Lesson Explainer: Simplifying Rational Functions | Nagwa Lesson Explainer: Simplifying Rational Functions | Nagwa

Lesson Explainer: Simplifying Rational Functions Mathematics • Third Year of Preparatory School

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In this explainer, we will learn how to simplify rational functions and how to find their domains.

Before we begin to simplify rational functions, let’s start by recalling that a rational function is the quotient of two polynomials. So, we say that 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯) is a rational function if both 𝑝(π‘₯) and π‘ž(π‘₯) are polynomials, where π‘ž(π‘₯) is not the zero polynomial. We can also recall that constant functions are examples of polynomials, so we can start by simplifying the quotient of constant functions, in other words, simplifying fractions.

We recall that to simplify a fraction, say π‘›π‘š, we find the common factors of 𝑛 and π‘š and cancel these out. More accurately, we divide the numerator and denominator by these shared factors. For example, to simplify 7260, we write the numerator and denominator as the product of their prime factors to get 7260=2Γ—32Γ—3Γ—5.

We can then divide both the numerator and denominator by 2Γ—3, which has the effect of canceling the shared factors out. This gives 2Γ—32Γ—3Γ—5=23Γ—322Γ—3Γ—5=2Γ—35=65.

The process of simplifying a rational function is similar; we want to find the common factors of the numerator and denominator and then cancel these factors out. However, there are a few important differences in this case. To see these differences, let’s simplify the rational function 𝑓(π‘₯)=π‘₯π‘₯.

Our first instinct may be to cancel out the shared factor of π‘₯ in the numerator and denominator; this would give π‘₯π‘₯=1.

Let’s call this function 𝑔(π‘₯), so 𝑔(π‘₯)=1. We can immediately see that 𝑔(π‘₯) and 𝑓(π‘₯) are not the exact same function by considering both functions when π‘₯=0. We have that 𝑔(0)=1,𝑓(0)=00.

So, 𝑓 is not defined when π‘₯=0. We can recall that the set of all inputs is called the domain of a function, so we can say that 0 is not in the domain of 𝑓(π‘₯). Therefore, we cannot just cancel out the shared factor of π‘₯ when π‘₯ is 0 since π‘₯π‘₯ is only equal to 1 when the denominator is nonzero.

In fact, we have shown that when π‘₯β‰ 0, 𝑓(π‘₯)=π‘₯π‘₯=1. So, we can say that 𝑓(π‘₯)=1 over the domain β„βˆ’{0}. We can also see this in a diagram by plotting the graphs of both functions.

Both graphs are equal everywhere except when π‘₯=0 since π‘₯π‘₯ is undefined at this point.

This process can be generalized to simplify any rational function.

We first need to determine the domain of the rational function. Over this domain, the function is defined, so we can cancel out shared factors in the numerator and denominator (the function cannot output 00 for inputs in the domain). Finally, the function is equal to the simplified expression over the domain of the original function.

To apply this process, we first need to determine the domain of a rational function. We can do this directly from the definition of a rational function. Say 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯), where 𝑝(π‘₯) and π‘ž(π‘₯) are polynomials. We know that polynomials are defined for all real values, and we note that if π‘ž(π‘Ž)β‰ 0, then we have the quotient of two real numbers. So, 𝑓(π‘Ž) is defined and π‘Ž is in the domain of 𝑓(π‘₯). However, if π‘ž(π‘Ž)=0, then 𝑓(π‘Ž)=𝑝(π‘Ž)π‘ž(π‘Ž)=𝑝(π‘Ž)0, which is undefined; hence, π‘Ž is not in the domain of 𝑓(π‘₯). We have shown the following result.

Definition: Domain of a Rational Function

If 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯) is a rational function, then the domain of 𝑓(π‘₯) is all real numbers except those where π‘ž(π‘₯)=0.

Therefore, we can determine the domain of a rational function by finding the roots of the denominator. For high-degree polynomials, we may need to use a technique such as the factor theorem, algebraic division, or factoring by grouping to determine these roots. We should also note that sometimes quadratic equations will have no real roots (e.g., π‘₯+1 has no real roots).

For example, let’s determine the domain of the rational function 𝑓(π‘₯)=π‘₯+1π‘₯+π‘₯. We want to find the roots of the denominator, and we do this by taking out the shared factor of π‘₯, giving us π‘₯+π‘₯=0π‘₯(π‘₯+1)=0.

Then, for the product of two real-valued expressions to be 0, one of the factors must be 0. Hence, either π‘₯=0 or π‘₯=βˆ’1. Finally, the domain of 𝑓(π‘₯) is all real values except these roots; we can write this in set notation as β„βˆ’{βˆ’1,0}.

