Lesson Explainer: Combining Functions | Nagwa Lesson Explainer: Combining Functions | Nagwa

Lesson Explainer: Combining Functions Mathematics • Second Year of Secondary School

In this explainer, we will learn how to add, subtract, multiply, or divide two given functions to create a new function and how to identify the domain of the new function.

Combining functions in this way is incredibly intuitive.

Consider the functions 𝑓(π‘₯)=3π‘₯+1𝑔(π‘₯)=2π‘₯.and

The sum of these functions will give us a new function, 𝑓(π‘₯)+𝑔(π‘₯)=3π‘₯+1+2π‘₯=5π‘₯+1.

We define the sum of these functions using the notation (𝑓+𝑔)(π‘₯), where (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯). The domain of this new function is the intersection of the domains of the functions 𝑓(π‘₯) and 𝑔(π‘₯).

In a similar way, (π‘“βˆ’π‘”)(π‘₯) is found by subtracting the function 𝑔 from 𝑓: (π‘“βˆ’π‘”)(π‘₯)=𝑓(π‘₯)βˆ’π‘”(π‘₯)=3π‘₯+1βˆ’2π‘₯=π‘₯+1.

The domain of π‘“βˆ’π‘” is the intersection of the domains of the functions 𝑓(π‘₯) and 𝑔(π‘₯).

Next, (𝑓⋅𝑔)(π‘₯) is the product of the functions (𝑓⋅𝑔)(π‘₯)=𝑓(π‘₯)⋅𝑔(π‘₯)=(3π‘₯+1)β‹…2π‘₯=6π‘₯+2π‘₯, and, once again, its domain is the intersection of the domains of the functions 𝑓(π‘₯) and 𝑔(π‘₯).

Finally, 𝑓𝑔(π‘₯) is the quotient of the functions: 𝑓𝑔(π‘₯)=𝑓(π‘₯)𝑔(π‘₯)=3π‘₯+12π‘₯.

The domain of 𝑓𝑔(π‘₯) is the intersection of the domains of the functions 𝑓(π‘₯) and 𝑔(π‘₯), but we exclude values of π‘₯ from the domain of the combined function that make 𝑔(π‘₯)=0.

These processes can be generalized for any two real-valued functions 𝑓(π‘₯) and 𝑔(π‘₯).

Definition: Combining Functions and Domain of a Combined Function

Two functions 𝑓(π‘₯) and 𝑔(π‘₯) can be combined as follows:

  • (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯)
  • (π‘“βˆ’π‘”)(π‘₯)=𝑓(π‘₯)βˆ’π‘”(π‘₯)
  • (𝑓⋅𝑔)(π‘₯)=𝑓(π‘₯)⋅𝑔(π‘₯)
  • 𝑓𝑔(π‘₯)=𝑓(π‘₯)𝑔(π‘₯)

The domain of the combined function is the intersection of the domains of the functions 𝑓(π‘₯) and 𝑔(π‘₯).

In the case of 𝑓𝑔(π‘₯), we exclude values of π‘₯ from the domain of the combined function such that 𝑔(π‘₯)=0.

Note

The domain of each combination is the intersection of the domains of 𝑓 and 𝑔. Both functions must be defined at a point for the combination to be defined and we cannot, therefore, adjust the domain of the combination by considering it as an entirely separate function.

In our first example, we will look at how to combine a pair of functions by finding their sum and consider how the domain of the combined function might restrict our solutions.

Example 1: Finding and Evaluating the Sum of a Rational and a Linear Function

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯)=π‘₯βˆ’1π‘₯+3π‘₯βˆ’4 and 𝑔(π‘₯)=π‘₯+3, determine the value of (𝑓+𝑔)(βˆ’4) if possible.

Answer

The sum of two real-valued functions 𝑓(π‘₯) and 𝑔(π‘₯) is given by (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯), where the domain of (𝑓+𝑔) is the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯).

To find (𝑓+𝑔)(βˆ’4), we will find the sum of the functions and then check that βˆ’4 is in the domain of (𝑓+𝑔). If it is, we can then evaluate (𝑓+𝑔) at βˆ’4: (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯)=π‘₯βˆ’1π‘₯+3π‘₯βˆ’4+π‘₯+3.

