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 !