Motivating Questions
- How to obtain, from the graph of a given one-to-one function , the graph of inverse function ?
- Knowing formulas for and , is there a nice formula for the inverse of the composition , whenever it makes sense?
Introduction
Recall that given a function from a set to a set , the graph of is the set of all points , where belongs to . When and are intervals of real numbers, we are able to draw the graph of in the -plane.
When is one-to-one, we may talk about the inverse function (from the range of inside , to ), which is characterized by the equivalence of the relations and . However, as we have seen in the previous section, when solving algebraically for the inverse , we switch the letters and after all calculations are done. This step, while a priori seemingly arbitrary, allows us to draw the graphs of and together to compare them.
With this in mind, we have that a point is in the graph of precisely when is in the graph of . But what is the geometric relation between the points and ? Or, in other words, what does it mean to switch the coordinates of a point? Let’s see what happens with a few points, say, , , , and . They will be indicated in black, while the corresponding points with the switched coordinates will be indicated in blue.
As the picture indicates, switching the coordinates of a point amounts to reflecting it about the principal diagonal line, whose line equation is .
To elaborate more on this conclusion, we note that it allows us to read values of even without actually graphing it! Consider the following graph, of a function :
Even though we do not know the actual formula defining , just from knowing that and are in the graph of , we may safely conclude that and are in the graph of , which is to say that
Practicing
Let’s explore several situations on what follows. To draw the graph of to begin with, we rely on what was discussed on previous chapters.
Consider the function . Draw the graph of from the graph of . Then, algebraically
find to confirm your work.
Since is a linear function, all we need to know to draw its graph are its
slope and -intercept. The formula immediately tells us that the slope is
and the -intercept is . The inverse of a one-to-one linear function is also
a linear function, and a line is determined by two points. So in this case,
all we have to do is to reflect two points in the graph of about the line ,
and draw the line passing through the reflected points: this is the graph of
.
Let’s make a few observations, based on the picture.
- The -intercept of became the -intercept of .
- The -intercept of became the -intercept of .
- The graphs of and meet precisely at the line .
- Both and are increasing functions.
The first and second points are actually true for any (non-horizontal, and hence one-to-one) linear function; the third point is true for any one-to-one function (with the caveat that the graphs may meet more than once, but still always an points lying on the line ); the fourth point illustrates a very general phenomenon, also true for any one-to-one function – namely, that a one-to-one function must necessarily be (strictly) increasing or decreasing, and will have the same behavior as (explictly, is increasing if is increasing, and is decreasing if is decreasing).
Finally, let’s double-check our work algebraically. If , then . Dividing everything by gives . Replacing and with and , respectively, yields .
Consider the function . Draw the graph of from the graph of , deduce the formula
for from the picture, and then double-check your work algebraically.
Consider the function , defined for all , and draw the graph of from the
graph of . What happens if you take to be defined for every real number
instead?
We know that the graph of is the classical parabola we are already familiar with. As
we’re only considering the domain to be the set of real numbers with , we must look
only at the right half of the parabola (which passes the Horizontal Line Test), and
then reflect it about the line .
If we were to consider the graph of , this time defined for all real numbers, the full parabola enters the picture again. Reflecting it, we obtain the following curve indicated in blue:
The problem here is that the reflected curve does not pass the Vertical Line Test, which means that it is not the graph of any function. To understand why this should be expected, it suffices to observe that the function defining the original parabola, defined for all real numbers , is not one-to-one. This time, we have the freedom to switch signs, leading us to things like , and, more generally, for all real numbers (you might recall this last property as saying that is an “even function”; in fact, every even function is not one-to-one).
Graphically, the original parabola does not pass the Horizontal Line Test. Algebraically, while for every (with undefined for negative ), we have that for every real , with only for .
Consider the function , defined for all , and draw the graph of from the
graph of . What happens if you take to be defined for every real number
instead?
Consider the function , defined for all numbers . Draw the graph of from the graph
of . Does it look similar to the graph of ? Algebraically find to double-check your
possible suspicions.
We start from the graph of the famous function , which we have already encountered
before, and shift it one unit to the right.
The graph of , also formed by branches of hyperbolas and having a vertical asymptote at the line , suggests that the actual formula for should contain some term like itself. To verify this, we proceed by starting with , which is equivalent to , and adding to both sides as to obtain . Replacing and with and , respectively, gives us that , as expected.
Consider the function , defined for all numbers . Draw the graph of from the graph
of , and then algebraically find a formula for .
Consider the function , defined for all real numbers . Draw the graph of from the
graph of . Then, algebraically find to confirm your work.
To draw the graph of , we start from the graph of the exponential , shift it units to
the right, and then apply a vertical stretch by a factor of .
To verify that the inverse function is indeed given by , we start with and solve for in terms of . First, we divide both sides by , to obtain . Then, we apply to both sides, to obtain . Finally, we add to both sides, getting to . With this in place, we replace and with and , respectively, concluding that , as required.
Consider the function , defined for all . Draw the graph of from the graph of . Then,
algebraically find to confirm your work.
We also observe that, as suggested by the last exploration exercise above, the strategy illustrated here also allows us to find the graph of the original one-to-one function , if we’re given the graph of the inverse function instead. The reason for this is that the inverse of is , by design.
Inverses of Compositions
Lastly, there’s a general phenomenon worth mentioning, which sometimes makes things simpler. To understand it, let’s go back to the underpinning idea of what is the inverse function ’s job: to undo what does. So, if we have two one-to-one functions and for which the composition makes sense, then should also be one-to-one. In other words, the composition of two “reversible” processes must also be “reversible”. The next question, then, is what should be the inverse of the full composition . Since first applies , and then , reversing it should be done on the opposite order of things, first applying , and then .
Theorem: The composition of two one-to-one functions and , whenever it makes
sense, is also one-to-one, and the relation
holds. More generally, the composition of any number of one-to-one functions is
one-to-one, and the inverse of the composition is the composition of the inverses, but
in the reverse order.
In practice, this result can be used to quickly find formulas for inverse function, provided one is able to express (or reverse engineer) the given function as a composition of simpler functions whose inverses are known. Properly applied, this also allows us to bypass the verification that the given function is one-to-one (again, if the simpler functions are more well-known to be one-to-one).
Let’s revisit the last example above as an application of this:
Consider again the function given by , for all , on Example 3. What does do to a
number ? First, it subtracts . Secondly, it takes the square root. Lastly, it
exponentiates it. Writing it mathematically, we have that , where
All of those functions are one-to-one, with inverses given by
Hence
which agrees with what was obtained before.