###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Proving Two Functions are Inverses - Problem 2

Carl Horowitz
###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Proving that functions are inverses algebraically, whenever we prove that functions are inverses algebraically what we have to do is prove that the composition of two functions in both directions come out with x.

So the way I have written our composition is f of g of x and g of f of x. But those two functions are pretty arbitrary. What we really could say is f of f inverse of x, f of inverse of f, basically two functions that are inverses can go in here, so f and g in this cases are inverses we could have done standard notation f and f inverse.

So for this particular example we are given two functions we want to prove that they are inverses. It doesn’t matter what order we do our compositions because we are going to have to do both of them. So for this one I’ll start out with f of f inverse of x. Taking f inverse and plugging into f, so we have 2, f inverse goes in for x 1/2 x plus 3 those are f inverse plus 3. And do I have an extra parenthesis there I do.

So we want to prove that this is equal to x, so combine like terms inside of our parenthesis we end up with 2 1/2x plus 6 distribute in that 2 through we end up with x plus 12.

So we found the composition of f and f inverse to be x plus 12. In order for these two functions to be inverses what we have is for them to be x. We found it not to be x, so because one of them fails by default neither of these can be an inverse. So we don’t even need to find the other order we know that right away that these two functions are actually not inverses. So we could go through f inverse of f but there’s really no point because this is already proving they're not an inverse. Composition of functions always check to see if you are dealing with inverses.