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Let $$f : (0, 1) \rightarrow R$$ be defined by$$f(x) = \frac{b - x}{1 - bx}$$, where b is a constant such that 0 < b < 1. Then
f is not invertible on (0, 1)
$$f \neq f^{-1}$$ on (0, 1) and $$f'(b) = \frac{1}{f'(0)}$$
$$f = f^{-1}$$ on (0, 1) and $$f'(b) = \frac{1}{f'(0)}$$
$$f^{-1}$$ is differentiable on (0, 1)
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