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

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