# Question:A question about inverse functions

## Question:A question about inverse functions

Maple

I am presently working on bivariate functions defined this way

```C := (u, v) -> (phi@@(-1))(phi(u)+phi(v));
```

phi is a function of specific expression named the "generator". Both u and v are assumed to be in the closed interval [0,1].

Here is an example:

```restart
phi := u -> (u^(-theta)-1)/theta:
C := (u, v) -> (phi@@(-1))(phi(u)+phi(v)):
C(u, v)
/ (-theta)        (-theta)    \
| u         - 1   v         - 1|
@@(phi, -1)| ------------- + -------------|
\    theta           theta    /
```

This definition of C is correct providing that  theta in [-1, +infinity) \ {0}.
As you can see, the display of C(u, v) contains the inverse function phi@@(-1) which Maple doesn't seem to know what to do with.

What I would like is to get rid of  phi@@(-1) and get

```C(u, v);
(-1+u^(-theta)+v^(-theta))^(-1/theta)```

The only way I found to get this is to do that:

```restart
phi := u -> (u^(-theta)-1)/theta:
(phi@@(-1)) := u -> solve(phi(x)=u, x): # explicit definition of (phi@@(-1))
C := (u, v) -> simplify((phi@@(-1))(phi(u)+phi(v))) assuming theta >= -1, theta <> 0:
C(u, v);
/    1  \
|- -----|
\  theta/
/      (-theta)    (-theta)\
\-1 + u         + v        /

```

As you see I have been forced to tell Maple what the inverse function of phi was.
Is there another way do get this result without writting the bold red line?

Maple knows several inverse functions (trigonometric functions for instance), but how does it know that?
As Maple does not seem to use a (f@@(-1)) := u -> solve(f(x)=u, x) like definition, does it uses a correspondence table between functions and their inverse?
If it is so can we augment it?