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Question Number 206458 by mathlove last updated on 15/Apr/24

$$iff\left(x\right)=\sqrt{x-x^2}then{f}^{-1}\left(x\right)=?$$

Answered by Skabetix last updated on 15/Apr/24

$$f^{-1}\left(x\right)=\frac{1\pm \sqrt{-4x^2+1}}{2}$$

Commented by Tinku Tara last updated on 15/Apr/24

$$\mathrm{If}\mathrm{one}\mathrm{value}\mathrm{gives}\mathrm{two}\mathrm{result}\mathrm{then}f\left(x\right)$$$$\mathrm{is}\mathrm{not}\mathrm{invertible}$$

Commented by mathlove last updated on 15/Apr/24

$$yessir$$

Answered by cortano21 last updated on 15/Apr/24

Answered by Frix last updated on 15/Apr/24

$$f\left(x\right)=\sqrt{x-x^2}$$$$0⩽x⩽1\wedge 0⩽f\left(x\right)⩽\frac{1}{2}$$$$\Rightarrow$$$$$$$$f\left(x\right)=\sqrt{x-x^2};0⩽x<\frac{1}{2}$$$$f^{-1}\left(x\right)=\frac{1}{2}-\frac{\sqrt{1-4x^2}}{2};0⩽x⩽\frac{1}{2}$$$$$$$$f\left(x\right)=\sqrt{x-x^2};\frac{1}{2}⩽x⩽1$$$$f^{-1}\left(x\right)=\frac{1}{2}+\frac{\sqrt{1-4x^2}}{2};0⩽x⩽\frac{1}{2}$$

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