# Ex 1.3 , 6

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex1.3 , 6 Show that f: [−1, 1] → R, given by f(x) = 𝑥𝑥 + 2 is one-one. Find the inverse of the function f: [−1, 1] → Range f. (Hint: For y ∈ Range f, y = f(x) = 𝑥𝑥 + 2 , for some x in [−1, 1], i.e., x = 2𝑦1 − 𝑦 ) f(x) = xx+2 Check one-one f(x1) = 𝑥1𝑥1 + 2 f(x2) = 𝑥2𝑥2 + 2 Putting f(x1) = f(x2) 𝑥1𝑥1 + 2 = 𝑥2𝑥2 + 2 x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 – x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 ⇒ x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Finding inverse f(x) = 𝑥𝑥 + 2 Putting f(x) = y y = 𝑥𝑥 + 2 y(x + 2) = x yx + 2y = x yx – x = –2y x(y – 1) = –2y x = −2𝑦 𝑦 −1 x = −2𝑦 −1−𝑦 + 1 x = 2𝑦 1 − 𝑦 Let g(y) = 2𝑦 1 − 𝑦 where g: R → [−1, 1] Thus, g is inverse of f Inverse of f = g(y) = 𝟐𝒚 𝟏 − 𝒚

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