The function $\mathrm{f}$ is defined by
$\mathrm{f} : x \mapsto x^2 + 4x - 5, \quad \text{for } x \le k, k \in \mathbb{R}.$

1. Complete the square for $\mathrm{f}(x)$ to determine its turning point.

$\mathrm{f}(x) = x^2 + 4x - 5$
$\mathrm{f}(x) = (x+2)^2 - 4 - 5$
$\mathrm{f}(x) = (x+2)^2 - 9$

The turning point is a minimum at $(-2, -9)$.

2. State the condition for $\mathrm{f}^{-1}$ to exist and deduce the largest value of $k$.

For $\mathrm{f}^{-1}$ to exist, $\mathrm{f}$ must be a one-to-one function.
On the domain $x \le k$, the graph must not contain points on both sides of the turning point.
Hence, the largest exact value of $k$ is $-2$.

3. Make $x$ the subject of the formula to find the rule for $\mathrm{f}^{-1}$.

Let $y = \mathrm{f}(x)$.
$y = (x+2)^2 - 9$
$(x+2)^2 = y + 9$
$x+2 = \pm\sqrt{y+9}$

4. Determine the correct sign and state the domain of $\mathrm{f}^{-1}$.

Since $x \le -2$, we have $x+2 \le 0$. Thus, we must reject the positive root:
$x = -2 - \sqrt{y+9}$

The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$. For $x \le -2$, $\mathrm{f}(x) \ge -9$.
Therefore, the domain of $\mathrm{f}^{-1}$ is $x \ge -9$.

5. Define the inverse function in the required format.

$\mathrm{f}^{-1} : x \mapsto -2 - \sqrt{x+9}, \quad \text{for } x \ge -9$

1. Sketch the primary period for $0 < x \le 4$.

For $0 < x \le 2$, the graph is a downward-facing parabola $y = 4 - x^2$. It approaches $(0, 4)$ and ends at $(2, 0)$.
For $2 < x \le 4$, the graph is a straight line $y = 2x - 4$. It starts from $(2, 0)$ and ends at $(4, 4)$.

2. Extend the graph to the left for $-1 < x \le 0$ using the periodic property.

Given $\mathrm{g}(x) = \mathrm{g}(x + 4)$, the graph has a period of $4$.
For $-1 < x \le 0$, the graph is a translation of the segment $3 < x \le 4$ to the left by $4$ units.
$\mathrm{g}(-1) = \mathrm{g}(-1 + 4) = \mathrm{g}(3) = 2(3) - 4 = 2$.
$\mathrm{g}(0) = \mathrm{g}(4) = 4$.
This forms a straight line segment from $(-1, 2)$ (open circle) to $(0, 4)$ (solid dot).

3. Extend the graph to the right for $4 < x \le 7$ using the periodic property.

For $4 < x \le 6$, the graph is a translation of $0 < x \le 2$ to the right by $4$ units. It is a downward-facing parabola from $(4, 4)$ to $(6, 0)$.
For $6 < x \le 7$, the graph is a translation of $2 < x \le 3$ to the right by $4$ units. It is a straight line segment from $(6, 0)$ to $(7, 2)$ (solid dot).

Key features on the final sketch:

• Open circle at $(-1, 2)$
• Maximum points at $(0, 4)$ and $(4, 4)$
• Minimum points ($x$-axis intersections) at $(2, 0)$ and $(6, 0)$
• Closed solid dot at the endpoint $(7, 2)$

1. Determine the domain of $\mathrm{f}^{-1}$ and the range of $\mathrm{g}$ using the results from parts (i) and (ii).

From part (i), by completing the square, $\mathrm{f}(x) = (x+2)^2 - 9$ for $x \le -2$.
The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$, so $D_{\mathrm{f}^{-1}} = R_{\mathrm{f}} = [-9, \infty)$.
From part (ii), the supremum of $4-x^2$ for $0 < x \le 2$ is $4$ (exclusive) and the minimum is $0$ (inclusive). The range of $2x-4$ for $2 < x \le 4$ is $(0, 4]$ (inclusive of 4). Thus, combining these intervals over a full period gives the overall range of $\mathrm{g}$ as $R_{\mathrm{g}} = [0, 4]$.

2. Verify the condition for the existence of the composite function $\mathrm{f}^{-1}\mathrm{g}$.

For the composite function $\mathrm{f}^{-1}\mathrm{g}$ to exist, the range of $\mathrm{g}$ must be a subset of the domain of $\mathrm{f}^{-1}$.
Since $R_{\mathrm{g}} = [0, 4] \subseteq [-9, \infty) = D_{\mathrm{f}^{-1}}$, the condition is satisfied and the composite function $\mathrm{f}^{-1}\mathrm{g}$ exists.

3. Evaluate $\mathrm{g}(6)$ using the periodic property of $\mathrm{g}$.

Since $\mathrm{g}(x) = \mathrm{g}(x+4)$ for all real $x$, the function repeats every $4$ units.
$\mathrm{g}(6) = \mathrm{g}(6-4) = \mathrm{g}(2)$.
Using the piecewise definition for $0 < x \le 2$:
$\mathrm{g}(2) = 4 - (2)^2 = 0$.

4. Find the expression for $\mathrm{f}^{-1}(x)$ and calculate the exact value of $\mathrm{f}^{-1}\mathrm{g}(6)$.

From part (i), let $y = (x+2)^2 - 9$ with $x \le -2$.
$(x+2)^2 = y + 9 \implies x = -2 - \sqrt{y+9}$ (taking the negative root since $x \le -2$).
So, $\mathrm{f}^{-1}(x) = -2 - \sqrt{x+9}$.
Therefore,
$\mathrm{f}^{-1}\mathrm{g}(6) = \mathrm{f}^{-1}(0)$
$= -2 - \sqrt{0+9}$
$= -2 - 3 = -5$.