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 .

> 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,
. 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** !

Tue Nov 28 15:50:23 EST 1995