The function f is defined by $f : x \mapsto e^{(x+c)^2}$, $x \in \mathbb{R}$, $x \le k$, $c>0$.

1. State the condition for the inverse function to exist

For $f^{-1}$ to exist, the function $f$ must be a one-to-one function.

2. Identify the turning point to find the largest valid domain

The function $f(x) = e^{(x+c)^2}$ has a minimum turning point where its exponent $(x+c)^2$ is minimum, which occurs at $x = -c$.
For $x \le -c$, the function is strictly decreasing and hence one-to-one.
For $x > -c$, the function starts to increase, which would make it a many-to-one function.
Therefore, the largest domain of the form $x \le k$ for which $f$ is one-to-one is $x \le -c$.
$k = -c$

1. Deduce the domain of $f$ from part (i).

From part (i), the minimum point of $y = e^{(x+c)^2}$ occurs at $x = -c$.
For $f$ to be one-to-one (so that its inverse exists) on the domain $x \le k$, the maximum value of $k$ is $-c$.
Thus, $k = -c$, and the domain of $f$ is $x \le -c$.

2. Set up the equation to find the inverse.

Let $y = f(x)$.
$y = e^{(x+c)^2}$
Taking the natural logarithm on both sides:
$\ln y = (x+c)^2$

3. Solve for $x$ and select the correct root based on the domain of $f$.

Taking the square root gives:
$x + c = \pm \sqrt{\ln y}$
Since the domain is $x \le -c$, we have $x + c \le 0$. Thus, we must reject the positive root:
$x + c = -\sqrt{\ln y}$
$x = -c - \sqrt{\ln y}$

4. State the inverse function.

Replacing $y$ with $x$ to express the inverse function in terms of $x$:
$f^{-1}(x) = -c - \sqrt{\ln x}$

5. Determine the domain of $f^{-1}$.

The domain of $f^{-1}$ is the range of $f$.
For the domain $x \le -c$, we have $(x+c)^2 \ge 0$, which implies:
$f(x) = e^{(x+c)^2} \ge e^0 = 1$
The range of $f$ is $[1, \infty)$.
Therefore, the domain of $f^{-1}$ is $x \ge 1$.

1. Identify the domain, range, and endpoints of $f$ and $f^{-1}$ to prepare for sketching.

From parts (i) and (ii), we use $k = -c$.
For $f(x) = e^{(x+c)^2}$, the domain is $x \le -c$. Since $c > 0$, $-c$ is negative.
At the boundary $x = -c$, $f(-c) = e^0 = 1$. As $x \to -\infty$, $f(x) \to \infty$.
Thus, the graph of $f$ is a curve in the second quadrant ending at $(-c, 1)$.

The domain of $f^{-1}$ is the range of $f$, which is $x \ge 1$.
The range of $f^{-1}$ is the domain of $f$, which is $y \le -c$.
Thus, the graph of $f^{-1}$ starts at $(1, -c)$ and goes towards $-\infty$ as $x \to \infty$.

2. Sketch the graphs of $f$ and $f^{-1}$ on the same diagram.

To sketch:

1. Draw the $x$ and $y$ axes and add a dotted line for $y = x$.
2. Mark the endpoint $(-c, 1)$ in the second quadrant and sketch $y=f(x)$ curving upwards and to the left.
3. Mark the endpoint $(1, -c)$ in the fourth quadrant and sketch $y=f^{-1}(x)$ curving downwards and to the right.
4. Ensure both curves appear as mirror images of each other across the line $y=x$.

3. State the relationship between the two graphs.

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$.

1. Determine $k$ from part (i) to establish the domain and evaluate the range of $f$.

For $f$ to have an inverse, it must be a one-to-one function. The curve $y = e^{(x+c)^2}$ is symmetrical about $x = -c$, so the domain must be restricted to $x \le -c$, giving $k = -c$.
For $x \le -c$, we have $x+c \le 0 \implies (x+c)^2 \ge 0$.
Therefore, $f(x) = e^{(x+c)^2} \ge e^0 = 1$.
The range of $f$ is $R_f = [1, \infty)$.

2. State the reason for the existence of the composite function $gf$.

The domain of $g$ is $D_g = (0, \infty)$.
Since $R_f \subseteq D_g$ (i.e. $[1, \infty) \subseteq (0, \infty)$), the composite function $gf$ exists.

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

$gf(x) = g(f(x)) = g(e^{(x+c)^2}) = \ln(e^{(x+c)^2}) = (x+c)^2$