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If f: Rvec(-1,1) is defined by f(x)=-(x|...

If `f: Rvec(-1,1)` is defined by `f(x)=-(x|x|)/(1+x^2),t h e nf^(-1)(x)` equals `sqrt((|x|)/(1-|x|))` (b) `-sgn(x)sqrt((|x|)/(1-|x|))` `-sqrt(x/(1-x))` (d) none of these

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Clearly, `f:R→(−1,1)` given by `f(x)=(−x|x|)/(1+x^2) `is a bijection.
Now, `fof^(−1)(x)=x `
`⇒f(f^(−1(x)))=x `
`⇒(f^(−1(x))∣f^(−1)(x)∣)/(1+(f^(−1)(x)))^2=x ` ...
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