Now that we can determine the domain of a rational function, we can also simplify rational functions by canceling the shared factors out, where the domain of the new function is inherited from the original function. We can describe this process as follows.

How To: Simplifying Rational Functions

To simplify a rational function 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯), we need to perform the following steps:

  1. Find the values of π‘₯ where π‘ž(π‘₯)=0. Then, the domain of 𝑓(π‘₯) is all real values except these roots.
  2. Fully factor both 𝑝(π‘₯) and π‘ž(π‘₯); this may require using the factor theorem or factoring by grouping.
  3. Cancel out the shared factors in the numerator and denominator, where we restrict the values of π‘₯ to be in the domain of 𝑓(π‘₯).
  4. Equate 𝑓(π‘₯) to the simplified expression restricted to the domain of 𝑓(π‘₯).

Let’s now see some examples of how to apply this process to simplify a rational function.

Example 1: Simplifying and Determining the Domain of Rational Functions

Simplify the function 𝑓(π‘₯)=π‘₯+2π‘₯π‘₯βˆ’4, and find its domain.

Answer

We first note that 𝑓(π‘₯) is the quotient of two polynomials, so it is a rational function. We recall that to simplify a rational function, we find its domain, factor the numerator and denominator, and then cancel out the shared factors over the domain.

Therefore, we need to start by finding the domain of 𝑓(π‘₯). We do this by recalling that the domain of a rational function is all real values of π‘₯ except those where the denominator is equal to zero. To find the roots of the denominator, we will factor by using the difference between two squares: π‘₯βˆ’4=(π‘₯βˆ’2)(π‘₯+2).

Hence, (π‘₯βˆ’2)(π‘₯+2)=0 when π‘₯=2 or when π‘₯=βˆ’2. Therefore, the domain of 𝑓(π‘₯) is β„βˆ’{βˆ’2,2}.

Next, we need to fully factor the numerator. We can do this by noting that both terms share a factor of π‘₯, giving us π‘₯+2π‘₯=π‘₯(π‘₯+2).

Therefore, 𝑓(π‘₯)=π‘₯+2π‘₯π‘₯βˆ’4=π‘₯(π‘₯+2)(π‘₯βˆ’2)(π‘₯+2).

Now, since π‘₯=βˆ’2 is not in the domain of 𝑓(π‘₯), we can cancel out the shared factor of π‘₯+2 to give us 𝑓(π‘₯)=π‘₯(π‘₯+2)(π‘₯βˆ’2)(π‘₯+2)=π‘₯π‘₯βˆ’2, where π‘₯βˆˆβ„βˆ’{βˆ’2,2}.

Hence, we have shown that 𝑓(π‘₯)=π‘₯π‘₯βˆ’2 and its domain is β„βˆ’{βˆ’2,2}.

In our next example, we will demonstrate how to find the domain of a rational function where both the numerator and denominator are quadratics with leading coefficients greater than 1.

Example 2: Simplifying a Rational Function

Simplify the function 𝑓(π‘₯)=7π‘₯+43π‘₯+67π‘₯+50π‘₯+7, and find its domain.

Answer

We first note that 𝑓(π‘₯) is the quotient of two polynomials, so it is a rational function. We then recall that to simplify a rational function, we find its domain, factor the numerator and denominator, and then cancel out the shared factors over the domain.

Therefore, we need to start by finding the domain of 𝑓(π‘₯). We do this by recalling that the domain of a rational function is all real values of π‘₯ except where the denominator is equal to zero.

There are a number of techniques to help us factor this sort of quadratic. One of these is sometimes called factoring by grouping. We will demonstrate this here, though we might alternatively choose to use inspection or another suitable method.

To find the roots of the denominator, we need to factor 7π‘₯+50π‘₯+7. We can do this by finding two numbers whose product is 7Γ—7=49 and sum to give 50. We see that these numbers are 1 and 49. We then use this to split the middle term so that we can use factoring by grouping: 7π‘₯+50π‘₯+7=7π‘₯+49π‘₯+π‘₯+7.

The first two terms share a factor of 7π‘₯, and the last two terms only share a factor of 1.

Taking these factors out, we have ο€Ή7π‘₯+49π‘₯+(π‘₯+7)=7π‘₯(π‘₯+7)+(π‘₯+7).

Then, we take out the factor of π‘₯+7 to get 7π‘₯(π‘₯+7)+(π‘₯+7)=(π‘₯+7)(7π‘₯+1).