The domain of (𝑓+𝑔)(π‘₯) is the intersection of the domains of 𝑓 and 𝑔. That is the set of possible inputs (π‘₯ values) that can be substituted into both functions and will output real values.

𝑔(π‘₯) is a polynomial, so its domain is the set of real numbers. 𝑓(π‘₯), however, is a rational function; the domain of a rational function is all real numbers excluding those where the denominator is equal to zero. To find these values, we set the denominator equal to zero and solve for π‘₯: π‘₯+3π‘₯βˆ’4=0(π‘₯+4)(π‘₯βˆ’1)=0π‘₯=βˆ’4,π‘₯=1.

The values of π‘₯ where the function 𝑓(π‘₯) is undefined are βˆ’4 and 1. The domain of 𝑓(π‘₯) is therefore β„βˆ’{βˆ’4,1}.

Since the domain of (𝑓+𝑔) is the intersection, or overlap, of the domains of 𝑓 and 𝑔, this means the domain of (𝑓+𝑔) is also β„βˆ’{βˆ’4,1}.

Since βˆ’4 is not in the domain of our new function, we cannot evaluate (𝑓+𝑔)(βˆ’4).

(𝑓+𝑔)(βˆ’4) is undefined.

In our first example, we found the sum of a rational and a polynomial function, and the domain of this new function. We saw that, since the domain of the sum of two functions is the intersection of the domains of both functions, the function will not necessarily be defined for all real values of π‘₯. In our next example, we will find the domain of the sum of a polynomial and a root function.

Example 2: Finding the Domain of the Sum of Two Real Functions

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯)=π‘₯βˆ’5π‘₯ and 𝑔(π‘₯)=√π‘₯+1, find the domain of the function (𝑓+𝑔).

Answer

The sum of two functions 𝑓(π‘₯) and 𝑔(π‘₯) is given by (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯), where the domain of (𝑓+𝑔) is the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯).

Since this question requires us to calculate the domain of (𝑓+𝑔), we will consider the domains of 𝑓 and 𝑔 and then find their intersection. In other words, we will find the set of possible π‘₯ values that can be substituted into both functions that will output real 𝑦 values.

𝑓(π‘₯) is a polynomial function, so its domain is the set of real numbers.

𝑔(π‘₯) is a radical function, and we know that the square root of a negative number is not real, so we need to consider the values of π‘₯ that ensure the expression inside the root is nonnegative: π‘₯+1β‰₯0π‘₯β‰₯βˆ’1.

In interval notation, the domain of 𝑔(π‘₯) is given by [βˆ’1,∞[.

The domain of (𝑓+𝑔) is the intersection, or overlap, of the two domains. That is, [βˆ’1,∞[βˆ©β„.

The intersection of these two sets is [βˆ’1,∞[.

Thus, the domain of (𝑓+𝑔) is given by [βˆ’1,∞[.

We will now look at how to evaluate the difference between two functions.

Example 3: Evaluating the Difference between Two Functions

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯)=π‘₯+9π‘₯+15π‘₯+54 and 𝑔(π‘₯)=π‘₯+8, determine the value of (π‘“βˆ’π‘”)(βˆ’6) if possible.

Answer

The difference of two functions 𝑓(π‘₯) and 𝑔(π‘₯) is given by (π‘“βˆ’π‘”)(π‘₯)=𝑓(π‘₯)βˆ’π‘”(π‘₯), where the domain of (π‘“βˆ’π‘”) is the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯).

To find (π‘“βˆ’π‘”)(βˆ’6), we will check that π‘₯=βˆ’6 is in the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯). If it is, we can then subtract the functions and substitute βˆ’6 into the new function (π‘“βˆ’π‘”).

𝑔(π‘₯) is a polynomial, so its domain is the set of real numbers. 𝑓(π‘₯) is a rational function, so we will need to be careful when considering its denominator.

Since it is the quotient of two polynomials, its domain will be the set of real numbers excluding those values of π‘₯ that make the denominator zero. To find these values, we set the denominator equal to zero and solve for π‘₯: π‘₯+15π‘₯+54=0(π‘₯+9)(π‘₯+6)=0π‘₯=βˆ’9,π‘₯=βˆ’6.

