Algebraic Operations on Functions: We are familiar with the basic algebraic operations on numbers; we know how to do addition, subtraction, multiplication, and division of two or more numbers. But do you know how to perform these operations on functions? Algebraic operations of a function are nothing but the arithmetic operations of the function. If you know to perform addition, subtraction, multiplication, and division of algebraic expression, you can do this for functions.

We know that a function is defined as \(f: x \rightarrow y\), where each element of set \(x\) is assigned to exactly one element of \(y\). The set \(x\) is called the domain of the function and the set \(y\) is called the codomain of the function. In this article, let us learn more about the algebraic operations on functions.

Domain of a Function

The domain of a function is the set of all possible values of \(x\) for which the function is defined.

Note: The formulas hold only when the domains of both the functions are the same.

Types of Algebraic Operations on Functions

The composition of functions is one way of combining two or more functions. But there are other ways as well that are usual algebraic operations on functions such as:

• Addition of functions
• Subtraction of functions
• Multiplication of functions
• Division of functions

We combine functions by performing the algebraic operations on the outputs of the given functions and defining the result as the output of the resulting function after applying the operations. After performing the operations on functions, we can find the domain and range of the resulting function.

1. Addition of Functions

Suppose we have two functions \(f(x)\) and \(g(x)\), then the sum of these two functions for the given input is equal to the sum of the independent functions for the same input. Mathematically, this can be written as shown below.

\((f+g)(x)=f(x)+g(x)\)

The domain of the sum of two functions \((f(x)+g(x))\) is the intersection of the domain of the independent functions \(f(x)\) and \(g(x)\)

Domain \([(f+g)(x)]=\) Domain \(f(x) \cap\) Domain \(g(x)\)

2. Subtraction of Functions

Suppose we have two functions \(f(x)\) and \(g(x)\) then to find the difference of two functions at a given input, find the difference between the independent functions at the same input. This can be represented as

\((f-g)(x)=f(x)-g(x)\)

The domain of the difference of the two functions \((f(x)-g(x))\) is the intersection of the domain of the independent functions.

Domain \([(f-g)(x)]=\) Domain \(f(x) \cap\) Domain \(g(x)\)

3. Multiplications of Functions

Suppose we have two functions \(f(x)\) and \(g(x)\) then to find the product of two functions at a given input multiply the independent functions at the same input. Multiplication of two functions is written as shown below.

\((f \cdot g)(x)=f(x) \cdot g(x)\)

The product of a constant function and a linear function is also a linear function. Similarly, the product of a linear function with another linear function is a binomial function.

Hence, we can say that the product of two functions results in linear, binomial, cubic , or polynomial functions.

Same as addition and subtraction, the domain of the product of two functions is the intersection of the domain of the independent functions.

Domain \([(f.g)(x)] = \) Domain \(f(x) \cap\) Domain \(g(x)\)

4. Division of Functions

Suppose we have two functions \(f(x)\) and \(g(x)\) then to find the division of two functions at a given input, find the quotient of the independent functions at the same input. i.e.,

\((\boldsymbol{f} / \boldsymbol{g})(\boldsymbol{x})=\boldsymbol{f}(\boldsymbol{x}) / \boldsymbol{g}(\boldsymbol{x})\)

The domain of the division of two functions is the intersection of the domain of the independent functions except for the points where \(g(x) = 0\)

Domain \({\rm{[(f/g)(x)] = }}\) Domain \(f(x) \cap\) Domain \(g(x)\) Such that \(g(x) \neq 0\)

Example: Find the domain of \(\frac{1}{x^{2}-4}\)

Solution: Here \(x^{2}-4\) is a quadratic polynomial which is defined for all real numbers.

Therefore, \(x^{2}-4\) is defined for all real numbers \(R\).

But, for \(x^{2}-4=0, \frac{1}{x^{2}-4}\) is not defined.

i.e, for \(x=\pm 2, \frac{1}{x^{2}-4}\) is not defined.

Hence, the domain of \(\frac{1}{x^{2}-4}\) is all numbers expect \(\pm 2\) or \(\mathbb{R} – \left\{ { \pm 2} \right\}\)

Note:

The function \(f(x) / g(x)\) is defined only when \(g(x) \neq 0\) because if the divisor is \(0\), then the division is mathematically undefined.

