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Question Number 13595 by Tinkutara last updated on 21/May/17

If f : R → (−1, 1) is defined by f(x) = ((−x∣x∣)/(1 + x^2 )) , then prove that f^(−1) (x) = −sgn(x)(√((∣x∣)/(1 − ∣x∣)))

$$\mathrm{If}\:{f}\::\:\mathbb{R}\:\rightarrow\:\left(−\mathrm{1},\:\mathrm{1}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$${f}\left({x}\right)\:=\:\frac{−{x}\mid{x}\mid}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:,\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:−\mathrm{sgn}\left({x}\right)\sqrt{\frac{\mid{x}\mid}{\mathrm{1}\:−\:\mid{x}\mid}} \\ $$

Answered by mrW1 last updated on 21/May/17

f(x)=y = ((−x∣x∣)/(1 + x^2 )) if y≥0⇒x≤0⇒∣x∣=−x ..(i) if y≤0⇒x≥0⇒∣x∣=x ..(ii) (i): y = (x^2 /(1 + x^2 )) (1−y)x^2 =y x=−(√(y/(1−y))) (for y≥0, x≤0) (ii): y = −(x^2 /(1 + x^2 )) (1+y)x^2 =−y x=(√((−y)/(1+y))) (for y≤0, x≥0) for both (i) and (ii): x=−sgn(y)(√((∣y∣)/(1−∣y∣))) ⇒f^(−1) (x)=−sgn(x)(√((∣x∣)/(1−∣x∣)))

$${f}\left({x}\right)={y}\:=\:\frac{−{x}\mid{x}\mid}{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$${if}\:{y}\geqslant\mathrm{0}\Rightarrow{x}\leqslant\mathrm{0}\Rightarrow\mid{x}\mid=−{x}\:\:\:\:..\left({i}\right) \\ $$$${if}\:{y}\leqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{0}\Rightarrow\mid{x}\mid={x}\:\:\:\:..\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right): \\ $$$${y}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}−{y}\right){x}^{\mathrm{2}} ={y} \\ $$$${x}=−\sqrt{\frac{{y}}{\mathrm{1}−{y}}}\:\:\:\:\:\left({for}\:{y}\geqslant\mathrm{0},\:{x}\leqslant\mathrm{0}\right) \\ $$$$\left({ii}\right): \\ $$$${y}\:=\:−\frac{{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+{y}\right){x}^{\mathrm{2}} =−{y} \\ $$$${x}=\sqrt{\frac{−{y}}{\mathrm{1}+{y}}}\:\:\:\:\:\:\left({for}\:{y}\leqslant\mathrm{0},\:{x}\geqslant\mathrm{0}\right) \\ $$$$ \\ $$$${for}\:{both}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$${x}=−{sgn}\left({y}\right)\sqrt{\frac{\mid{y}\mid}{\mathrm{1}−\mid{y}\mid}}\: \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=−{sgn}\left({x}\right)\sqrt{\frac{\mid{x}\mid}{\mathrm{1}−\mid{x}\mid}}\: \\ $$

Commented by Tinkutara last updated on 21/May/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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