Question

Let $f:(0,1)\to R$ be defined by$f\left(x\right)=\frac{b-x}{1-bx}$, where $b$ is a constant such that $0<b<1$, then:

• A. $f$ is not invertible on $(0,1)$
• B. $f\ne f^{-1}$ on $(0,1)$ and $f^{\prime }\left(b\right)=-\frac{1}{f^{\prime }\left(0\right)}$
• C. $f=f^{-1}$ on $(0,1)$ and $f^{\prime }\left(b\right)=\frac{1}{f^{\prime }\left(0\right)}$
• D. $f^{-1}$ is differentiable on $(0,1)$

Answer 1

Answer: (C), (D)

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

$f\left(x\right)=\frac{b-x}{1-bx}$

$f^{\prime }\left(x\right)=\frac{-(1-bx)+b(b-x)}{{(1-bx)}^2}=\frac{b^2-1}{{(1-bx)}^2}<0$ for $0<b<1$

Therefore, $f$ is always decreasing function and thus it is invertible

let $f\left(x\right)=\frac{b-x}{1-bx}=y$

$\Rightarrow x=\frac{b-y}{1-by}$

Therefore, $f^{-1}\left(x\right)=\frac{b-x}{1-bx}=f\left(x\right)$

and $f^{\prime }\left(b\right)=\frac{1}{b^2-1}$ and $f^{\prime }\left(0\right)=b^2-1$