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(a)
(i)
$f(x) = \frac{1}{1-x^2}, x \in \mathbb{R}, x>1$

$f'(c) = \frac{2x}{1-x^2} > 0$ for all $x \in \mathbb{R}, x>1$

Since $f(x)$ is increasing function, it has no turning points, f is 1-1, its inverse will exist.

(a)
(ii)
Let $f(x) = y$

$x = \pm \sqrt{1- \frac{1}{y}}$ Reject $-\sqrt{1- \frac{1}{y}}$ since $x>1$

$\Rightarrow f^{-1}(x) = \sqrt{1- \frac{1}{x}}$

$D_{f^{-1}} = \{x \in \mathbb{R} | x ~ \textless ~ 0 \}$

(b)
Let $g(x) = y = \frac{2+x}{1-x^2}$

We need $b^2 - 4ac \ge 0$ for all real values of g.

$(1)^2 - 4(y)(2-y) \ge 0$

$1 - 8y + 4y^2 \ge 0$

$y = \frac{8 \pm \sqrt{48}}{8} = 1 \pm \frac{\sqrt{3}}{2}$

$y \ge 1 + \frac{\sqrt{3}}{2} \mathrm{~or~} y \le 1 - \frac{\sqrt{3}}{2}$

$R_g = (-\infty, 1- \frac{\sqrt{3}}{2} ] \cup [1 + \frac{\sqrt{3}}{2}, \infty )$