In this section, we will practice how to actually algebraically determine whether the inverse for a given function exists, and how to find it.
Let be the function given by , for every real number . Is one-to-one? If so, what is
the inverse ? What is its domain?
Recall here that is one-to-one if always implies that . What happens in this case?
We start with the equality
and if we can arrive at , then our function is one-to-one. Subtracting from both
sides gives . Dividing both sides by , we have that . As the function “taking the third
power” is one-to-one (namely, its inverse is “taking cube roots”), it follows that .
This means that one can start with and solve for as a function of . When this is
finished, one replaces and with and , respectively, for the sake of conventions. We
have that
Therefore, we conclude that
Observe that the domain of is precisely the set of all real numbers, as there is no
positivity restriction for taking “odd roots”. This turns out to be exactly the range of
the original function , suitably read in the variable: namely, the range of consists of
all real numbers , as the graph shows: 
Observe that the graph also shows that passes the Horizontal Line Test, a useful
sanity check.
What happened with the domain of on the previous example is a very general
phenomenon: if a given function is one-to-one, so that exists, the domain of is
exactly the range of the original function .
Let be the rational function given by for every real number not equal to . Is
one-to-one? If so, what is the inverse ? What is its domain?
The formula defining is different from the formula defining the function from the
previous example: there, we had a polynomial, while here we have a rational
function. The process to study it, however, is the same. To decide whether is
one-to-one or not, we will start with , and try to conclude that . So, we start
with
By cross multiplying, we have that
Distributing the products on both sides, we obtain
Since , adding this quantity to both sides gives us that . Finally, dividing everything
through by , it follows that . Therefore, is one-to-one. With this in place, we can
attempt to find . Starting with
the only thing it seems we might be able to try is to cross multiply terms, and
distribute the products. So
As our goal is to solve for in terms of , let’s move everything that contains in it to
one side, and leave the rest which does not contain it on the other side. We obtain .
Factoring , it follows that . Finally, we conclude that
by replacing and with and , respectively. We observe that the domain of
consists of all real numbers different from , in the same way that the range
of consists of all real numbers different from (to wit, if we had , then
cross multiplying and simplifying yields , leading to a terribly nonsensical
).
Let be the function given by for every real number greater or equal to . Is
one-to-one? If so, what is the inverse ? What is its domain?
Let’s start verifying whether is one-to-one or not. So, as before, we start with and
try to obtain that . In other words, we start with . From arguing either that the
exponential function itself is one-to-one, or applying on both sides of this equality, it
follows that . Raising both sides to the square, we have that , and adding to
everything yields , as desired. Therefore, is one-to-one. To find a formula for , we
start with and solve for in terms of . First, we apply on both sides of the equality,
as to obtain . Next, we take squares on both sides, so . Finally, we add to both
sides:
Note that is not the same thing as (to wit, the latter equals , by properties of the
logarithm function we shall soon see). The inverse is defined for all , due to the
presence of in its formula.