Content text METHODS OF FINDING DOMAIN OF FUNCTIONS.pdf
What you already know What you will learn • Functions • Types of functions • Logarithmic and exponential functions • Domain of a function • Greatest integer function Domain All the values of x for which the function is defined is known as the domain of the function. For example, fx x ( )= , where + x R ∈ . Here, R+ is the domain of the function x . Domain of a function can be obtained by following two methods: (a) Solving the basic types of functions (b) Using graph of the functions Find the domain of x - 3 Let f(x) = x - 3 For the function fx fx ( ), ( ) 0 ≥ , Therefore, x x ≥ ⇒∈ ∞ 3 [3, ) (a) ( ) ( ) 1 , 0 f x ≠ f x where (b) fx fx ( ), where 0 ( ) ≥ (c) ( ) ( ) 1 , 0 f x > f x where Types of basic functions Example NOTES RELATIONS AND FUNCTIONS MATHEMATICS METHODS OF FINDING DOMAIN OF FUNCTIONS © 2020, BYJU'S. All rights reserved
Find the domain of the following. Find the domain of the following functions: Solution Solution (a) Let f(x) = 1 - 5x For the function fx fx ( ), ( ) 0 ≥ , 1 - 5x ≥ 0, x ≤ 1 5 ( 5 1 x - , ∈ ∞ (a) ( ) ( ) ( ) 2 2 +2 +1 = - -6 x x f x x x = ( ) ( ) g x h x , where g(x) = x2 + 2x + 1 and h(x) = (x2 - x - 6) = (x + 2)(x - 3) The domain of g(x) is R The domain of h(x) is R - {-2, 3} , as (x2 - x - 6) is in the denominator. Hence cannot be 0. Taking the intersection of both cases, the domain of f(x) is R - {-2, 3}. (c) Let f(x) = x For the function ( ) ( ) 1 , f x ≠ 0 f x , Therefore, x ∈ R - {0} (e) Let f(x) = 2x - x2 For the function fx fx ( ), ( ) 0 ≥ , 2x - x2 ≥ 0 ⟹ x(2 - x) ≥ 0 x(x - 2) ≤ 0 Therefore, x ∈ [0, 2] (b) Let f(x) = 1 - 5x For the function ≠ 1 , ( ) 0 ( ) f x f x , 1 - 5x ≠ 0, Therefore, 1 5 x ∈ R - (d) Let f(x) = x For the function fx fx x ( ), ( ) 0 = ≥ , Therefore, x ∈ R+ (f) Let f(x) = 3x - 2 For the function > 1 , ( ) 0 ( ) f x f x , 3x - 2 > 0 ⟹ x > 2 3 Therefore, 2 , 3 ∈ ∞ x (a) (b) (c) (d) (e) (f) y x y y x x y x y x x y x 2 1 1 = 1 - 5 = = 1-5 1 = = 2 - = 3 -2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 +2 +1 1 - +5 = = 4 - + = = 3 8 5 - -6 - 1 + 3 3 -6 1 = = | 2x - 9 | = + 3 - 10 = +2 -8 (a) (b) (c) (d) (e) (f) (g) (h) x x x f x fx x f x fx x x x x x x f x f x ln f x ln x x f x x x ln - - x © 2020, BYJU'S. All rights reserved 02
(b) ( ) 2 1 = 4 - + - 1 fx x x , = g(x) + h(x), where g(x) = 4 - x and h(x) = 2 1 x - 1 Here, g(x) = 4 - x ⟹ 4 - x ≥ 0 ⟹ -x ≥ -4 ⟹ x ≤ 4 So, the domain of g(x) is x ≤ 4 Similarly h(x) = 2 1 x - 1 ⟹ x2 - 1 > 0 ⟹ (x + 1) (x - 1) > 0 The domain of h(x) using wavy curve; x ∈ ( -∞, -1) U (1, ∞). Taking the intersection of both the cases, the domain of f(x) is x ∈ ( -∞, -1) U (1, 4] (c) ( ) - +5 = + 3 - +5 0 + 3 - 5 0 + 3 (-3, 5] x f x x x x x x x ≥ ≤ ∈ Using wavy curve method (f) f(x) = ln (|2x - 9|), where base = e > 0 Argument: (|2x - 9|) > 