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Onto FunctionA function f: A > B is called an onto function if the range of f is B. In other words, if each b ∈ B there exists at least one a ∈ A such that.f(a) = b, then f is an onto function. An onto function is also called surjective function. Let A = {a _{1} , a _{2} , a _{3} } and B = {b _{1} , b _{ 2} } then f : A > B. Examples on onto functionExample 1: Let A = {1, 2, 3}, B = {4, 5} and let f = {(1, 4), (2, 5), (3, 5)}. Show that f is an surjective function from A into B.Solution: Domain = {1, 2, 3} = A Range = {4, 5} The element from A, 2 and 3 has same range 5. So f : A > B is an onto function. Example 2: State whether the given function is onto or not. f : R > R defined by f(x) = 1 + x ^{2} . Solution: f(x) = 1 + x ^{2 } Let x = 1 f(1) = 1 + 1 ^{ 2} f(1) = 1 + 1 f(1) = 2 (equation 1) Now, let x = 1 f(1) = 1+ (1) ^{2} = 1 + 1 f(1) = 2 (equation 2) but 1 ≠ 1 ∴ f is not one to one function. If f is onto then O ∈ R f(x) = 0 there4; 1 + x ^{2} = 0 there4; x ^{2} = 1 x = ~+mn~ √1 but x ∈ R ∴ f is not onto function Example 3: Determine which of the following functions f : R > R are onto i. f(x) = x + 1 Solution: f(x) = x + 1 x = 1 ; f(x) = 1 + 1 f(1) = 2 x = 1 ; f(1) = 1 + 1 f(1) = 0 x = 0; f(x) = x + 1 f(0) = 0 + 1 f(0) = 1 From the above we can see that domain = {1, 0, 1} and range = {2, 0, 1} So, the given function is onto. ii) f(x) = x + x Solution: f(x) = x + x For x = 0 ; f(0) = 0 For x = 1; f(1) = 1 + 1 = 1 + 1 f(1) = 2 for x = 1; f(1) = 1 + 1 = 1 + 1 f(1) = 2 Domain = {0, 1, 1.} Range = {0, 2, 2. } So, the given function is not onto. From onto function to Home page
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