Let `f: RrarrR` be defined by `f(x)=(e^x-e^(-x))//2dot` then find its inverse.

Explanation

Let `p,q in R " and let " f(p)=f(q)`
`impliese^(p)-e^(-p)=e^(q)-e^(-q)`
`implies e^(p)-e^(q)=e^(-p)-e^(-q)`
`implies e^(p)-e^(q)=(1)/(e^(p))-(1)/(e^(q))`
`implies e^(p)-e^(q)=(e^(p)-e^(q))/(e^(p)e^(q))`
`implies e^(p)-e^(q)=0 " or " e^(p)e^(q)= -1 ` (not possible)
`implies e^(p)=e^(q)`
`implies p=q`
Thus, f(x) is one-one.
Lets us find the range of the function.
Clearly when x approaches to infinity `(e^(x)-e^ (-x))` approaches to infinity and when x approaches to negative infinity, `(e^(x)-e^ (-x))` approaches to negative infinity.
Also, `(e^(x)-e^ (-x))` continuously exists.
So, range of function is R.
Thus f(x) bijective and so invertible.
Let `e^(x)-e^(-x)=y`
`implies e^(2x)-ye^(x)-1=0`
`impliese^(x)=(y+-sqrt(y^(2)+4))/(2)`
`impliese^(x)=(y+sqrt(y^(2)+4))/(2) " " ("as "y-sqrt(y^(2)+4) lt 0` for all y and `e^(x)` is always positive)
`implies x=log_(e)((y+-sqrt(y^(2)+4))/(2))`
` implies f^(-1)(x)=log_(e)((x+-sqrt(x^(2)+4))/(2))`