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Question:

Let f:(0,1)→ℝ be defined by f(x) = (b-x)/(1-bx), where b is a constant such that 0 < b < 1, then:

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)

Solution:

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