Inverse functions, Algebra

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In previous section we looked at the two functions  f ( x) = 3x - 2 and g ( x )= x/3 + 2/3 and saw that

                                         ( f o g ) ( x ) =(g o f )( x ) = x

and as noted in that section it means that these are very special functions. Let's see what makes them so special.  Assume the following evaluations.

f ( -1) = 3( -1) - 2 = -5 ⇒     g ( -5) = -5/3 + 2/3 = -3/3 = -1

g ( 2) = 2/3 +2/3 = 4 /3       ⇒ f ( 4 /3)=3(4/3 ) - 2 = 4 - 2 = 2

In first one we plugged x = -1 into f ( x ) and got a value of -5. Then we turned around and

Plugged x = -5 into g ( x ) and got a value of -1, the number that we begin with.

In the second one we did something similar.  Here we plugged x = 2 into g ( x ) and got a value of 4/3 , we turned around & plugged this into f ( x ) & got a value of 2, that is again the number that we begin with.

Note that here we actually are doing some function composition.

The first one is actually,

                                        ( g o f ) ( -1) = g [f ( -1)]=  g [-5] = -1

and the second one is,

                          ( f o g ) ( 2) =f [g ( 2)]= f [ 4/3 ] = 2

So, just what is going on here?  In some manner we can think of these two functions as undoing what the other did to number.  In the first one we plugged x = -1 into f ( x ) and then plugged the result from this function evaluation back into g ( x ) and in some way g( x ) undid what f ( x ) had done to x = -1 and gave us back the original x which we started with.

Function pairs which exhibit this behavior are called inverse functions. Before formally explaining inverse functions and the notation which we're going to employ for them we have to get a definition out of the way.


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