Range of a Function

If \(f: D_{f}(\subseteq X) \rightarrow Y\) is a function with domain \(D_{f}\), then the set of images \(y\) (output \(\in Y)\) generated corresponding to the input \(x \in D_{f}\) is called the range of the function. It is denoted by \(R_{f}\)

After getting a new function, we can find the range of the obtained function by using the definition of the range.

Formulas and Domains

The formulas of algebraic operations on two functions say \(f(x)\) and \(g(x)\) and their domains are given below.

No.	Operation	Formula	Domain
1.	Addition	\((f+g)(x)=f(x)+g(x)\)	\(\mathrm{D}[(f+g)(x)]=\mathrm{D} f(x) \cap \mathrm{D} g(x)\)
2.	Subtraction	\((f-g)(x)=f(x)-g(x)\)	\(\mathrm{D}[(f-g)(x)]=\mathrm{D} f(x) \cap \mathrm{D} g(x)\)
3.	Multiplication	\((f \cdot g)(x)=f(x) \cdot g(x)\)	\(\mathrm{D}[(f \cdot g)(x)]=\mathrm{D} f(x) \cap \mathrm{D} g(x)\)
4.	Division	\((f / g)(x)=f(x) / g(x)\) where \(g(x) \neq 0\)	\(\mathrm{D}[(f / g)(x)]=\mathrm{D} f(x) \cap \mathrm{D} g(x)\) such that \(g(x) \neq 0\)

Solved Examples on Algebraic operations on functions

Q.1. If \(f(x)=x^{2}\) and \(g(x)=10-8 x\),then find the domain and range of the function \(h(x)\) such that \(h(x)=f(x)+g(x)\).
Sol:
Given: \(f(x)=x^{2}\)
Here,
Domain is a set of real numbers as it is a polynomial
Range is \([0, \infty) \forall x \in R\)
\(g(x)=10-8 x\)
Here,
Domain and range are set of real numbers as it is a linear polynomial.
Given: \(h(x)=f(x)+g(x)\)
\(\therefore h(x)=x^{2}+10-8 x\)
\(\therefore h(x)=x^{2}-8 x+10\) is a quadratic equation
Since, \(D_{h}=D_{f} \cap D_{g}\)
Hence, Domain is a set of real numbers.
The range of quadratic equation is given by \(\left[\frac{-D}{4 a}, \infty\right)\) where \(D\) is the discriminant of the quadratic equation.
Hence, the range is \(\left[\frac{-\left((-8)^{2}-4 \times 1 \times 10\right)}{4 \times 1}, \infty\right)=\left[\frac{-24}{4}, \infty\right)=[-6, \infty)\)

Q.2. Find the domain and range of the function \(h(x)=f(x)-g(x)\) where \(f(x)=\cos 2 x\) and \(g(x)=\sin 2 x\)
Sol:
Given:
\(f(x)=\cos 2 x\)
Here,

• Domain is a set of real numbers
• Range is \([-1,1] \forall x \in R\)

Given: \(g(x)=\sin 2 x\)
Here,

• Domain is a set of real numbers
• Range is \([-1,1] \forall x \in R\)

Given: \(h(x)=f(x)-g(x)\)
\(\Rightarrow h(x)=\cos 2 x-\sin 2 x\)
We know that, \(D_{h}=D_{f} \cap D_{g}\)
Hence, the domain is a set of real numbers.
\(h(x)=\cos 2 x-\sin 2 x\)
\(=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos 2 x-\frac{1}{\sqrt{2}} \sin 2 x\right)\)
\(=\sqrt{2}\left(\cos \left(2 x+\frac{\pi}{4}\right)\right)\)
We know that, \(\cos \left(2 x+\frac{\pi}{4}\right) \in[-1,1]\)
\(\Rightarrow \sqrt{2} \operatorname{Cos}\left(2 x+\frac{\pi}{4}\right) \in[-\sqrt{2}, \sqrt{2}]\)
Hence, the range is \([-\sqrt{2}, \sqrt{2}]\)

Q.3. Find the area of the rectangle whose length and breadth are given by the functions \(p(x)=8-x, x>8\) and \(g(x)=x-1, x>1\)
Sol:
Length of the rectangle is given by \(p(x)=8-x\)
Breadth of the rectangle is given by \(g(x)=x-1\)
Let area of the rectangle be \(A(x)\)
\(\therefore A(x)=\) length \(\times\) breadth
\(\Rightarrow A(x)=(8-x) \times(x-1)\)
\(\therefore A(x)=-x^{2}+9 x-8\) for \(x>8\)
\(\left[\because D_{A}=D_{f} \cap D_{g}\right]\)