We can now find the roots of the denominator: (π‘₯+7)(7π‘₯+1)=0.

These are π‘₯=βˆ’7 and π‘₯=βˆ’17.

Therefore, the domain of 𝑓(π‘₯) is β„βˆ’{βˆ’7,βˆ’17}. We can now simplify 𝑓(π‘₯) by factoring the numerator and canceling out the shared factors over this domain.

To factor 7π‘₯+43π‘₯+6, we need two numbers that multiply to give 7Γ—6=42 and add to give 43. We can see that these numbers are 1 and 42, so we use this to split the middle term and then factor: 7π‘₯+43π‘₯+6=7π‘₯+42π‘₯+π‘₯+6=7π‘₯(π‘₯+6)+(π‘₯+6)=(7π‘₯+1)(π‘₯+6).

Therefore, 𝑓(π‘₯)=(7π‘₯+1)(π‘₯+6)(π‘₯+7)(7π‘₯+1)=(7π‘₯+1)(π‘₯+6)(π‘₯+7)(7π‘₯+1)=π‘₯+6π‘₯+7, where π‘₯βˆˆβ„βˆ’{βˆ’7,βˆ’17}.

Hence, 𝑓(π‘₯)=π‘₯+6π‘₯+7 and its domain is β„βˆ’{βˆ’7,βˆ’17}.

In our next example, we will use a given simplification of a rational function to determine the value of an unknown in the function.

Example 3: Finding the Value of an Unknown in a Rational Function given Its Simplified Form

Given that 𝑛(π‘₯)=π‘₯+12π‘₯+36π‘₯βˆ’π‘ŽοŠ¨οŠ¨ simplifies to 𝑛(π‘₯)=π‘₯+6π‘₯βˆ’6, what is the value of π‘Ž?

Answer

We recall that to simplify a rational function, we find its domain, factor the numerator and denominator, and then cancel out the shared factors over the domain.

Therefore, since the simplified function has a linear denominator and the original function has a quadratic denominator, we must be able to factor π‘₯βˆ’π‘ŽοŠ¨ into two linear factors one of which is π‘₯βˆ’6. We can note that if π‘Ž<0, then π‘₯βˆ’π‘ŽοŠ¨ has no factors since the roots are Β±βˆšπ‘Ž and we cannot take the root of a negative number in the set of real numbers. Hence, π‘Žβ‰₯0.

Then, we can factor this expression by using a difference between squares: π‘₯βˆ’π‘Ž=ο€Ίπ‘₯βˆ’βˆšπ‘Žο†ο€Ίπ‘₯+βˆšπ‘Žο†.

Since π‘₯βˆ’6 is a factor of the denominator, we know that βˆšπ‘Ž=6, so π‘Ž=36. This means the denominator of 𝑛(π‘₯) is ο€Ήπ‘₯βˆ’36=(π‘₯βˆ’6)(π‘₯+6), which has zeros at π‘₯=Β±6.

Hence, the domain of 𝑛(π‘₯) is β„βˆ’{βˆ’6,6}.

Next, we factor the numerator of π‘₯+12π‘₯+36 by noticing that 6Γ—6=36 and 6+6=12, so π‘₯+12π‘₯+36=(π‘₯+6)(π‘₯+6).

Therefore, if π‘₯βˆˆβ„βˆ’{βˆ’6,6}, we have 𝑛(π‘₯)=π‘₯+12π‘₯+36π‘₯βˆ’36=(π‘₯+6)(π‘₯+6)(π‘₯βˆ’6)(π‘₯+6)=(π‘₯+6)(π‘₯+6)(π‘₯βˆ’6)(π‘₯+6)=π‘₯+6π‘₯βˆ’6.

This confirms that π‘Ž=36.

In our previous examples, we have only dealt with quadratic polynomials. Let’s now see an example involving a cubic polynomial.

Example 4: Simplifying and Determining the Domain of Rational Functions

Simplify the function 𝑓(π‘₯)=π‘₯βˆ’81π‘₯+729 and find its domain.

Answer

We recall that to simplify a rational function, we find its domain, factor the numerator and denominator, and then cancel out the shared factors over the domain.

Therefore, we need to start by finding the domain of 𝑓(π‘₯). We do this by recalling that the domain of a rational function is all real values of π‘₯ except where the denominator is equal to zero.

We can recall the sum of cubes formula, which states that π‘₯+π‘Ž=(π‘₯+π‘Ž)ο€Ήπ‘₯βˆ’π‘Žπ‘₯+π‘Žο…οŠ©οŠ©οŠ¨οŠ¨. Setting π‘Ž=9, we have π‘₯+729=(π‘₯+9)ο€Ήπ‘₯βˆ’9π‘₯+81.