0 x ∈ R - 9 2 (h) f(x) = logx e, where argument > 0 and base x > 0 and x ≠ 1 ⟹ The domain of f(x); x ∈ (0, ∞) - {1} (e) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 = +2 - 8 > 0 + 4 - 2 > 0; - , -4 2, . , x = 3 - 6 x R 3 -6 = = +2 -8 . x g x f x h x x x x h x x x hx x g x x ∈∞ ∞ ⇒ ∈ ⇒ ⇒ The domain of is Since is a linear funct U ion. So, its domain is So, the domain of is x - , -4 2, . f x( ) ∈∞ ∞ ( )U( ) (d) ( ) ) 13 1 3 13 , 3 = 3 - 8 - 5 3 - 8 5 3 - 8 5 or 3 - 8 -5 or (- , 1] fx x x x x x x x ≥ ≥ ≤ ≥ ≤ ⇒∈ ∞ ∞ U (g) f(x) = ln (x2 + 3x - 10), where base = e > 0 Argument: (x2 + 3x - 10) > 0 ⟹ (x + 5)(x - 2) > 0 ⟹ x ∈ (-∞, -5) U (2, ∞) © 2020, BYJU'S. All rights reserved 03
Find the domain of ( ) f x 2 x - x 2x + 4 1 1 = - 3 3 Solution Let the function f(x) = g x( ), therefore, g(x) ≥ 0 ⟹ 2x - x 2x + 4 1 1 0 3 - 3 ≥ Base of the given exponential function is < 1. Therefore, the function will be decreasing. 2x - x 2x + 4 1 1 3 3 ≥ Therefore, x2 - x ≤ 2x + 4 x2 - 3x - 4 ≤ 0 (x - 4)(x + 1) ≤ 0 So, the domain of f(x) will be x ∈ [-1, 4]. Note If the base of exponential function is less than 1, then the function will be decreasing in nature and the sign of inequalities will get reversed while solving. Concept Check 1. Find the domain of y = |x| - 2 2. Find the domain of 1 y = 1 - x 3. Find the domain of y = log3 (3x - 2) 4. Find the domain of f(x) = 2 1 log ( + 1) x x Greatest integer function The real function f : R → R defined by f(x) = [x], x ∈ R assumes the value of the greatest integer less than or equal to x is known as the greatest integer function. f(x) = [x], where [x] denotes the greatest integer less than or equal to x. [2.7] = 2, [-3.2] = -4, [0.75] = 0, [5] = 5 Example © 2020, BYJU'S. All rights reserved 04
Note [x] = -2 when -2 ≤ x < -1; [x] = -1 when -1 ≤ x < 0; [x] = 0 when 0 ≤ x < 1; [x] = 1 when 1 ≤ x < 2; [x] = 2 when 2 ≤ x < 3; and so on. To plot the graph use this definition of [x]. The graph would be in the form of steps. f(x) = [x] = -1; -1 ≤ x < 0 f(x) = [x] = 0; 0 ≤ x < 1 f(x) = [x] = 1; 1 ≤ x < 2 f(x) = [x] = 2; 2 ≤ x < 3 and so on............ 1 X 1 -1 -1 2 2 -2 -2 3 3 -4 -3 4 Greatest integer function Y D( f ) = R, and R( f ) = Z Property 1: [x] ≤ x < [x] + 1 Property 2: x - 1 < [x] ≤ x Property 3: [x + m] = [x] + m, m ∊ Z Property 4: [x] ≥ n, then x ≥ n, n ∊ Z Property 5: [x] > n, then x ≥ n + 1, n ∊ Z Property 6: [x] < n, then x < n, n ∊ Z Property 7: [x] + [-x] = 0, x ∊ Z, [x] + [-x] = -1, x ∉ Z Property 8: [x] ≤ n, then x < n + 1, n ∊ Z Property 9: [x] = 2 2 ( + 1) + x x Observations Find the value of the following: (a) [1] (b) [1.999] (c) [-2.05] (d) [2.7] (e) [-3.0] (a) [1] = 1 (b) [1.999] = 1 (c) [-2.05] = -3 (d) [2.7] = 2 (e) [-3.0] = -3 Solution 0 Properties of greatest integer function are as follows © 2020, BYJU'S. All rights reserved 05