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

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)


Graph of 3ii
Graph of 3ii

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

KS Comments:

Students should be careful to draw the curves according to the given domain. And since the question requests for equations of asymptotes, we should have label them too.

The last part, students can also use the inverse function to solve.

This question, in general did not pose much of a problem for students.

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