State the coordinates of the minimum point of the graph of the function $h$ defined in $\mathbb { R }$ with $h ( x ) = - g ( x - 3 )$.

Subtask Part A 4b $( 3 \mathrm { marks } )$\\
The graph of an antiderivative of $g$ passes through $P$. Sketch this graph in Figure 2.

Given is the function $f : x \mapsto 2 e ^ { - \frac { 1 } { 8 } x ^ { 2 } }$ defined in $\mathbb { R }$. Figure 3 shows the graph $G _ { f }$ of $f$, which has the x-axis as a horizontal asymptote.

\begin{figure}[h]
\begin{center}
  \textit{[Figure]}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

\textbf{(1a)} [2 marks] Calculate the coordinates of the intersection point of $G _ { f }$ with the y-axis and prove by calculation that $G _ { f }$ is symmetric with respect to the y-axis.

\textbf{(1b)} [5 marks] The point $W \left( - 2 \left\lvert \, 2 e ^ { - \frac { 1 } { 2 } } \right. \right)$ is one of the two inflection points of $G _ { f }$. The tangent to $G _ { f }$ at point $W$ is denoted by $w$. Determine an equation of $w$ and calculate the point where $w$ intersects the x-axis.\\
(for verification: $f ^ { \prime } ( x ) = - \frac { 1 } { 2 } x \cdot e ^ { - \frac { 1 } { 8 } x ^ { 2 } }$ )

For each value $c \in \mathbb { R } ^ { + }$, consider the rectangle with vertices $P ( - c \mid 0 ) , Q ( c \mid 0 )$, $R ( c \mid f ( c ) )$ and $S$.

\textbf{(1c)} [1 marks] Draw the rectangle PQRS in Figure 3 for $c = 2$.

\textbf{(1d)} [3 marks] Calculate the value of $c$ for which $\overline { \mathrm { QR } } = 1$ holds.

\textbf{(1e)} [3 marks] State the side lengths of rectangle PQRS as a function of $c$ and justify that the area of the rectangle is given by the term $A ( c ) = 4 c \cdot e ^ { - \frac { 1 } { 8 } c ^ { 2 } }$.

\textbf{(1f)} [4 marks] There is a value of $c$ for which the area $A ( c )$ of rectangle PQRS is maximal. Calculate this value of $c$.

For $k \in \mathbb { R }$, consider the functions $f _ { k } : x \mapsto f ( x ) + k$ defined in $] - \infty ; 0 ]$. Thus $f _ { 0 } ( x ) = f ( x )$, where $f _ { 0 }$ and $f$ differ in their domain.

\textbf{(1g)} [4 marks] Justify using the first derivative of $f _ { k }$ that $f _ { k }$ is invertible for every value of $k$. Sketch the graph of the inverse function of $f _ { 0 }$ in Figure 3.

\textbf{(1h)} [2 marks] State all values of $k$ for which the graph of $f _ { k }$ and the graph of the inverse function of $f _ { k }$ have no common point.

\begin{figure}[h]
\begin{center}
  \textit{[Figure]}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

Figure 4 shows a house with a roof dormer, whose front is shown schematically in Figure 5. The front is described by a model as the region enclosed by the graph $G _ { f }$ of the function $f$ from Part B Subtask 1, the x-axis, and the lines with equations $x = - 4$ and $x = 4$. Here, one unit of length in the coordinate system corresponds to one meter in reality.

\begin{figure}[h]
\begin{center}
  \textit{[Figure]}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}

\textbf{(2a)} [2 marks] State the width and height of the front of the roof dormer.

In the front of the roof dormer there is a window. In the model, the window corresponds to the region enclosed by the graph of the function $g$ with $g ( x ) = a x ^ { 4 } + b$ and suitable values $a , b \in \mathbb { R }$ with the x-axis (see Figure 5).

\textbf{(2b)} [2 marks] Justify that $a$ is negative and $b$ is positive.

To determine the area of the front of the roof dormer, an antiderivative $F$ of $f$ is considered.

\textbf{(2c)} [2 marks] One of the graphs I, II and III is the graph of $F$. Justify that this is Graph I by giving one reason each for why Graph II and Graph III do not apply.\\
\textit{[Figure]}\\
\textit{[Figure]}\\
\textit{[Figure]}

\textbf{(2d)} [5 marks] Now determine the area of the entire front of the roof dormer (including the window) using the graph of $F$ from Part B Subtask 2c.\\
Describe, incorporating this area, the essential steps of a solution method by which the value of $a$ could be calculated so that with a window height of 1.50 m, the part of the front of the roof dormer shown shaded in Figure 5 has an area of $6 \mathrm {~m} ^ { 2 }$.

\textbf{(2e)} [5 marks] In order to calculate an approximate value for the length of the upper profile line of the front of the roof dormer, $G _ { f }$ in the range $- 4 \leq x \leq 4$ is approximated by four circular arcs that transition seamlessly into one another and are congruent to each other. One of these circular arcs extends in the range $0 \leq x \leq 2$ and is part of the circle with center $M ( 0 \mid - 1 )$ and radius 3. Calculate the central angle of the circular sector corresponding to this circular arc and use it to determine the desired approximate value.