If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.
Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.
Restrict the domain by determining a domain on which the original function is one-to-one.
Replace
Interchange
Solve for
and rename the function or pair of function
Revise the formula for
by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.
Restricting the domain to find the inverse of a polynomial function
Find the inverse function of
The original function
is not one-to-one, but the function is restricted to a domain of
or
on which it is one-to-one. See
[link] .
To find the inverse, start by replacing
with the simple variable
This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of
and
for the original
we looked at the domain: the values
could assume. When we reversed the roles of
and
this gave us the values
could assume. For this function,
so for the inverse, we should have
which is what our inverse function gives.
The domain of the original function was restricted to
so the outputs of the inverse need to be the same,
and we must use the + case:
The domain of the original function was restricted to
so the outputs of the inverse need to be the same,
and we must use the – case:
Finding the inverse of a quadratic function when the restriction is not specified
Restrict the domain and then find the inverse of
We can see this is a parabola with vertex at
that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to
To find the inverse, we will use the vertex form of the quadratic. We start by replacing
with a simple variable,
then solve for
Now we need to determine which case to use. Because we restricted our original function to a domain of
the outputs of the inverse should be the same, telling us to utilize the + case
If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.