The functions f and g are defined by $f : x \mapsto x^{2} + 2x + 4, \; x \le -1$ and $g : x \mapsto \ln(2x-1), \; x > \tfrac{1}{2}$.

1. Set $y = f(x)$ and complete the square to make $x$ the subject.

$y = x^2 + 2x + 4$
$y = (x^2 + 2x + 1) - 1 + 4$
$y = (x+1)^2 + 3$
$(x+1)^2 = y - 3$
$x + 1 = \pm\sqrt{y-3}$
$x = -1 \pm \sqrt{y-3}$

2. Select the correct branch using the domain of $f$ to write the expression for $f^{-1}(x)$.

Since the domain of $f$ is $x \le -1$, we must have $x + 1 \le 0$.
Therefore, we reject the positive root and take the negative root:
$x = -1 - \sqrt{y-3}$
Replacing $y$ with $x$:
$f^{-1}(x) = -1 - \sqrt{x-3}$

3. Determine the domain of $f^{-1}$ by finding the range of $f$.

The domain of $f^{-1}$ is exactly the range of $f$.
From the completed square form $f(x) = (x+1)^2 + 3$, and given the domain $x \le -1$:
The minimum value occurs at $x = -1$, where $f(-1) = 3$.
Since $(x+1)^2 \ge 0$ for all real $x$, $f(x) \ge 3$.
Thus, the range of $f$ is $[3, \infty)$.
The domain of $f^{-1}$ is $x \ge 3$.

1. Sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$.

• Draw the coordinate axes and the dashed line $y = x$.
• For $y = f(x)$: plot the starting vertex at $(-1, 3)$. Since the domain is $x \le -1$, draw a concave-up curve going towards the top-left (e.g., passing through $(-3, 7)$).
• For $y = f^{-1}(x)$: plot the starting vertex at $(3, -1)$. Since its domain is $x \ge 3$, draw a concave-down curve going towards the bottom-right (e.g., passing through $(7, -3)$).
• Visually verify that the two curves are reflections of each other across the line $y = x$.

2. Deduce the nature of the solution for the equation $f(x) = f^{-1}(x)$ and state the reason.

The equation $f(x) = f^{-1}(x)$ has no solution.

Reason:
The graphs of $y = f(x)$ and $y = f^{-1}(x)$ do not intersect.
(Alternative acceptable reason: The domain of $f$ is $x \le -1$ and the domain of $f^{-1}$ is $x \ge 3$. Since the domains are mutually exclusive, there is no common value of $x$ for which both functions are defined.)

1. Determine the range of $f$ and check the condition for the existence of $gf$.

By completing the square, $f(x) = x^2 + 2x + 4 = (x+1)^2 + 3$.
For the domain $x \le -1$, $(x+1)^2 \ge 0$, so the range of $f$ is $R_f = [3, \infty)$.

The domain of $g$ is $D_g = (\frac{1}{2}, \infty)$.
Since $R_f \subseteq D_g$ (i.e. $[3, \infty) \subseteq (\frac{1}{2}, \infty)$), the composite function $gf$ exists.

2. Find the algebraic expression for $gf(x)$.

$gf(x) = g(f(x))$
$= g(x^2 + 2x + 4)$
$= \ln\left(2(x^2 + 2x + 4) - 1\right)$
$= \ln(2x^2 + 4x + 8 - 1)$
$= \ln(2x^2 + 4x + 7)$

3. Determine the exact range of $gf$.

The range of $gf$ is given by substituting the range of $f$ into $g$.
We evaluate $g(x)$ over the domain corresponding to $R_f = [3, \infty)$.

Since $g(x) = \ln(2x - 1)$ is a strictly increasing function, the minimum value occurs at the lower bound $x = 3$:
$g(3) = \ln(2(3) - 1) = \ln 5$
As $x \to \infty$, $g(x) \to \infty$.
Therefore, the exact range of $gf$ is $[\ln 5, \infty)$.