Q.4. If \(f(x)=x-1\) and \(g(x)=x-4\) then find \(\frac{f(x)}{g(x)}\) and its domain
Sol:
Given: \(f(x)=x-1\) and \(g(x)=x-4\)
Let \(h(x)=\frac{f(x)}{g(x)}\)
\(\Rightarrow h(x)=\frac{x-1}{x-4}\)
For \(h(x)=\frac{x-1}{x-4}\) to be meaningful \(x-4\) must be non-zero, which means that the domain must consist of all real numbers \(x\) such that \(x-4 \neq 0\), or \(x \neq 4\). That is domain is \(\mathbf{R} \sim 4\)

Q.5. A circle has its centre at the origin and radius \(3\). Let t denote the positive angle between \(x\) axis and the radius vector at some point \(P(x, y)\) of the circle. Find the parametric equation and then the standard form of the circle.
Sol:
Given: Centre of the circle \((0,0)\)
Positive angle between the \(x-\) axis and the radius vector at some point \(P(x, y)\) of the circle is \(t ; 0 \leq t \leq 2 \pi\)
Parametric equations of the circle are \(x=3 \cos t, y=3 \sin t, 0 \leq t \leq 2 \pi\)
For the standard form of the circle, we have to eliminate the parameter t from the parametric equations.
On squaring the parametric equations, we get
\(x^{2}=9 \cos ^{2} t\) (i)
\(y^{2}=9 \sin ^{2} t\) (ii)
Adding equations (i) and (ii), we get
\(x^{2}+y^{2}=9\left(\cos ^{2} t+\sin ^{2} t\right)\)
\(x^{2}+y^{2}=9\)
Hence, the standard form of the circle is \(x^{2}+y^{2}=9\)

Summary

A function is defined as \(f: x \rightarrow y\), where each element of set \(x\) is assigns to exactly one element of \(y\). And the set \(x\) is called the domain of the function and the set \(y\) is called the codomain of the function. The domain of a function is the set of all possible values of \(x\) for which function is defined. There are four basic algebraic oprartions on function which is also known as arithmetic oprations on functions such as addition of functions, subtraction of functions, multiplication of functions, and division of functions.

The domain of resulting function is intersection of domain of given functions, for division domain is intersection of given function except the point where divisor is zero as if divisior is zero then mathematically division is undefined. The range of resulting function can be calculated by using any of the standard methods of finding range.

FAQs on Algebraic Operations on Functions

Q.1. What are the operations of functions?
Ans: The four arithmetic operations of functions are addition, subtraction, multiplication, and division. When these operations combine any two or more functions, then the domain of the resulting function is only the element that was shared by the domains of the original functions.

Q.2. How do you solve operations with functions?
Ans: The formulas for the various operations on functions are given below.

No.	Operation	Formula
1.	Addition	\((f+g)(x)=f(x)+g(x)\)
2.	Subtraction	\((f-g)(x)=f(x)-g(x)\)
3.	Multiplication	\((f \cdot g)(x)=f(x) \cdot g(x)\)
4.	Division	\((f / g)(x)=f(x) / g(x)\), where \(g(x) \neq 0\)

Q.3. What is an addition in operation on function?
Ans: Addition is an algebraic operation that we use to get one resulting function when two or more functions are given. To solve the addition of two functions, we use the following steps.
Step 1: Write both the functions and combine them with the help of addition \((+)\)
Step 2: Write like terms together.
Step 3: Add the like terms
Step 4: Write them in the simplest form.

Q.4. How do you find the domain of function operations?
Ans: Steps to find the domain of function operations:
Step 1: Find the resultant function by applying the operation given.
Step 2: Write the resulting function in its simplest form.
Step 3: Find all the possible values of the independent variable for which function is defined.
Step 4: Values of the independent variable constitute the domain of the function.

Q.5. How do you solve operations with division functions?
Ans: Here the meaning of division of functions is nothing but dividing one polynomial by another polynomial. When we divide two polynomials, we get it in a fraction form, and this is called rational expression. So a rational expression is the division of two functions. We divide the degree of the divisor is less than or equal to the degree of dividend.

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