(i) To find \( f^{-1}(x) \), let \( y = f(x) = e^{3x + 1} \).

Switch variables:

\[ x = e^{3y + 1} \]

Take the natural logarithm:

\[ \ln x = 3y + 1 \]

\[ 3y = \ln x – 1 \]

\[ y = \frac{1}{3} (\ln x – 1) \]

Thus, \( f^{-1}(x) = \frac{1}{3} (\ln x – 1) \).

(ii) Since \( f(x) = e^{3x + 1} > 0 \), the domain of \( f^{-1} \) is the range of \( f \):

\[ x \in \mathbb{R}^+ \text{ or } x > 0 \]

Answer: (i) \( f^{-1}(x) = \frac{1}{3} (\ln x – 1) \), (ii) \( x \in \mathbb{R}^+ \).

The graphs of \( y = g(x) = \ln x \) and \( y = f^{-1}(x) = \frac{1}{3} (\ln x – 1) \) intersect at point \( P \).

Solve \( \ln x = \frac{1}{3} (\ln x – 1) \):

\[ \ln x – \frac{1}{3} \ln x = -\frac{1}{3} \]

\[ \frac{2}{3} \ln x = -\frac{1}{3} \]

\[ \ln x = -\frac{1}{2} \]

\[ x = e^{-\frac{1}{2}} \]

Find the y-coordinate:

\[ y = \ln \left( e^{-\frac{1}{2}} \right) = -\frac{1}{2} \]

Coordinates of \( P \):

\[ \left( e^{-\frac{1}{2}}, -\frac{1}{2} \right) \]

Answer: \( \left( e^{-\frac{1}{2}}, -\frac{1}{2} \right) \).

The graph of \( y = g(x) = \ln x \) intersects the x-axis at \( Q \):

\[ \ln x = 0 \Rightarrow x = 1 \]

Coordinates of \( Q \): \( (1, 0) \).

Find the tangent at \( Q \):

\[ g'(x) = \frac{1}{x} \]

At \( x = 1 \), \( g'(1) = 1 \).

Equation of the tangent at \( (1, 0) \):

\[ y – 0 = 1 (x – 1) \]

\[ y = x – 1 \]

Answer: The equation of the tangent \( T \) is \( y = x – 1 \).

The region \( R \) is bounded by \( y = g(x) = \ln x \), \( y = x – 1 \), and \( x = e \). Since \( \ln x \leq x – 1 \) for \( x \in [1, e] \), compute the area as the difference:

\[ A = \int_1^e (x – 1) \, dx – \int_1^e \ln x \, dx \]

First integral:

\[ \int_1^e (x – 1) \, dx = \left[ \frac{x^2}{2} – x \right]_1^e \]

\[ = \left( \frac{e^2}{2} – e \right) – \left( \frac{1}{2} – 1 \right) = \frac{e^2}{2} – e – \frac{1}{2} + 1 = \frac{e^2 – 2e + 1}{2} \]

Second integral, using integration by parts (\( u = \ln x \), \( dv = dx \)):

\[ du = \frac{1}{x} \, dx, \quad v = x \]

\[ \int \ln x \, dx = x \ln x – \int x \cdot \frac{1}{x} \, dx = x \ln x – x + C \]

\[ \int_1^e \ln x \, dx = \left[ x \ln x – x \right]_1^e \]

\[ = (e \ln e – e) – (1 \ln 1 – 1) = (e – e) – (0 – 1) = 0 + 1 = 1 \]

Area:

\[ A = \left( \frac{e^2 – 2e + 1}{2} \right) – 1 = \frac{e^2 – 2e + 1 – 2}{2} = \frac{e^2 – 2e – 1}{2} \]

Answer: The area of region \( R \) is \( \frac{e^2 – 2e – 1}{2} \).

(i) Show \( g(x) \le x – 1 \), \( x \in \mathbb{R}^+ \).

Method 1:

Consider \( h(x) = x – 1 – \ln x \).

\[ h(1) = 1 – 1 – \ln 1 = 0 \]

\[ h'(x) = 1 – \frac{1}{x} \]

For \( x \ge 1 \), \( h'(x) \ge 0 \), so \( h(x) \ge 0 \) for \( x \ge 1 \).

For \( 0 < x \le 1 \), \( h'(x) \le 0 \), so \( h(x) \ge 0 \) for \( 0 < x \le 1 \).

Thus, \( h(x) \ge 0 \), so \( g(x) = \ln x \le x – 1 \).

Method 2:

\[ g”(x) = -\frac{1}{x^2} < 0 \text{ for } x \in \mathbb{R}^+ \]

Since \( g(x) = \ln x \) is concave down, its graph lies below its tangent at \( x = 1 \), which is \( y = x – 1 \).

Thus, \( g(x) \le x – 1 \).

Method 3:

The graphs of \( y = x – 1 \) and \( y = \ln x \) for \( x > 0 \):

The graph of \( \ln x \) is below its tangent at \( x = 1 \), so \( \ln x \le x – 1 \).

(ii) Replace \( x \) with \( \frac{1}{x} \) in \( \ln x \le x – 1 \):

\[ \ln \left( \frac{1}{x} \right) \le \frac{1}{x} – 1 \]

\[ -\ln x \le \frac{1 – x}{x} \]

\[ \ln x \ge \frac{x – 1}{x} \]

Thus, \( \frac{x – 1}{x} \le g(x) \), \( x \in \mathbb{R}^+ \).

Answer: (i) \( g(x) \le x – 1 \), (ii) \( \frac{x – 1}{x} \le g(x) \).