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11 June, 04:16

Prove the function f: R - {1} to R - {1} defined by f (x) = ((x+1) / (x-1)) ^3 is bijective.

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  1. 11 June, 05:23
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    See explaination

    Step-by-step explanation:

    given f:R-/left / { 1 / right / }/rightarrow R-/left / { 1 / right / } defined by f (x) = / left (/frac{x+1}{x-1} / right) ^{3}

    let f (x) = f (y)

    /left (/frac{x+1}{x-1} / right) ^{3}=/left (/frac{y+1}{y-1} / right) ^{3}

    taking cube roots on both sides, we get

    /frac{x+1}{x-1} = / frac{y+1}{y-1}

    /Rightarrow (x+1) (y-1) = (x-1) (y+1)

    /Rightarrow xy-x+y-1=xy+x-y-1

    /Rightarrow - x+y=x-y

    /Rightarrow x+x=y+y

    /Rightarrow 2x=2y

    /Rightarrow x=y

    Hence f is one - one

    let y/in R, such that f (x) = / left (/frac{x+1}{x-1} / right) ^{3}=y

    /Rightarrow / frac{x+1}{x-1} = / sqrt[3]{y}

    /Rightarrow x+1=/sqrt[3]{y}/left (x-1 / right)

    /Rightarrow x+1=/sqrt[3]{y} x - / sqrt[3]{y}

    /Rightarrow / sqrt[3]{y} x-x=1 + / sqrt[3]{y}

    /Rightarrow x/left (/sqrt[3]{y} - 1 / right) = 1 + / sqrt[3]{y}

    /Rightarrow x=/frac{/sqrt[3]{y}+1}{/sqrt[3]{y}-1}

    for every y/in R-/left / { 1 / right / }/exists x/in R-/left / { 1 / right / } such that x=/frac{/sqrt[3]{y}+1}{/sqrt[3]{y}-1}

    Hence f is onto

    since f is both one - one and onto so it is a bijective
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