Since the denominator of the rational function is zero when π‘₯=βˆ’9 and π‘₯=βˆ’6, we need to exclude these values of π‘₯ from the domain of 𝑓(π‘₯).

The domain of 𝑓(π‘₯) is therefore β„βˆ’{βˆ’9,βˆ’6}.

The domain of (π‘“βˆ’π‘”) is the intersection, or overlap, of the two domains. The intersection of the set of real numbers and the domain of 𝑓 is just the domain of 𝑓; it is β„βˆ’{βˆ’9,βˆ’6}.

Since π‘₯=βˆ’6 is not included in the domain of (π‘“βˆ’π‘”), we cannot substitute it into the function.

(π‘“βˆ’π‘”)(βˆ’6) is undefined.

Example 4: Determining the Product and Resulting Domain of Two Functions

Given that π‘“β„βŸΆβ„π‘“(π‘₯)=π‘₯βˆ’4:suchthat and 𝑓]βˆ’9,1]βŸΆβ„π‘“(π‘₯)=5π‘₯βˆ’2,:suchthat find (𝑓⋅𝑓)(π‘₯) and state its domain.

Answer

The product of two functions 𝑓(π‘₯) and 𝑔(π‘₯) is given as (𝑓𝑔)(π‘₯) or (𝑓⋅𝑔)(π‘₯). Then, the domain of the combined function is the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯).

We can therefore say that (𝑓⋅𝑓)(π‘₯)=𝑓(π‘₯)⋅𝑓(π‘₯)=(π‘₯βˆ’4)(5π‘₯βˆ’2)=5π‘₯βˆ’22π‘₯+8.

The domain of π‘“β‹…π‘“οŠ§οŠ¨ will be the intersection of the domains of π‘“οŠ§ and π‘“οŠ¨.

We are given the domains of the two functions in the question. We are told that π‘“οŠ§ maps numbers from the set of positive real numbers onto the set of real numbers, so its domain is β„οŠ°. The domain of π‘“οŠ¨ is ]βˆ’9,1].

The intersection of these two sets is the set of positive real numbers up to and including 1, β„βˆ©]βˆ’9,1]οŠ°β€”in other words, the interval ]0,1].

Therefore, (𝑓⋅𝑓)(π‘₯)=5π‘₯βˆ’22π‘₯+8, π‘₯∈]0,1].

Finding the domain of a combined function has, until this stage, only involved finding the domains of the given functions. When we are finding the quotient of two functions, we also need to consider the values of π‘₯ that make the divisor equal to zero and exclude those from the domain of the combined function, as in our next example.

Example 5: Evaluating the Division of Two Functions

Given that 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯)=π‘₯βˆ’1 and 𝑔(π‘₯)=√π‘₯+5, find the value of 𝑔𝑓(βˆ’2) if possible.

Answer

We can combine two functions by finding their quotient, such that 𝑔𝑓(π‘₯)=𝑔(π‘₯)𝑓(π‘₯), where the domain of the new function is the intersection of the domains of 𝑓 and 𝑔, excluding any values of π‘₯ that make 𝑓(π‘₯)=0.

Since we are looking to evaluate 𝑔𝑓(βˆ’2), we will check that π‘₯=βˆ’2 is defined for 𝑔𝑓 by calculating the domain of this combined function.

Since 𝑓(π‘₯) is a polynomial, the domain of 𝑓(π‘₯) is the set of real numbers. The domain of 𝑔(π‘₯) will be the set of π‘₯ values such that the expression inside the square root is nonnegative.

In other words, π‘₯+5β‰₯0π‘₯β‰₯βˆ’5.

The intersection of the domains of 𝑓 and 𝑔 is [βˆ’5,∞[. To find the domain of 𝑔𝑓, we need to remove values of π‘₯ that make 𝑓(π‘₯)=0 from this set. Since 𝑓(π‘₯)=π‘₯βˆ’1, we set this expression equal to zero and solve for π‘₯: π‘₯βˆ’1=0π‘₯=Β±1.

