We say that two functions and are inverses if for all in the domain of and for all in the domain of . A function can only have an inverse if it is one-to-one, i.e. if we never have for different elements and of the domain. This is equivalent to saying that the graph of the function passes the horizontal line test. The inverse of is denoted , which should not be confused with the function .
For a one-to–one function ,
The domain of is the range of ; the range of is the domain of .
for all in the domain of .
for all in the domain of .
The graph of is the reflection across of the graph of .
if and only if and is in the domain of .
Watch the video:
Finding the Inverse of a Function or Showing One Does not Exist, Ex 3 from https://youtu.be/BmjbDINGZGg
To determine whether or not and are inverses for each other, we check to see whether or not for all in the domain of ,and for all in the domain of .
Determine whether or not the following pairs of functions are inverses:
To check the composition we plug in for in the definition of as follows:
So for all in the domain of . Likewise, you can check that for all in the domain of , so and are inverses.
If we try to proceed as before, we find that:
This doesn’t seem to be the same as the identity function . To verify this, we find a number in the domain of and show that for that value. Let’s try . Since , we find that . Since , these functions are not inverses.
Unfortunately, not every function we would like to find an inverse for is one-to-one. For example, the function is not one-to-one because . If is an inverse for , then implies that . On the other hand, , so . We cannot have it both ways if is a function, so no such inverse exists. We can find a partial solution to this dilemma by restricting the domain of . There are many possible choices, but traditionally we restrict the domain to the interval . The function is now an inverse for this restricted version of .
We consider the function , which is not one-to-one. A piece of the graph of is in Figure 7.1.3(a). In order to find an appropriate restriction of the domain of , we look for consecutive critical points where takes on its minimum and maximum values. In this case, we use the interval . We define the inverse of on this restricted range by if and only if and . The graph is a reflection of the graph of across the line , as seen in Figure 7.1.3(b).
Next we consider the function , which is also not one-to-one. A piece of the graph of is given in Figure 7.1.4(a). In order to find an interval on which the function is one-to-one and on which the function takes on all values in the range, we use an interval between consecutive vertical asymptotes. Traditionally, the interval is chosen. Note that we choose the open interval in this case because the function is not defined at the endpoints. So we define if and only if and . The graph of is shown in Figure 7.1.4(b). Also note that the vertical asymptotes of the original function are reflected to become horizontal asymptotes of the inverse function.
The other inverse trigonometric functions are defined in a similar fashion. The resulting domains and ranges are summarized in Figure 7.1.5.
Find exact values for the following:
Sometimes, is used instead of . Similar “arc” functions are used for the other inverse trigonometric functions as well.
(a) (b) (c) (d)
Since is not in the range of the inverse sine function, we should be careful with this one.
Since we don’t know the value of , we let stand for this value. We know that is an angle between and and that . In Figure 7.1.6, we use this information to construct a right triangle with angle , where the adjacent side over the hypotenuse must equal 11/15. Applying the Pythagorean Theorem we find that ††margin: Λ
Finally, we have:
T/F: Every function has an inverse.
In your own words explain what it means for a function to be “one to one.”
If lies on the graph of , what can be said about the graph of ?
If a function doesn’t have an inverse, what can we do to help it have an inverse?
In Exercises 5–6., given the graph of , sketch the graph of .
In Exercises 7–10., verify that the given functions are inverses.
, and ,
, and ,
In Exercises 11–14., find a restriction of the domain of the given function on which the function will have an inverse.
In Exercises 15–18., find the inverse of the given function.
In Exercises 19–28., find the exact value.
In Exercises 29–32., simplify the expression.
Show that for any in the domain of we have .
Show that for any we have .
Show that for we have .
A mass attached to a spring oscillates vertically about the equilibrium position according to the function . Find the first positive time for which .