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VEDANTU COMMERCE Function | Domain, Codomain, Range Function | Domain, Codomain, Range Function | Domain, Codomain, Range

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VEDANTU COMMERCE Function | Domain, Codomain, Range Function | Domain, Codomain, Range Function | Domain, Codomain, Range Function | Domain, CoDomain, Range VEDANTU COMMERCE Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Thought of the Day Function | Domain, CoDomain, Range Save Yourself From Covid Function | Domain, CoDomain, Range Today’s Schedule Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Like Share Subscribe Function | Domain, CoDomain, Range Are You Ready For it…??? Function | Domain, CoDomain, Range Q. If function f = {(-2, -11), (-1, -7), (1, 1), (2, 5)} is described by a linear relation, then f(x) = A 5x - 2 B 4x + 3 C 2x + 3 D 4x - 3 Function | Domain, CoDomain, Range Well Done Ninjas…!!! Function | Domain, CoDomain, Range Well Done Ninjas…!!! Function | Domain, CoDomain, Range Are You Excited To Take Udaan…??? Function | Domain, CoDomain, Range Till Now We Covered... ● Sets ● Ordered Pairs ● Cartesian Product of two Sets ● Relation ● Domain & Range ● Functions Function | Domain, CoDomain, Range Today’s Objective ● Domain of a Function ● Problems ● Range of a Function ● Problems ● Codomain of a Function ● Problems Function | Domain, CoDomain, Range Domain, Codomain & Range -2 a -1 b 0 c 1 d Range 2 e A B Domain Codomain Function | Domain, CoDomain, Range Domain If f: A ⟶ B be a function, then all the input values that satisfy the function or the elements in set A, are known as domain. Ex: The function is not defined when x = 0. So, A = R - {0} Function | Domain, CoDomain, Range Q. The domain of the function f: A ⟶ B, is A R B R - {1} C [-3, 3] - {1} D R - [-3, 3] Function | Domain, CoDomain, Range Q. The domain of the function f: A ⟶ B, is A R B R - {1} C [-3, 3] - {1} D R - [-3, 3] Solution: The function is defined when the denominator of the rational function is not zero. The numerator is a radical function which should not have negative value inside the radical. So, Function | Domain, CoDomain, Range Solution: Function | Domain, CoDomain, Range Q. The domain of the function f: A ⟶ B, is A R B R - {-1, 0, 1} C R - {1} D R - [-1, 1] Function | Domain, CoDomain, Range Q. The domain of the function f: A ⟶ B, is A R B R - {-1, 0, 1} C R - {1} D R - [-1, 1] Solution: The function is defined when the denominator of the rational function is not zero. The denominator of the function is a radical function, which is defined when the value inside the radical is greater than or equal to zero. So, Function | Domain, CoDomain, Range Solution: Function | Domain, CoDomain, Range Range If f: A ⟶ B be a function, then the set of values of functions which it attains at points in its domain, is called Range. Ex: f: A ⟶ B; f(x) = 3x + 2 The function will return all the real values. So,Range(f), B = R Function | Domain, CoDomain, Range Q. The range of the function f: A ⟶ R, f(x) = x2 + 1, where A ={-1, 0, 2, 4} A {-1, 0, 2, 4} B {1, 2, 5, 17} C {0, 2, 5, 17} D {1, 2, 4, 17} Function | Domain, CoDomain, Range Q. The range of the function f: A ⟶ R, f(x) = x2 + 1, where A ={-1, 0, 2, 4} A {-1, 0, 2, 4} B {1, 2, 5, 17} C {0, 2, 5, 17} D {1, 2, 4, 17} Solution: So, the range is, {2, 1, 5, 17} Function | Domain, CoDomain, Range Q. If f: A ⟶ R, f(x) = x2 + x - 2, where A ={-2, -1, 1, 2}, then the range of the function is, A {-2, 0, 4} B {-2, 0, 4, 0} C {-2, -1, 0, 1} D {-2, -1, 0, 4} Function | Domain, CoDomain, Range Q. If f: A ⟶ R, f(x) = x2 + x - 2, where A ={-2, -1, 1, 2}, then the range of the function is, A {-2, 0, 4} B {-2, 0, 4, 0} C {-2, -1, 0, 1} D {-2, -1, 0, 4} Solution: Function | Domain, CoDomain, Range Procedure to calculate Range ● Put y = f(x) and solve for x in terms of y, x = g(y). Function | Domain, CoDomain, Range Procedure to calculate Range ● Interchange x & y, find the values of x for which y is not defined. y is not defined when x = -1. y ∈ R - {-1}. ● The set of values obtained in step II in the range of f(x). So, range = R - {-1} Function | Domain, CoDomain, Range Q. function is A [0, 4] B (-∞, -4) ∪ (4, ∞) C [-4, 4] D None Function | Domain, CoDomain, Range Q. function is A [0, 4] B (-∞, -4) ∪ (4, ∞) C [-4, 4] D None Solution: Function | Domain, CoDomain, Range Solution: Function | Domain, CoDomain, Range Codomain If f: A ⟶ B be a function, then the set B is known as the codomain of the function. Function | Domain, CoDomain, Range Q. The codomain of the function f: A ⟶ R, f(x) = 2x2 + x + 3, where A ={-1, 0, 2, 4} is A {-1, 0, 2, 4} B {1, 2, 5, 17} C R D {1, 2, 4, 17} Function | Domain, CoDomain, Range Q. The codomain of the function f: A ⟶ R, f(x) = 2x2 + x + 3, where A ={-1, 0, 2, 4} is A {-1, 0, 2, 4} B {1, 2, 5, 17} C R D {1, 2, 4, 17} Solution: The given function is, f: A ⟶ R, f(x) = 2x2 + x + 3 It is defined for set A and R. So, the codomain for the function is all the real numbers, R. Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Q. The range of the function is A [1, ∞) B [2, ∞) C D None of these Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range MON-TUE-THUR-FRI @ 1:00 PM Function | Domain, CoDomain, Range START YOUR PREPARATION OF CLASS 11 NOW. Mega Offer Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Function | Domain, CoDomain, Range Like Share Subscribe Function | Domain, CoDomain, Range V C E O D M M A E N R T C U E.
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