Consider the following set
For all , we have
Since if and only if (Exercise-1),
From (1-2) and (1-3), we conclude:
i.e., defines a function.
Alternatively, (1-1) can be expressed as
It shows that is the inverse of . Therefore, we name the function defined by (1-1) and write it as
Let’s look at another set:
For a pair , we have
For another pair ,
Since does not implie . It means does not define a function.
However, modification of gives
It defines a function.
Rewrite (2-3) as
Then , we have
Notice that by (2-3),
Cleary, (2-5) is true if and only if . For if , by ,
We name the function defined by (2-3) as (2-4) shows that it is the inverse of
See “Deriving Two Inverse Functions” for the explicit expressions of and .
Without loss of generality, we assume that By Lagrange’s mean-value theorem (see “A Sprint to FTC“),
Exercise-1 Show that if and only if .