We cannot factor the quadratic. So, there are no π‘₯-intercepts and hence no roots to this quadratic. Therefore, the domain of 𝑓(π‘₯) is β„βˆ’{βˆ’9}.

We can then factor the numerator by using the difference between squares; we have π‘₯βˆ’81=(π‘₯βˆ’9)(π‘₯+9).

Now, for π‘₯β‰ βˆ’9, 𝑓(π‘₯)=π‘₯βˆ’81π‘₯+729=(π‘₯βˆ’9)(π‘₯+9)(π‘₯+9)(π‘₯βˆ’9π‘₯+81)=(π‘₯βˆ’9)(π‘₯+9)(π‘₯+9)(π‘₯βˆ’9π‘₯+81)=π‘₯βˆ’9π‘₯βˆ’9π‘₯+81.

Hence, 𝑓(π‘₯)=π‘₯βˆ’9π‘₯βˆ’9π‘₯+81 and its domain is β„βˆ’{βˆ’9}.

Before we move onto our final examples, we can now discuss what it means for two rational functions to be equal and equivalent.

First, we say that two rational functions are equal if the functions are equal. In other words, they must have the same domain, and the outputs must be equal across the entire domain. We can write this formally as follows.

Definition: Equality of Rational Functions

If 𝑛(π‘₯) and 𝑛(π‘₯) are rational functions, we say that 𝑛=π‘›οŠ§οŠ¨ if they have the same domain and are equal on this entire domain.

We note that this is equivalent to saying the zeros of the denominators of both rational functions are equal and 𝑛=π‘›οŠ§οŠ¨ on this domain.

Sometimes, rational functions will simplify to give the same rational function; however, their domains may not be the same. In this case, we say that we have equivalent rational functions.

We say that two rational functions are equivalent if they are equal on the intersection of their domains. We can write this more formally as follows.

Definition: Equivalence of Rational Functions

If 𝑛(π‘₯) and 𝑛(π‘₯) are rational functions, we say that π‘›οŠ§ is equivalent to π‘›οŠ¨ if they are equal on their shared domain. We note that the shared domain of the two rational functions is all real numbers excluding the zeros of each denominator.

In our final two examples, we will look at how to compare rational functions with the same simplified form and determine if they are equivalent.

Example 5: Finding the Domain for Two Rational Functions to Be Equal

Given the functions 𝑛(π‘₯)=π‘₯π‘₯βˆ’10π‘₯ and 𝑛(π‘₯)=1π‘₯βˆ’10, what is the set of values on which 𝑛=π‘›οŠ§οŠ¨?

Answer

We want to determine the set of values on which the two given functions are equal. We can note that if 𝑛=π‘›οŠ§οŠ¨ on their entire shared domain, then they are equivalent. We can do this by noticing that both of the given functions are rational, so we can simplify these functions by finding their domains and canceling the shared factors out.

We observe that the numerator and denominator of 𝑛(π‘₯) share no common factors, so 𝑛(π‘₯) cannot be simplified. So, let’s start with 𝑛(π‘₯). We recall that the domain of a rational function is the set of all real numbers excluding those where the denominator is zero. We can determine when the denominator is zero by factoring. We have π‘₯βˆ’10π‘₯=0π‘₯(π‘₯βˆ’10)=0.

So, the denominator is zero when π‘₯=0 or when π‘₯=10. This means that the domain of 𝑛(π‘₯) is β„βˆ’{0,10}. We can then cancel the shared factor of π‘₯ out to see that 𝑛(π‘₯)=π‘₯π‘₯βˆ’10π‘₯=π‘₯π‘₯(π‘₯βˆ’10)=1π‘₯π‘₯(π‘₯βˆ’10)=1π‘₯βˆ’10, when π‘₯βˆˆβ„βˆ’{0,10}. We can then see that this is the same expression as 𝑛(π‘₯).

Since we can only simplify 𝑛(π‘₯) by excluding the values π‘₯=0 and π‘₯=10, we have demonstrated that 𝑛(π‘₯) is only equal to 𝑛(π‘₯) over this set of values, in other words, over the domain of 𝑛(π‘₯).

It is worth noting that the domain of π‘›οŠ¨ is β„βˆ’{10} and that the domain of π‘›οŠ§ is β„βˆ’{0,10}. So, the shared domain of these functions is the intersection of these sets: β„βˆ’{0,10}. We have shown that π‘›οŠ§ and π‘›οŠ¨ are equal on this set, so they are equivalent.

