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
.