Consider the functions *f*, *g* defined for any real *x* by:

You can plot the functions to get a hint to whether they are invertible or not. on the interval [-1, 1].

> f:=x->E^x+E^(-x);

> plot(f(x),x=-1..1);

> g:=x->E^x-E^(-x);

> plot(g(x),x=-1..1);

We observe that *f* is not invertible since it does not satisfy the
horizontal line test. Indeed *f* is not one-to-one since, for instance,
*f*(.5) = *f*(-.5). From the plot, it seems that the function *g*
satisfies the horizontal line test. In order to determine its inverse,
we solve for x:

> solve(E^x-E^(-x)=y,x);

We observe that one of the solutions is not defined since the argument of the logarithm can only be positive. Thus:

Let us check that we computed the right inverse. By definition

Indeed if we denote the inverse function by *ginv* and compose the
functions we get:

> ginv:=y->ln(1/2*y+1/2*sqrt(y^2+4));

> (g@ginv)(y);

> simplify((g@ginv)(y));

> (ginv@g)(x);

> simplify((ginv@g)(x));

You have do some manipulations in the last output to obtain *x* !

Thu Apr 10 08:50:01 EDT 1997