Algebra of functions: Additions, Subtraction, Multiplication and Quotient

Algebra of functions, practice problems and FAQs

We all are familiar with the fact that we can perform algebraic operations on numbers but now the question arises “ Can we perform algebraic operations on functions too? What are the conditions required to perform these operations?Can you visualize the resultant function?”

You will be able to answer all these questions after going through this article.

So in this article, we will learn about algebra of functions i.e., how to add two real functions, subtract a real function from another, multiply a real function by a scalar(real number), multiply two real functions and divide one real function by the other.

Table of contents

Addition of two real functions
Subtraction of one real function from another
Multiplication by a scalar
Multiplication of two real functions
Quotient of two real functions
Practice Problems
FAQs

Addition of two real functions

Let h: $h : X \to$ and $p : X \to$ be any two real functions, where $X \subseteq$ . Then, we can define $(h + p) : X \to$ by $(h + p) (x) = h (x) + p (x)$ , for all $x \in X$ .

The domain of the function $(h + p) (x)$ is $D (h + p) = D (h) \cap D (p)$

Example : Let $h (x) = x + 4$ and $p (x) = 3 x$

Then $(h + p) (x) = h (x) + p (x)$

$\Rightarrow (h + p) (x) = x + 4 + 3 x = 4 x + 4$

The domain of the function $(h + p) (x)$ is $D (h + p) = D (h) \cap D (p)$ . Now, the domain of both the functions is ,

$\Rightarrow D (h + p) =$ $\cap$ = .

Subtraction of one real-valued function from another

Let $h : X \to$ and $p : X \to$ be any two real functions, where $X \subseteq$ . Then, we can define $(h - p) : X \to$ by $(h - p) (x) = h (x) - p (x)$ , for all $x \in X$ .

The domain of the function (h-p)(x) is $D (h - p) = D (h) \cap D (p)$

Example:Let $h (x) = x^{2} + 1$ and $p (x) = \frac{1}{x}$

Then $(h - p) (x) = h (x) - p (x)$

$\Rightarrow (h - p) (x) = x^{2} + 1 - \frac{1}{x}$

Since D(h)= and D(p)= -{0}, $D (h - p) = D (h) \cap D (p)$

D(h-p)= $\cap$ -{0} = -{0}.

Multiplication by a scalar

Let $h : X \to$ be a real valued function and $β \in$ be a scalar. Then, the product $β h$ is a function from $X \to$ defined by $(β h) (x) = β . h (x),$ , for all $x \in X$ .

Note: The domain of function remains same as that of the original function in this case.

Example : Let $f (x) = 1 + x^{2}$ and $α = 3$

$\Rightarrow (α f) (x) = 3 (1 + x^{2}) = 3 + 3 x^{2}$

Obviously domain of f is also .

Multiplication of two real functions

The product of two real functions $h : X \to$ and $p : X \to$ is a function $(h p) : X \to$ defined by $(h p) (x) = h (x) p (x)$ , for all $x \in X$ . It is also called pointwise multiplication.

Note : If D(h) and D(p) are domains of the two functions h and p respectively then the domain of the functions (h+p), (h-p), hp will be $D (h) \cap D (p)$

Example: Let f(x)=x+1 and $g (x) = \frac{1}{x}$

Then $(f g) (x) = f (x) g (x)$

(fg)(x)= $<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow><mi>x</mi></mfrac></math>$

Since D(f)= and D(g)= -{0}, $D (f g) = D (f) \cap D (g)$

D(fg)= $\cap$ -{0} = -{0}.

Quotient of two real functions

Let $h : X \to$ and $p : X \to$ be any two real functions, where $X \subseteq$ . Then, the quotient of h by p denoted by, $\frac{h}{p}$ is a function defined as $\frac{h}{p} (x) = \frac{h (x)}{p (x)}$ , provided $p (x) \neq 0$ , $x \in X$ .

