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 $y = \mathrm{f}(x)$ and complete the square to prepare for making $x$ the subject.

Let $y = \mathrm{f}(x)$.
$y = x^2 + 2x + 4$
$y = (x^2 + 2x + 1) - 1 + 4$
$y = (x + 1)^2 + 3$

2. Rearrange to make $x$ the subject, carefully selecting the correct square root sign based on the domain of $\mathrm{f}$.

$(x + 1)^2 = y - 3$
$x + 1 = \pm\sqrt{y - 3}$
Since the domain of $\mathrm{f}$ is given as $x \le -1$, we have $x + 1 \le 0$. Therefore, we must reject the positive root and choose the negative root:
$x + 1 = -\sqrt{y - 3}$
$x = -1 - \sqrt{y - 3}$
Replacing $y$ with $x$ gives the inverse function:
$\mathrm{f}^{-1}(x) = -1 - \sqrt{x - 3}$

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

The domain of $\mathrm{f}^{-1}$ is precisely the range of $\mathrm{f}$.
From the completed square form $\mathrm{f}(x) = (x + 1)^2 + 3$ and knowing $x \le -1$:
$(x + 1)^2 \ge 0$
$(x + 1)^2 + 3 \ge 3$
So, the range of $\mathrm{f}$ is $[3, \infty)$.
Therefore, the domain of $\mathrm{f}^{-1}$ is $x \ge 3$.

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

• Draw a set of $x$ and $y$ axes.
• For $y=f(x)$, plot the starting endpoint at $(-1, 3)$. Since it is a quadratic curve for $x \le -1$, draw a smooth half-parabola opening upwards and curving to the left.
• For $y=f^{-1}(x)$, plot the starting endpoint at $(3, -1)$. Draw a smooth half-parabola opening downwards and curving to the right.
• (Optional but recommended) Draw the dashed line $y=x$ to indicate the line of reflection symmetry between the two curves.

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

From the sketch, the curve $y=f(x)$ lies entirely above the line $y=x$ (in the region $x \le -1, y \ge 3$), and the curve $y=f^{-1}(x)$ lies entirely below the line $y=x$ (in the region $x \ge 3, y \le -1$).

Since the two graphs are completely separated and do not intersect, they share no common points.
Therefore, the equation $f(x) = f^{-1}(x)$ has no real solution.

1. Find the range of $f(x)$.

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

2. Check the condition for the existence of $gf$.

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

3. Find the expression for $gf(x)$.

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

4. Find the exact range of $gf$.

The range of $gf$ is the set of values of $g(y)$ for $y \in R_f$.
Since $R_f = [3, \infty)$ and $g(x) = \ln(2x - 1)$ is a strictly increasing function,
the minimum value of $gf(x)$ occurs when $f(x) = 3$.
Minimum value $= \ln(2(3) - 1) = \ln 5$.
As $f(x) \to \infty$, $gf(x) \to \infty$.
Therefore, the exact range of $gf$ is $[\ln 5, \infty)$.