Hence, they are equal on the entire domain of 𝑛(π‘₯), which is β„βˆ’{0,10}.

In the previous example, we saw that if we have two rational functions that simplify to give the same expression, then they agree everywhere where both functions are defined. In other words, if two rational functions have the same simplified expression, then they are equal on the intersection of their domains. This gives us a useful result.

Property: Equivalence of Rational Functions Using Simplified Forms

If 𝑛(π‘₯) and 𝑛(π‘₯) are rational functions where the simplified forms of π‘›οŠ§ and π‘›οŠ¨ are equal, then π‘›οŠ§ and π‘›οŠ¨ are equivalent.

Therefore, we can determine the equivalence of rational functions by simplifying each rational function by canceling shared factors out.

Let’s look at another example where we identify the set on which two functions are equal by considering their domains. This will then allow us to discuss the equivalence and equality of the two rational functions.

Example 6: Finding the Domain for Two Rational Functions to Be Equal

Given that 𝑛(π‘₯)=π‘₯βˆ’5π‘₯+5 and 𝑛(π‘₯)=π‘₯βˆ’5π‘₯π‘₯+5π‘₯, find the largest set on which the functions π‘›οŠ§ and π‘›οŠ¨ are equal.

Answer

In order to determine the set of values on which the two given functions are equal, we notice that both of the given functions are rational, so we can simplify these functions by finding their domains and canceling the shared factors out.

We note that 𝑛(π‘₯) has distinct linear functions in its numerator and denominator, so they share no common nonconstant polynomial factors. Hence, we cannot simplify 𝑛(π‘₯) any further.

So, let’s instead start with 𝑛(π‘₯). We recall that the domain of a rational function is the set of all real numbers except those that make the denominator zero. We can determine when the denominator is zero by factoring, so we have π‘₯+5π‘₯=0π‘₯(π‘₯+5)=0.

So, the denominator is zero when π‘₯=0 or when π‘₯=βˆ’5. This means that the domain of 𝑛(π‘₯) is β„βˆ’{βˆ’5,0}. We can then factor the numerator to get π‘₯βˆ’5π‘₯=π‘₯(π‘₯βˆ’5).

Then, on the domain of 𝑛(π‘₯), we can cancel the shared factors out to get 𝑛(π‘₯)=π‘₯βˆ’5π‘₯π‘₯+5π‘₯=π‘₯(π‘₯βˆ’5)π‘₯(π‘₯+5)=π‘₯βˆ’5π‘₯+5, for π‘₯βˆˆβ„βˆ’{βˆ’5,0}.

We can then see that this is the same expression as 𝑛(π‘₯).

Since both rational functions have the same simplified expression, we have shown that they are equivalent. However, we can note that these functions are not equal since 𝑛(0)=0βˆ’50+5=βˆ’1 but 0 is not in the domain of π‘›οŠ¨.

Hence, we have shown that they are equal on the entire domain of 𝑛(π‘₯), which is β„βˆ’{βˆ’5,0}. We cannot extend this set any further since 𝑛(π‘₯) is not defined when π‘₯=βˆ’5 or when π‘₯=0. Thus, the functions cannot be equal at these values. Therefore, the largest set on which these functions are equal is β„βˆ’{βˆ’5,0}.

Let’s finish by recapping some of the important points in this explainer.

Key Points

  • The domain of a rational function is all real values except those that make the denominator equal to zero.
  • To simplify a rational function 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯), we need to perform the following steps:
    1. Find the domain of 𝑓(π‘₯) by finding the roots of π‘ž(π‘₯).
    2. Fully factor both 𝑝(π‘₯) and π‘ž(π‘₯).
    3. Cancel out the shared factors in the numerator and denominator, where we restrict the values of π‘₯ to be in the domain of 𝑓(π‘₯).
    4. Equate 𝑓(π‘₯) to the simplified expression over the domain of 𝑓(π‘₯).
  • If two rational functions have the same simplified expressions, then they are equal across the intersections of their domains.
  • If 𝑛(π‘₯) and 𝑛(π‘₯) are rational functions, we say that 𝑛=π‘›οŠ§οŠ¨ if they have the same domain and are equal on this entire domain.
  • If 𝑛(π‘₯) and 𝑛(π‘₯) are rational functions, we say that π‘›οŠ§ is equivalent to π‘›οŠ¨ if they are equal on their shared domain.
  • If the simplified forms of two rational functions π‘›οŠ§ and π‘›οŠ¨ are equal, then π‘›οŠ§ and π‘›οŠ¨ are equivalent.

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