Consider the real function f : R → R : f ( x ) = x + 5 for all x ∈ R . Find its domain and range. Draw the graph of this function.

Given f ( x ) = x + 5 ∀ x ∈ R The domain of the given function is all real numbers expect where the expression is undefined. In this case, there is no real number which makes the expression undefined. As f ( x ) is a polynomial function, we can have any value of x . Therefore, Domain ( f ) = R Now L e t y = f ( x ) y = x + 5 x = y - 5 The range is set of all valid values of y Therefore, Range ( f ) = R

Consider the function f : R → R defined by f ( x ) = 1 - x when x < 0 1 when x = 0 1 + x when x > 0 Write its domain and range. Also, draw the graph of f ( x ) .

Given: f ( x ) = 1 - x when x < 0 1 when x = 0 1 + x when x > 0 To find: Domain f ( x ) When f ( x ) = 1 - x , x < 0 In this case there is no value of x ( x < 0 ) which makes the above expression undefined. Therefore, Domain ( f ) = ( - ∞ , 0 ) . . . 1 When f ( x ) = x x = 0 In this case there is no value other than 0 which makes the above expression undefined. Therefore, Domain ( f ) = 0 . . . 2 When f ( x ) = x + 1 x > 0 In this case there is no value of x ( x > 0 ) which makes the above expression undefined. Therefore Domain f = 0 , ∞ . . . 3 Therefore, from equations ( 1 ) , ( 2 ) , ( 3 ) we can say that the domain of f ( x ) as a whole: Domain ( f ) = ( - ∞ , ∞ ) Now when f ( x ) = 1 - x As x ranges from - ∞ , 0 then f ( x ) ranges from 1 , ∞ Therefore Range ( f ) = ( 1 , ∞ ) . . . 4 Now when f ( x ) = x as x = 0 Therefore, Range ( f ) = 0 . . . 5 Now when f ( x ) = x + 1 ⇒ x = f ( x ) - 1 As x ranges from 0 , ∞ then f ( x ) ranges from 1 , ∞ Therefore, Range ( f ) = ( 1 , ∞ ) . . . 6 From ( 4 ) , ( 5 ) , ( 6 ) the range of f ( x ) as whole: Range ( f ) = 0 ∪ ( 1 , ∞ )

Find the domain and the range of the square root function, f : R + ∪ { 0 } → R f ( x ) = x for all non-negative real numbers.

Given, f x = x we know that function is not defined if expression under even root is negative. So, f x will be defined if x ≥ 0 Therefore, the domain of function is D f = [ 0 , ∞ ) . Let f x = y = x ∵ x ≥ 0 ⇒ y ≥ 0 Therefore, the range of function is R f = [ 0 , ∞ ) .

Find the domain and the range of the cube root function, f : R → R : f ( x ) = x 1 3 for all x ∈ R

Given f ( x ) = x 1 3 ∀ x ∈ R Here f ( x ) = x 1 3 Because all real numbers have a cube root. There is no value of which makes the function undefined. The domain of the above function would be, Domain ( f ) = R Consider f ( x ) = y Then y = x 1 3 ⇒ y 3 = x Since f ( x ) is continuous, it follows that for every value of y there would be a cube of that value. Range ( f ) = R

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Set and Functions>Chapter 3 - Functions>EXERCISE 3D

R S Aggarwal Solutions for Exercise 4: EXERCISE 3D

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 4: EXERCISE 3D

Attempt the practice questions from Exercise 4: EXERCISE 3D with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.