All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let $y = \mathrm{ln} (2x+1) + 3$

$x = \frac{e^{y-3}-1}{2}$

$\therefore, f^{-1}(x) = \frac{e^{x-3}-1}{2}, x \in \mathbb{R}$

$D_f = \mathbb{R}$

$R_{f^{-1}} = D_f = (-\frac{1}{2}, \infty)$

(ii)

(iii)
$f(x) = f^{-1}(x) = x$

$\Rightarrow f(x) = x$

$\mathrm{ln}(2x+1) + 3 = x$

$\mathrm{ln}(2x+1) = x - 3$

Using Graphing calculator, $x = -0.4847 \mathrm{~or~} 5.482$