 4. Sof 1(x) =14x(divide by 4).There is an important point about notation here. You should notice thatf 1(x)does not mean1/f(x). For this example,1/f(x)would be1/4xwith thexin the denominator, and that is notthe same is a slightly more complicated example.

3 Suppose we havef(x) = 3x+ can break up this function into a series of operations. First the function multiplies by 3, andthen it adds on 33x + 23x+ mathcentre 2009To get back toxfromf(x), we would need to reverse these operations. So we would need totake away 2, and then divide by 3. When we undo the operations,we have to reverse the orderas (x 2)/3 3 33xx 2 + 2 23x+ 2xNow we have reversed all the operations carried out byf, and so we are left withf 1(x) =x is one more example of how we can reverse the operations of a function to find its we havef(x) = 7 is easier to see the sequence of operations to be carried out onxif we rewrite the function asf(x) = x3+ the first operation performed byftakesxtox3; then the result is multiplied by 1; andfinally 7 is added (cube)x3 ( 1) x3 + 7 x3+ 7So to get fromf(x)tox, we need to start by taking away 7.

4 Then we need to undo the operation multiply by 1 , so we divide by 1. And finally we undo the first operation by taking the 7 x (cube) (cube root)x37 x ( 1) ( 1) x3x 7 + 7 7 x3+ 7xNow we have reversed every operation carried out byf. Sof 1(x) =3 7 x . mathcentre 2009 Key PointWe can work outf 1by reversing the operations off. If there is more than one operation thenwe must reverse the order as well as reversing the Work out the inverses of the following functions :(a)f(x) = 6x, (b)f(x) = 3 + 4x3, (c)f(x) = 1 Using algebraic manipulation to work out inversefunctionsAnother way to work out Inverse functions is by using algebraic manipulation. We can demon-strate this using our second example,f(x) = 3x+ the Inverse function takes us fromf(x)back tox. If we sety=f(x) = 3x+ 2,thenf 1is the function that takesytox. So to work outf 1we need to know how to get toxfromy.

5 If we rearrange the expression forywe obtainy= 3x+ 2,y 2 = 3xso thatx=y we wantf 1(y) = (y 2)/3, and this is exactly the same as saying that the functionf 1isgiven byf 1(x) = (x 2) can use the method of algebraic manipulation to work out inverses when we have slightlytrickier functions than the ones we have seen so far. Let us takef(x) =xx 1,x > have made the restrictionx >1because atx= 1the function does not have a value. Thisis because the denominator is zero whenx= mathcentre 2009 Now we sety=x/(x 1). Multiplying both sides byx 1we gety(x 1) =x ,and then multiplying out the bracket givesyx y=x .We want to rearrange this equation so that we can expressxas a function ofy, and to do thiswe take the terms involvingxto the left-hand side, givingyx x=y .Now we can then take outxas a factor on the left-hand side to getx(y 1) =y ,and dividing throughout byy 1we finally obtainx=yy the Inverse function isf 1(y) =y/(y 1), and this is exactly the same as saying that thefunctionf 1is given byf 1(x) =x/(x 1).

6 So in this casef 1happens to be the same the last example, it would not have been possible to work out the Inverse function by trying toreverse the operations off. This example shows how useful it is to have algebraic manipulationto work out PointAlgebraic manipulation is another method that can be used towork out Inverse Use algebraic manipulation to work outf 1for each of the following functions :(a)f(x) =34x 4forx >1, (b)f(x) =x+ 1x+ 2forx > mathcentre 20094. Restricting domainsNot all functions have inverses. For example, let us see whathappens if we try to find an inverseforf(x) = (x)x ?x ?f(x) = x2 When we define an Inverse function forf, we look for another function that takes the valuesf(x)and gives us backx. But in the case off(x) =x2there are two values ofxthat give thesamef(x). This is because bothf(x) =x2and alsof( x) =x2. We cannot definef 1ofsomething to be two different get around this problem, we restrict the domain of the function.

7 So for example withf(x) =x2, if we define the function only forx 0then the graph looks like ?f(x)f(x) = x2x 0So now we have exactly one value ofxgiving each value off(x). This restricted version off(x)can have an Inverse . The Inverse isf 1(x) = + mathcentre 2009We could instead have restrictedf(x)tox 0. This still gives only one value ofxfor eachvalue ofx2. This time the Inverse is ?xf(x) = x2x 0f(x)Here is another example of a function that we need to restrictin order to define an Inverse . Letus look atf(x) = sinx. The graph of the function looks like (x)f(x) = sin xThis time if we try to define the Inverse function, we see that there are many possible values ofxfor eachf(x). We need to restrict the domain of our function so that we are looking at asection with only a single value ofxfor each value off(x). We do this by setting the domain ofsinxto be 90 x 90.

8 Mathcentre 2009f(x)f(x) = sin x 90 x 90 It is particularly important here to remember thatsin 1xis the Inverse function tosinx, andthat it does not mean(sinx) 1, even though the similar expressionsin2xdoes mean(sinx) this reason, the Inverse functionsin 1xis sometimes is also possible to define the Inverse functionscos 1xandtan 1xby restricting the domainsof the functionscosxandtanx. These Inverse functions are also calledarccosxandarctanx,and you can find out more about them in the unit on Trigonometric functions cannot have inverses, even if we restrict their domains. For example, a constantfunction cannot have an (x)f(x) = 4x ?However small we make the domain, there are always lots of values ofxgiving the same value off(x). The only way we can get a single value ofxis by restricting the domain to a single we say that this function has no Which of these functions need to have their domains restricted in order to define an Inverse ?

9 How would you restrict their domains?(a)f(x) =x2+ 2x+ 1, (b)f(x) =x3, (c)f(x) = 5 x2, (d)f(x) = sin mathcentre 20095. The graph off 1 There is an easy way to work out the graph of an Inverse functionf 1, using the graph of theoriginal that we have the graph of some functionf. Then a point on the graph offwill haveco-ordinates(x, f(x)).(x, f(x))Remember that an Inverse function sendsf(x)back tox. So the graph off 1must containthe points(f(x), x). So if we interchange thexandyaxes, we will get the graph of the inversefunction.(x, f(x))(f(x), x)By going from the graph offto the graph off 1, we are reflecting the graph in the diagonalline. But this diagonal line is the liney=x. So the graph off 1is just the graph offreflectedin the liney= For each of the following, sketch the graph off(x)and use it to sketch the graph off 1(x):(a)f(x) = 3x,(b)f(x) = 2x+ 5, (c)f(x) =x2forx 0,(d)f(x) = 1/xforx >0, (e)f(x) = mathcentre 2009 Answers1.

10 (a)f 1(x) =x6(b)f 1(x) =3 x 34(c)f 1(x) =x 1 3=1 x32.(a)f 1(x) =3 + 4x4x(b)f 1(x) =1 2xx 13.(a)x 1orx 1(b) no restriction needed (c)x 0orx 0(d) 45 x 45 4.(a)f(x)f 1(x)(b)f(x)f 1(x)(c)f(x)f 1(x)(d)f(x)f 1(x)(e)f(x)f 1(x) mathcentre 2009