Note: Domain of $<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>h</mi><mi>p</mi></mfrac><mo>=</mo><mfenced open="{" close="}"><mrow><mi>x</mi><mo> </mo><mo>|</mo><mo> </mo><mi>x</mi><mo>∈</mo><mi>D</mi><mfenced><mi>h</mi></mfenced><mo>∩</mo><mi>D</mi><mfenced><mi>p</mi></mfenced><mo>,</mo><mi>p</mi><mfenced><mi>x</mi></mfenced><mo>≠</mo><mn>0</mn></mrow></mfenced></math>$

Example : If $f (x) = x^{2} + 1$ and g(x)=x+2 then find $<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>f</mi><mi>g</mi></mfrac></math>$ and its domain.

Solution: We have, $<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>f</mi><mi>g</mi></mfrac><mfenced><mi>x</mi></mfenced><mo>=</mo><mfrac><mrow><mi>f</mi><mfenced><mi>x</mi></mfenced></mrow><mrow><mi>g</mi><mfenced><mi>x</mi></mfenced></mrow></mfrac></math>$

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mfrac><mi>f</mi><mi>g</mi></mfrac><mfenced><mi>x</mi></mfenced><mo>=</mo><mfrac><mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>2</mn></mrow></mfrac></math>$

Since D(f)= and D(g)= ,

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>D</mi><mfenced><mfrac><mi>f</mi><mi>g</mi></mfrac></mfenced><mo>=</mo><mfenced open="{" close="}"><mrow><mi>x</mi><mo> </mo><mo>|</mo><mo> </mo><mi>x</mi><mo>∈</mo><mi>D</mi><mfenced><mi>f</mi></mfenced><mo>∩</mo><mi>D</mi><mfenced><mi>g</mi></mfenced><mo>,</mo><mi>g</mi><mfenced><mi>x</mi></mfenced><mo>≠</mo><mn>0</mn></mrow></mfenced></math>$

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>D</mi><mfenced><mfrac><mi>f</mi><mi>g</mi></mfrac></mfenced><mo>=</mo><mi mathvariant="normal">ℝ</mi><mo>-</mo><mfenced open="{" close="}"><mrow><mo>-</mo><mn>2</mn></mrow></mfenced></math>$ as g(x)=0 at x=-2.

Note: The concept of the algebra of functions is used along with other concepts of mathematics like limits, continuity, differentiability, etc. where we need to combine two or more functions. Also, sometimes we break bigger functions into smaller ones with the help of the algebra of functions. Let us understand this with the help of some solved examples.

Practice Problems

Example : Let f: $\to$ be the function defined by $f (x) = x^{3} - x^{2} + (x - 1) \sin x$ &

g: $\to$ be an arbitrary function. Let fg: $\to$ be the product function defined by (fg)(x) = f(x)g(x). Then figure out which of the following statements are true ?

(A) fg is differentiable at x = 1 if g is continuous at x = 1
(B) g is continuous at x = 1 if fg is differentiable at x = 1
(C) fg is differentiable at x = 1 if g is differentiable at x = 1
(D) g is differentiable at x = 1 if fg is differentiable at x = 1

Solution :

