Mathematics FUNCTIONS -Some functions and their graphs and Algebra of real functions
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### ✎ Topics Covered

star Functions
star Some functions and their graphs
star Algebra of real functions

### Functions

● We study a special type of relation called function. It is one of the most important concepts in mathematics.

● We can, visualise a function as a rule, which produces new elements out of some given elements. There are many terms such as ‘map’ or ‘mapping’ used to denote a function.

\color{blue} ☛\color{blue} ul("Function")
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

\color{green} ✍️ In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element.

\color{green} ✍️  If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the image of a under f and a is called the preimage of b under f.

\color{green} ✍️  The function f from A to B is denoted by f: A -> B.

\color{blue} ☛ color{blue} ul ("Real valued function")
A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function.
Q 3017291189

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)}
(iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)}

Solution:

(i) Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function.
(ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function.
(iii) Since every element has one and only one image, this relation is a function.
Q 3067491385

Let N be the set of natural numbers. Define a real valued function f : N → N by f (x) = 2x + 1. Using this definition, complete the table given below.

x 1 3 4 5 6 7
y f(1) = ......... f(2) = ........ f(3) = ...... f(4) = ... f(5) = ..... f(6) = ...... f(7) = ....

Solution:

The completed table is given by

x 1 3 4 5 6 7
y f(1) = 3 f(2) = 5 f(3) = 7 f(4) =11 f(5) = 13 f(6) = 15 f(7) = 15

Q 3008401308

Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).

Solution:

Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R,
f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1.

### Identity function

● Let R be the set of real numbers.

● Define the real valued function f : R → R by y = f(x) = x for each x ∈ R. Such a function is called the identity function.

● Here the domain and range of f are R. The graph is a straight line as shown in Fig . It passes through the origin.

### Constant function

● Define the function f: R → R by y = f (x) = c, x ∈ R where c is a constant and each x ∈ R.

● Here domain of f is R and its range is {c}.

● The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its graph will be a line as shown in the Fig .

### Polynomial function

● A function f : R→R is said to be polynomial function if for each x in R,

● y = f (x) = a_0 + a_1x + a_2x^2 + ...+ a_n x^n, where n is a non-negative integer and a_0, a_1, a_2,...,a_n ∈R.

●The functions defined by f(x) = x^3 – x^2 + 2, and g(x) = x^4 + 2 x are some examples of polynomial functions,
whereas the function h defined by h(x) = x^(2/3) +2x is not a polynomial function.
Q 3017591489

Define the function f: R → R by y = f(x) = x^2, x ∈ R. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f.

Solution:

The completed Table is given below:

Domain of f = {x : x∈R}. Range of f = {x : x ≥ 0, x ∈ R}. The graph of f is given by Fig
Q 3057691584

Draw the graph of the function f : R → R defined by f (x) = x^3, x ∈ R.

Solution:

We have
f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc.
Therefore, f = {(x,x^3): x∈R}.
The graph of f is given in fig

### Rational functions

● Rational functions are functions of the type (f( x))/ (g (x)) ,

where f(x) and g(x) are polynomial functions of x defined in a domain, where g(x) ≠ 0.
Q 3028201101

Define the real valued function f : R – {0} → R defined by f (x) = 1 /x , x ∈ R –{0}. Complete the Table given below using this definition. What is the domain and range of this function?

Solution:

The completed Table is given by The domain is all real numbers except 0 and its range is also all real numbers except 0. The graph of f is given in

### The Modulus function

● The function f: R→R defined by f(x) = |x| for each x ∈R is called modulus function.

● For each non-negative value of x, f(x) is equal to x. But for negative values of x, the value of f(x) is the negative of the value of x, i.e.,

f(x)= { tt (( x , if , x >= 0),(-x, if, x< 0))

The graph of the modulus function is given in Fig .

### Signum function

● Signum function The function f:R→R defined by

f(x)= { tt (( 1, if , x > 0),(0, if, x=0),( -1, if , x < 0))

is called the signum function.

● The domain of the signum function is R and the range is the set {–1, 0, 1}.

● The graph of the signum function is given by the Fig .

### Greatest integer function

● The function f: R → R defined by

● f(x) = [x], x ∈R assumes the value of the greatest integer, less than or equal to x. Such a function is called the greatest integer function.

From the definition of [x], we can see that

tt (( [x]= -1, "for" \ \ –1 ≤ x < 0 ),([x]=0 , "for" \ \ 0 ≤ x < 1 ),( [x] = 1 , "for" \ \ 1 ≤ x < 2 ),([x] = 2, "for" \ \ 2 ≤ x < 3))

and so on.
The graph of the function is shown in Fig .

### Algebra of real functions

(i) "Addition of two real functions" Let f : X → R and g : X → R be any two real functions, where X ⊂ R.

Then, we define (f + g): X → R by
●(f + g) (x) = f (x) + g (x), for all x ∈ X.

(ii) "Subtraction of a real function from another" Let f :X → R and g:X → R be any two real functions, where X ⊂ R.

Then, we define (f – g) : X→R by
● (f–g) (x) = f(x) –g(x), for all x ∈ X.

(iii) "Multiplication by a scalar" Let f : X→R be a real valued function and α be a scalar. Here by scalar, we mean a real number.
Then the product α f is a function from X to R defined by
● (α f ) (x) = α f (x), x ∈X.

(iv) "Multiplication of two real functions" The product (or multiplication) of two real functions f:X→R and g:X→R is a function
● fg:X→R defined by (fg) (x) = f(x) g(x), for all x ∈ X.
● This is also called "pointwise multiplication."

(v) "Quotient of two real functions" Let f and g be two real functions defined from X→R where X ⊂R.
The quotient of f by g denoted by f/g is a function defined by ,
● (f/g)(x)=(f(x))/(g(x)) provided g(x) ≠ 0, x ∈ X
Q 3058201104

Let f(x) = x^2 and g(x) = 2x + 1 be two real functions.Find (f + g) (x), (f –g) (x), (fg) (x), (f/g) (x)

Solution:

We have,
(f + g) (x) = x^2 + 2x + 1, (f –g) (x) = x^2 – 2x – 1,

(fg)(x) = x^2 (2x+1) = 2x^3 + x^2 , (f/g) (x) = x^2/(2x+1) , x ne -1/2
Q 3038301202

Let f(x) = x and g(x) = x be two functions defined over the set of non negative real numbers. Find (f+g) (x) , (f-g) (x) , (fg) (x) and (f/g) (x)

Solution:

We have
(f + g) (x) = sqrtx + x, (f – g) (x) = sqrtx-x

(fg) x = sqrtx (x) = x^(3/2) and (f/g) (x) = sqrtx/x = x^(-1/2) , x ne 0