Let f:(0,1) rightarrow R be defined by f(x)=f | vedclass

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(C) Given $f(x) = \frac{b-x}{1-bx}$.To find the inverse,let $y = f(x) = \frac{b-x}{1-bx}$.$y(1-bx) = b-x \Rightarrow y - bxy = b - x \Rightarrow x(1-b

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$f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$

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(C) Given $f(x) = \frac{b-x}{1-bx}$.To find the inverse,let $y = f(x) = \frac{b-x}{1-bx}$.$y(1-bx) = b-x \Rightarrow y - bxy = b - x \Rightarrow x(1-by) = b-y \Rightarrow x = \frac{b-y}{1-by}$.Thus,$f^{-1}(y) = \frac{b-y}{1-by}$,which implies $f(x) = f^{-1}(x)$,so $f = f^{-1}$.Now,$f^{\prime}(x) = \frac{(1-bx)(-1) - (b-x)(-b)}{(1-bx)^2} = \frac{-1+bx+b^2-bx}{(1-bx)^2} = \frac{b^2-1}{(1-bx)^2}$.$f^{\prime}(0) = \frac{b^2-1}{(1-0)^2} = b^2-1$.$f^{\prime}(b) = \frac{b^2-1}{(1-b^2)^2} = \frac{b^2-1}{(b^2-1)^2} = \frac{1}{b^2-1}$.Therefore,$f^{\prime}(b) = \frac{1}{f^{\prime}(0)}$.

Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then

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