The domain of 𝑔𝑓 is [βˆ’5,∞[βˆ’{βˆ’1,1}.

Since π‘₯=βˆ’2 is within the domain of the combined function, we will be able to evaluate 𝑔𝑓(βˆ’2): 𝑔𝑓(π‘₯)=√π‘₯+5π‘₯βˆ’1𝑔𝑓(βˆ’2)=βˆšβˆ’2+5(βˆ’2)βˆ’1=√33.

𝑔𝑓(βˆ’2) is defined and is equal to √33.

Let’s once again consider the functions 𝑓(π‘₯)=π‘₯βˆ’1 and 𝑔(π‘₯)=√π‘₯+5.

Substituting π‘₯=βˆ’2 into both functions, we find, 𝑓(βˆ’2)=(βˆ’2)βˆ’1=3 and 𝑔(βˆ’2)=βˆšβˆ’2+5=√3.

Next, finding their quotient, 𝑔(βˆ’2)𝑓(βˆ’2)=√33=𝑔𝑓(βˆ’2).

We observe that evaluating the functions at βˆ’2 and then combining them yields the same result as evaluating the combined function at βˆ’2. In general, this is true for all combinations of functions.

Note

For values of π‘₯ in the domain of the combined function, the result will be the same if we evaluate the combined function at some π‘₯ as if we evaluate the functions at this π‘₯ and then combine them.

In our final example, we will look at how to apply the rules for combining functions when one of the functions is piecewise defined.

Example 6: Finding the Domain of Piecewise-Defined Rational Functions

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯)=2π‘₯+2π‘₯<βˆ’3,π‘₯βˆ’4βˆ’3≀π‘₯<0ifif and 𝑔(π‘₯)=5π‘₯, determine the domain of the function 𝑔𝑓.

Answer

The function 𝑔𝑓(π‘₯) is the quotient of the functions 𝑓(π‘₯) and 𝑔(π‘₯) such that 𝑔𝑓(π‘₯)=𝑔(π‘₯)𝑓(π‘₯).

The domain of this new function is the intersection of the domains of 𝑓(π‘₯) and 𝑔(π‘₯), excluding the values of π‘₯ that make 𝑓(π‘₯)=0.

Let’s begin with 𝑔(π‘₯). 5π‘₯ is a polynomial, so the domain of this function is the set of real numbers.

𝑓(π‘₯) is a piecewise defined function, and its subfunctions are both linear. Its domain is therefore the union of the subdomains. This is the interval ]βˆ’βˆž,0[.

The intersection of the domains of 𝑓 and 𝑔 is therefore the interval ]βˆ’βˆž,0[. To find the domain of 𝑔𝑓, we need to exclude the values of π‘₯ that make the function 𝑓(π‘₯)=0 from this interval. We will set each subfunction equal to zero and solve for π‘₯, considering the domains of each subfunction as we do so: 2π‘₯+2=0π‘₯=βˆ’1.

Since βˆ’1 is not in the subdomain of this subfunction, which is ]βˆ’βˆž,βˆ’3[, we do not need to exclude it from the domain of 𝑔𝑓: π‘₯βˆ’4=0π‘₯=4.

We would need to exclude this from the domain of 𝑔𝑓; however, it is already outside of the subdomain of this subfunction.

The domain of 𝑔𝑓 is therefore ]βˆ’βˆž,0[.

Key Points

  • Two real-valued functions 𝑓 and 𝑔 can be combined by finding their sum, difference, product, or quotient.
  • For values of π‘₯ in the domains of both 𝑓 and 𝑔,
    • (𝑓+𝑔)(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯),
    • (π‘“βˆ’π‘”)(π‘₯)=𝑓(π‘₯)βˆ’π‘”(π‘₯),
    • (𝑓⋅𝑔)(π‘₯)=𝑓(π‘₯)⋅𝑔(π‘₯)
  • For values of π‘₯ in the domains of both 𝑓 and 𝑔, and where 𝑔(π‘₯)β‰ 0, 𝑓𝑔(π‘₯)=𝑓(π‘₯)𝑔(π‘₯).

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