$f (x) = x^{3} - x^{2} + (x - 1) \sin x$ and g: $\to$

$h (x) = f (x) g (x) = (x^{3} - x^{2} + (x - 1) \sin x) . g (x)$

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>'</mo><mfenced><msup><mn>1</mn><mo>+</mo></msup></mfenced><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mfenced open="{" close="}"><mrow><msup><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced><mn>3</mn></msup><mo>-</mo><msup><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced><mn>2</mn></msup><mo>+</mo><mi>h</mi><mo> </mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow><mi>h</mi></mfrac><mspace linebreak="newline"/><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mfenced open="{" close="}"><mrow><mn>1</mn><mo>+</mo><msup><mi>h</mi><mn>3</mn></msup><mo>+</mo><mn>3</mn><mi>h</mi><mo>+</mo><mn>3</mn><msup><mi>h</mi><mn>2</mn></msup><mo>-</mo><mn>1</mn><mo>-</mo><msup><mi>h</mi><mn>2</mn></msup><mo>-</mo><mn>2</mn><mi>h</mi><mo>+</mo><mi>h</mi><mo> </mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow><mi>h</mi></mfrac><mspace linebreak="newline"/><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mfenced open="{" close="}"><mrow><msup><mi>h</mi><mn>3</mn></msup><mo>+</mo><mn>2</mn><msup><mi>h</mi><mn>2</mn></msup><mo>+</mo><mi>h</mi><mo>+</mo><mi>h</mi><mo> </mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow><mi>h</mi></mfrac><mspace linebreak="newline"/><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfenced open="{" close="}"><mrow><msup><mi>h</mi><mn>2</mn></msup><mo>+</mo><mn>2</mn><mi>h</mi><mo>+</mo><mn>1</mn><mo>+</mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>+</mo><mi>h</mi></mrow></mfenced><mspace linebreak="newline"/></math>$

……(i)

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>'</mo><mfenced><msup><mn>1</mn><mo>-</mo></msup></mfenced><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mfenced open="{" close="}"><mrow><msup><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced><mn>3</mn></msup><mo>-</mo><msup><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced><mn>2</mn></msup><mo>+</mo><mfenced><mrow><mo>-</mo><mi>h</mi></mrow></mfenced><mo> </mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></mrow><mrow><mo>-</mo><mi>h</mi></mrow></mfrac><mspace linebreak="newline"/><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mfenced open="{" close="}"><mrow><mo>-</mo><msup><mi>h</mi><mn>3</mn></msup><mo>+</mo><mn>3</mn><msup><mi>h</mi><mn>2</mn></msup><mo>-</mo><mn>3</mn><mi>h</mi><mo>-</mo><msup><mi>h</mi><mn>2</mn></msup><mo>+</mo><mn>2</mn><mi>h</mi><mo>-</mo><mi>h</mi><mo> </mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></mrow><mrow><mo>-</mo><mi>h</mi></mrow></mfrac><mspace linebreak="newline"/><mo>=</mo><munder><mrow><mi>l</mi><mi>i</mi><mi>m</mi></mrow><mrow><mi>h</mi><mo>→</mo><mn>0</mn></mrow></munder><mfenced open="{" close="}"><mrow><msup><mi>h</mi><mn>2</mn></msup><mo>-</mo><mn>3</mn><mi>h</mi><mo>+</mo><mn>3</mn><mo>+</mo><mi>h</mi><mo>-</mo><mn>2</mn><mo>+</mo><mi>sin</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></mrow></mfenced><mi>g</mi><mfenced><mrow><mn>1</mn><mo>-</mo><mi>h</mi></mrow></mfenced></math>$

……(ii)

We know that sin x is a continuous function $\Rightarrow \sin (1 + h) = \sin (1 - h)$

Now,if the function g(x) is continuous at x= $\Rightarrow g (1 + h) = g (1 - h) = g (1)$

Hence from (i) & (ii), h'(1+)=h'(1-)=(1+sin 1)g(1)

Therefore, h(x) is differentiable at x=1.

Also if g(x) is differentiable at $x = 1 \Rightarrow g (x)$ is continuous at x=1

Therefore, h(x) will still be differentiable at x=1

Hence, options (A) and (C) are the correct answers.

Example : If $F (n + 1) = \frac{2 F (n) + 1}{2}, n = 1,2, 3 . . . .$ and F(1)=2, then find F(101).

