what is the image of 4 in the function f(x) = 2x+3 ?
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Answer:
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Step-by-step explanation:
A = {–2, –1, 0, 1, 2} f : A → Z such that f(x) = x2 – 2x – 3 (i) A is the domain of the function f. Thus, range is the set of elements f(x) for all x ∈ A. By substituting x = –2 in f(x), we get f(–2) = (–2)2 – 2(–2) – 3 = 4 + 4 – 3 = 5 By substituting x = –1 in f(x), we get f(–1) = (–1)2 – 2(–1) – 3 = 1 + 2 – 3 = 0 By substituting x = 0 in f(x), we get f(0) = (0)2 – 2(0) – 3 = 0 – 0 – 3 = – 3 By substituting x = 1 in f(x), we get f(1) = 12 – 2(1) – 3 = 1 – 2 – 3 = – 4 By substituting x = 2 in f(x), we get f(2) = 22 – 2(2) – 3 = 4 – 4 – 3 = –3 Hence, the range of f is {-4, -3, 0, 5}. (ii) pre-images of 6, –3 and 5 Suppose x be the pre-image of 6 ⇒ f(x) = 6 x2 – 2x – 3 = 6 x2 – 2x – 9 = 0 x = [-(-2) ± √ ((-2)2 – 4(1) (-9))] / 2(1) = [2 ± √ (4+36)] / 2 = [2 ± √40] / 2 = 1 ± √10 However, 1 ± √10 ∉ A Hence, there exists no pre-image of 6. Then, suppose x be the pre-image of –3 ⇒ f(x) = –3 x2 – 2x – 3 = –3 x2 – 2x = 0 x(x – 2) = 0 x = 0 or 2 It is clear, both 0 and 2 are elements of A. Hence, 0 and 2 are the pre-images of –3. Then, suppose x be the pre-image of 5 ⇒ f(x) = 5 x2 – 2x – 3 = 5 x2 – 2x – 8= 0 x2 – 4x + 2x – 8= 0 x(x – 4) + 2(x – 4) = 0 (x + 2)(x – 4) = 0 x = –2 or 4 However, 4 ∉ A but –2 ∈ A Hence, –2 is the pre-images of 5. Thus, Ø, {0, 2}, -2 are the pre-images of 6, -3, 5Read more on Sarthaks.com - https://www.sarthaks.com/651195/let-a-1-2-and-f-a-z-be-a-function-defined-by-f-x-x-2-2x-3-find-i-range-of-f-i-e-f-a-ii-pre-images-of-6-3-and
Answer:
Step-by-step explanation:
Given as A = {–2, –1, 0, 1, 2} f : A → Z such that f(x) = x2 – 2x – 3 (i) A is the domain of the function f. Thus, range is the set of elements f(x) for all x ∈ A. By substituting x = –2 in f(x), we get f(–2) = (–2)2 – 2(–2) – 3 = 4 + 4 – 3 = 5 By substituting x = –1 in f(x), we get f(–1) = (–1)2 – 2(–1) – 3 = 1 + 2 – 3 = 0 By substituting x = 0 in f(x), we get f(0) = (0)2 – 2(0) – 3 = 0 – 0 – 3 = – 3 By substituting x = 1 in f(x), we get f(1) = 12 – 2(1) – 3 = 1 – 2 – 3 = – 4 By substituting x = 2 in f(x), we get f(2) = 22 – 2(2) – 3 = 4 – 4 – 3 = –3 Hence, the range of f is {-4, -3, 0, 5}. (ii) pre-images of 6, –3 and 5 Suppose x be the pre-image of 6 ⇒ f(x) = 6 x2 – 2x – 3 = 6 x2 – 2x – 9 = 0 x = [-(-2) ± √ ((-2)2 – 4(1) (-9))] / 2(1) = [2 ± √ (4+36)] / 2 = [2 ± √40] / 2 = 1 ± √10 However, 1 ± √10 ∉ A Hence, there exists no pre-image of 6. Then, suppose x be the pre-image of –3 ⇒ f(x) = –3 x2 – 2x – 3 = –3 x2 – 2x = 0 x(x – 2) = 0 x = 0 or 2 It is clear, both 0 and 2 are elements of A. Hence, 0 and 2 are the pre-images of –3. Then, suppose x be the pre-image of 5 ⇒ f(x) = 5 x2 – 2x – 3 = 5 x2 – 2x – 8= 0 x2 – 4x + 2x – 8= 0 x(x – 4) + 2(x – 4) = 0 (x + 2)(x – 4) = 0 x = –2 or 4 However, 4 ∉ A but –2 ∈ A Hence, –2 is the pre-images of 5. Thus, Ø, {0, 2}, -2 are the pre-images of 6, -3, 5Read more on Sarthaks.com - https://www.sarthaks.com/651195/let-a-1-2-and-f-a-z-be-a-function-defined-by-f-x-x-2-2x-3-find-i-range-of-f-i-e-f-a-ii-pre-images-of-6-3-and