f is not invertible on (0,1)
f = f⁻¹ on (0,1) and f'(b) = f'(0)
f ≠ f⁻¹ on (0,1) and f'(b) = -f'(0)
f⁻¹ is differentiable on (0,1)
f(x) = (b-x)/(1-bx)
f'(x) = -(1-bx) + b(b-x)(1-bx)² = b² - 1/(1-bx)² < 0 for 0 < b < 1
Therefore, f is always a decreasing function and thus it is invertible
Let f(x) = (b-x)/(1-bx) = y ⇒ x = (b-y)/(1-by)
Therefore, f⁻¹(x) = (b-x)/(1-bx) = f(x)
and f'(b) = 1/b² and f'(0) = b²