Consider the following set

For all , we have

That is,

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 ,

Therefore,

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 .

Prove

Without loss of generality, we assume that By Lagrange’s mean-value theorem (see “A Sprint to FTC“),

where

We have

and

Consequently,

i.e.,

*Exercise-1 *Show that if and only if .