# Thread: Rule of riciprocacy of inverse functions?

1. ## Rule of riciprocacy of inverse functions?

I'm in a debate with my math teacher over this, and she just doesn't understand what I'm asking. It's actually a pretty simple question. Is there a rule, law, or algebraic solution that states:

if $j(x)=h^-1(x)$ then $h(x)=j^-1(x)$

I was marked incorrectly on an assignment for refusing to make that assumption, and I refused to make that assumption because I was not able to find such a rule anywhere online or in the textbook. I don't mind being wrong if I actually am incorrect, I just want proof. Thanks in advance.

2. Originally Posted by ChrisEffinSmith
I'm in a debate with my math teacher over this, and she just doesn't understand what I'm asking. It's actually a pretty simple question. Is there a rule, law, or algebraic solution that states:

if $j(x)=h^-1(x)$ then $h(x)=j^-1(x)$

I was marked incorrectly on an assignment for refusing to make that assumption, and I refused to make that assumption because I was not able to find such a rule anywhere online or in the textbook. I don't mind being wrong if I actually am incorrect, I just want proof. Thanks in advance.

Your teacher is correct:

$j(x) = ln(x)$
$j^{-1}(x) = e^x$

$h(x) = e^x$
$h^{-1}(x) = ln(x)$

There may be a written rule for this somewhere, but it's actually mathematically true.

My homemade rule

If the function f(x) is the inverse function of g(x), then g(x) is the inverse function of f(x).

3. I assumed there was an answer somewhere. It just frustrates me to no end that instead of being taught (or even told), we were just expected to assume. I guess I'll just have to accept the lost marks, though I believe the fault still lies elsewhere... Thanks for your reply, at least now I know.

4. In fact, the definition of "inverse" function is that g is the inverse function to f if and only if f(g(x))= x and g(f(x))= x, and swapping "f" and "g" the first equation becomes g(f(x))= x, and the second equation becomes g(f(x))= x so the f is also the inverse of g.