- Can we find the inverse of a polynomial?
- What does it mean to take the root of a value?
Consider the polynomial . A good question to ask would be whether the function is invertible. To help us decide, here is the graph of .
Notice that this graph does not pass the Horizontal Line Test, so the function is not one-to-one, and therefore not invertible.
Our polynomial has many terms, so to simplify the situation, we’ll look only at polynomials of the form , where is a positive integer.
Odd Roots
Recall that every polynomial , where is odd, has the same basic shape. This is demonstrated in the figure below by the graphs of for .
To see more of how these graphs change with , follow the .
Now, are these functions invertible? Looking at the graphs, we see that these functions pass the Horizontal Line Test. Thus, the functions are one-to-one, and therefore invertible.
Let’s delve more deeply into the example. We have now established that it is invertible, and it’s inverse is .
Recall that a function is even if for all in its domain, and is odd if for all . Otherwise, the function is neither. Let’s consider an example. Given , and , since . Based on this example, do you think is even, odd, or neither?
If you guessed odd, then you are correct! All odd-index root functions are odd functions.
Even Roots
We again begin by recalling the general shape of , but this time for even. These functions also have the same basic shape for all even . This is demonstrated by the graphs of for given below.
To see more of how these graphs change with , follow the .
Now, are these functions invertible? All of the graphs in the figure above are symmetric about the -axis (example ), so they do not pass the horizontal line test. Thus, these functions are not one-to-one, and therefore not invertible.
So, how can we define an even root function? For example, what does really mean, and how is it related to ?
Consider the polynomial , graphed below with its inverse relation .
Observe that by restricting the domain of to , we now have a function which passes the horizontal line test, and can thus be inverted. The following picture illustrates the situation.
Now, our inverse relation is actually a function, since it passes the Horizontal Line Test. Therefore, if we let for , we can then define as the inverse function of on this restricted domain.
One question we might ask is whether and are truly inverses. The answer may seem like an obvious “yes!”, but since we restricted the domain of in order to define , we need to check. To check whether and are inverses, we need to confirm that and . That is, when we plug in a number to and , we should get the same number as the output. Let’s try plugging in to . This gives us which is not the same as . If we repeat this process with a few more numbers, we find that , , , and . We can conclude that is a function that takes its input and returns its absolute value. That is, . Since , we conclude that does not output its input, and therefore, and are not inverses. This is something that will be extremely important when solving equations using even roots.
Now, what if we instead restricted our domain to ? Consider defined for . The graph of this function is below.
- (a)
- , since .
- (b)
- , since .
- (c)
- , since . However, the root of -16 is not defined.
- (d)
- , since zero times any number is always zero. This is the example of an even root that has only one solution.
- (e)
- Likewise, .
Using Roots to Solve Equations
If we are asked to find all values such that , then the question is asking which values of multiplied by themselves give 4. In other words, find such that is equal to 4. It is simple to see that there are two values which make this true: In solving an equation, it is common to express this as follows.
Since is not one-to-one, there are two values of which make it equal to any positive number, as demonstrated in the following graph.- (a)
- Solve the equation .
Taking the cube root on both sides, we find that
Note that , since 3 is odd, and odd roots are really inverses to their corresponding power functions.
- (b)
- Solve the equation .
Taking the cube root on both sides, we find that
- (c)
- Solve the equation .
Taking the square root on both sides, we find that
Therefore, there are two solutions to this equation: and . - (d)
- Solve the equation .
First, rearrange the equation. Add 4 to both sides to find . Divide both sides by 2 to find . Taking the 4th root on both sides, we find that
Therefore, there are two solutions to this equation: and .Recall that for any even integer , .
- (e)
- Solve the equation .
First, rearrange the equation by dividing both sides by -3. This yields . Taking the 4th root on both sides, we find that
Since any even-index root of a negative number is not defined, there are no solutions to this equation.
These examples illustrate a general principle that is good to have in your toolbox for solving equations. If , then we know that . This is true for any odd powers. However, if , then either or . This is true for any even powers.
Finding -intercepts of a Quadratic in Vertex Form
Now, we can use our understanding of the squareroot function to find the -intercepts of a quadratic given in vertex form.
Notice that this gives us a third method for finding the roots (-intercepts) of a quadratic in general. We can use any of these methods to solve a quadratic.
- (a)
- Factor the quadratic and write it in Root Form
- (b)
- Use the quadratic formula to find the roots
- (c)
- Write the quadratic in vertex form and then solve using a squareroot
Mathematically, these last two methods are actually related. The quadratic formula is just what happens when you rewrite the general quadratic in vertex form and then solve for !
Composing and
This final example is going to be a very important one that comes up often so we we will give it its own section.
- a.
- Find the domain and range of and compare this function to and .
- b.
- Find the domain and range of and compare this function to and .
- a.
- First we consider and compare this function to and . We have and so
Let’s consider the domain of this function. Recall that the domain of a composite function is the set of those inputs in the domain of for which is in the domain of . In this case, this means that the domain of is the set of those inputs in the domain of for which is in the domain of . The implied domain of is since we cannot take the square root of a negative number. Therefore, since the domain of the composition has to be only values in the domain of , this means the largest our domain can be is . Now, the only additional limiting factor is that the values must be in the domain of but since the domain of is all real numbers, that will not limit the domain of the composition. Therefore, the domain of is .
Now that we know the domain and we know that squaring and squarerooting undo each other for nonnegative values of , we can conclude that is the identity function, but restricted to the domain . That is,
Since the absolute value function is the same as the identity function when . Therefore, we could also say that
From this information, we also know that the range of will also be , since .
Here is a graph of .
- b.
- Now we consider and compare this function to and . The domain of is the set
of those inputs in the domain of for which is in the domain of . The domain
of is all real numbers, so this does not reduce the domain of the composite
function. The range of is since the square of every number will be
greater than or equal to zero. The implied domain of is . Therefore,
every output from is in the domain of . Therefore, the domain of is
.
Now, let’s consider the range of . We know that the range of must be contained in the range of . The range of is , so that is the largest range possible for . We know that for values of , squaring and squarerooting undo one another so we know that all the values of are contained in the range of . More precisely, for any value in , so will be in the range of . Thus, the range of is .
Now, since we know that this function only outputs positive numbers, we know it cannot equal the identity function for inputs of . Let’s explore what this function does for values of by considering .
Notice, that we input but the output was positive 2. In fact, for all values of , .
Since the absolute value function is the same as the identity function when but the negative of the idenity function for , we have that
Here is a graph of .
However, when the polynomial is for a positive odd integer , the polynomial is invertible as the root function .
When is even, it is possible to define an inverse function on a restricted domain of . The root is defined as the inverse of on the restricted domain .