Solution :

Given, $F (n + 1) = \frac{2 F (n) + 1}{2}$

$\Rightarrow 2 F (n + 1) = 2 F (n) + 1 \Rightarrow F (n + 1) - F (n) = \frac{1}{2}$

$n = 1; F (2) - F (1) = \frac{1}{2}$

$n = 2; F (3) - F (2) = \frac{1}{2}$

$n = 3; F (4) - F (3) = \frac{1}{2}$

$n = 100; F (101) - F (100) = \frac{1}{2}$

On adding the above equations, we get

$F (101) - F (1) = 100 \times \frac{1}{2} = 50$

$\Rightarrow F (101) - 2 = 50 \Rightarrow F (101) = 52$

Example : If f (x) = cos (log x), then find the value of $<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo> </mo><mo>(</mo><mi>x</mi><mo>)</mo><mo>.</mo><mi>f</mi><mo> </mo><mo>(</mo><mn>4</mn><mo>)</mo><mo> </mo><mo>−</mo><mfenced open="[" close="]"><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mo> </mo><mfenced open="[" close="]"><mrow><mi>f</mi><mo> </mo><mfenced><mfrac><mi>x</mi><mn>4</mn></mfrac></mfenced><mo> </mo><mo>+</mo><mi>f</mi><mo> </mo><mo>(</mo><mn>4</mn><mi>x</mi><mo>)</mo></mrow></mfenced></math>$ .

Solution :

f (x) = cos (log x)

Now let , $<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>y</mi><mo>=</mo><mi>f</mi><mo> </mo><mo>(</mo><mi>x</mi><mo>)</mo><mo>.</mo><mi>f</mi><mo> </mo><mo>(</mo><mn>4</mn><mo>)</mo><mo> </mo><mo>−</mo><mfenced open="[" close="]"><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mo> </mo><mfenced open="[" close="]"><mrow><mi>f</mi><mo> </mo><mfenced><mfrac><mi>x</mi><mn>4</mn></mfrac></mfenced><mo> </mo><mo>+</mo><mi>f</mi><mo> </mo><mo>(</mo><mn>4</mn><mi>x</mi><mo>)</mo></mrow></mfenced></math>$

$<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>y</mi><mo>=</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>)</mo><mo>.</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>−</mo><mfenced open="[" close="]"><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mfenced open="[" close="]"><mrow><mi>cos</mi><mo> </mo><mi>log</mi><mo> </mo><mfenced><mfrac><mi>x</mi><mn>4</mn></mfrac></mfenced><mo>+</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mn>4</mn><mi>x</mi><mo>)</mo></mrow></mfenced><mspace linebreak="newline"/><mo>⇒</mo><mi>y</mi><mo>=</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>)</mo><mo> </mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>−</mo><mfenced open="[" close="]"><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mo>[</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>-</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>+</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>+</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>]</mo><mspace linebreak="newline"/><mo>⇒</mo><mi>y</mi><mo>=</mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>)</mo><mo> </mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>−</mo><mfenced open="[" close="]"><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mo>[</mo><mn>2</mn><mo> </mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mi>x</mi><mo>)</mo><mo> </mo><mi>cos</mi><mo> </mo><mo>(</mo><mi>log</mi><mo> </mo><mn>4</mn><mo>)</mo><mo>]</mo></math>$

y = 0.

Example : If . Find f(x).g(x).

Solution :

Drawing the graphs of f(x) & g(x)

Taking different values of f(x) & g(x) under the common interval using graphs of f and g

FAQs

Q1. Can we add any two real functions?
Answer: Any two real functions having common elements in their domains can be added.

Q2.Where do we use algebra of real functions?
Answer:It is used in different mathematical concepts involving 2 or more functions like, Matrices, derivatives, integrals, continuity and differentiability, etc.

Q3.What is the algebra of functions?
Answer: Algebra of functions basically deals with the addition, subtraction, multiplication, and division of functions.

NCERT Class 11 Maths Chapters

Sets	Relations and Functions	Triginometric Functions
Mathematical Induction	Numbers and Quadriatic Equations	Linear Inequalities
Premutations and Combinations	Binomial Theorem	Sequence and Series
Straight Lines	Conic Sections	3 D Geometry
Limits and Derivatives	Mathematical Reasoning	Statistics
Probability

NCERT Class 12 Maths Chapters

Relations and Functions	Continuity and Differentiability	Differential Equations
Inverse Trigonometric Functions	Applications of Derivatives	Vector Algebra
Matrices	Integrals	Three Dimensional Geometry
Determinants	Applications of Integrals	Linear Programming
Probability