# Inverse Functions

What is the general solution to the equations f(g(x))=h(x) and g(f(x))=h(x), or how do you find the particular soltuion to the equations given a function f (given any f what is the general form of h(x) such that g(x) exists)?

My thoughts on the topic:

Specific example:
Suppose we have f(x)=x+1 and h(x)=x. Then g(x)=x-1.
f(g(x))=g(x)+1=x+2
g(x)=x+1
g(f(x))=x+1+1=x+2 so it checks.
Now suppose h(x)=3x
f(g(x))=g(x)+1=3x
g(x)=3x-1
g(f(x))=3(x+1)-1=3x+2 doesn't check.

More general case:
Suppose now f(x)=x2
f(g(x))=[g(x)]2=h(x)
g(x)=sqrt(h(x))
g(f(x))=[squ]h(x2)=h(x)
So g(x) exists for a given h(x) if
h(x2)=[h(x)]2
I don't think I can solve this any further. I feel resigned to the fact that I must check to see that this holds for some specific h(x) rather than finding the general solution.
Certainly, h(x)=xk works. Any further thoughts?

Edit:
Oh by the way, h(x)=x of course works for all f provided that f has an inverse. So any general form should be able to reduce to h(x)=x.

Last edited:

Hurkyl
Staff Emeritus
Gold Member
Try making the substitution x = f(z)

On the verge of tears... can't make substitution work... must change major....no more hints....

Hurkyl
Staff Emeritus
Gold Member
Hrm, while that substitution is interesting, it doesn't seem to help as much as I had hoped...

(it only lets me prove that f, g, and h all commute)

Ok, here's the next suggestion! See if you can apply some interesting operations and substitutions to solve each of the initial two equations formally for g-1(x). I get:

g-1(x) = h-1(f(x)) = f(h-1(x))

and this solution, if it exists, checks out.

With the information given, I doubt you can make a much stronger statement than this.

HallsofIvy
Homework Helper
You are asking "given any f what is the general form of h(x) such that g(x) exists? " (such that f(g(x))= h(x)).

It doesn't depend of h so much as f: If f has an inverse, then

f-1(f(g(x))= f-1(h(x)) so
g(x)= f-1(h(x)).

Originally posted by HallsofIvy
You are asking "given any f what is the general form of h(x) such that g(x) exists? " (such that f(g(x))= h(x)).

It doesn't depend of h so much as f: If f has an inverse, then

f-1(f(g(x))= f-1(h(x)) so
g(x)= f-1(h(x)).
Your answer doesn't check out. I've come up with several expressions that did something similar. The tricky part is that you have to utilize both f(g(x))=g(f(x)) and f(g(x))=h(x). In your answer you don't utilize the first one.
Choose f(x)=x3, h(x)=2x
f-1(x)=x1/3
g(x)=(2x)1/3
but g(f(x))=x*21/3

Hurkyl
Staff Emeritus
Gold Member
(false therefore true) is a true implication.

Note that Ivy's conclusion began with "If f has an inverse".

But f(x)=x3 does have an inverse. In fact I purposely chose this one rahter than
f(x)=x2 for that exact reason.

Say we have f(x)=ex and h(x)=2x. Then g(x)=ln(2x) but g(f(x))=ln(2ex)=ln2+x.

Anyway the point is you can reach Ivy's conclusion from just the fact that f(g(x))=h(x). Simply apply the inverse of f on both sides. But this is true regardless of the fact that g(f(x))=f(g(x)). A solution should be dependent on this fact.