$y = |\mathrm{f}(x)|$ 的图像在 $(0,2)$ 处有一个转向点，并经过点 $(-4,0)$ 和 $(4,0)$。$y = \mathrm{f}'(x)$ 的图像在原点处有一个拐点。 在不同的图示中，分别画出下列函数的图像草图，并指出任何渐近线的方程、任何驻点的坐标以及与坐标轴交点的坐标。

1. Identify the geometrical transformations mapping the original graph to the new graph.

To sketch $y = \mathrm{f}'(-2x)$ from $y = \mathrm{f}'(x)$, we apply two transformations:

1. A horizontal scaling by a scale factor of $\frac{1}{2}$ parallel to the $x$-axis.
2. A reflection in the $y$-axis.

2. Determine the transformed asymptotes and axis intersections.

Under the transformation $x \mapsto -\frac{x}{2}$:

• The vertical asymptotes $x = -2$ and $x = 2$ map to $x = 1$ and $x = -1$.
• The horizontal asymptote $y = 0$ (the $x$-axis) remains unchanged at $y = 0$.
• The origin $(0,0)$, which is the only point of intersection and a point of inflexion, remains at $(0,0)$.

3. Determine the shape and position of the new curve branches.

Due to the reflection in the $y$-axis and scaling:

• The branch initially in $x < -2$ ($y > 0$) moves to $x > 1$ ($y > 0$).
• The branch initially in $x > 2$ ($y < 0$) moves to $x < -1$ ($y < 0$).
• The central branch ($-2 < x < 2$) passing through $(0,0)$ with a negative gradient reflects horizontally to become an increasing curve on $-1 < x < 1$, passing through $(0,0)$ with a positive gradient.

1. Determine the coordinates and nature of the stationary point for the middle branch.

From the graph of $y = \mathrm{f}'(x)$, we have $\mathrm{f}'(x) > 0$ for $-2 < x < 0$ and $\mathrm{f}'(x) < 0$ for $0 < x < 2$, with $\mathrm{f}'(0) = 0$.
This indicates that $\mathrm{f}(x)$ is increasing then decreasing, giving a local maximum at $x = 0$.
Since the graph of $y = |\mathrm{f}(x)|$ has a minimum at $(0, 2)$ and goes to $+\infty$ at $x = \pm 2$, $\mathrm{f}(x)$ must be negative in this interval to reflect into a minimum on the modulus graph.
Thus, the turning point is a local maximum at $(0, -2)$, and the curve approaches $-\infty$ as $x \to -2^+$ and $x \to 2^-$.

2. Determine the properties of the outer branches.

For $x < -2$, $\mathrm{f}'(x) > 0$, so $\mathrm{f}(x)$ is strictly increasing. Since it passes through the $x$-axis at $(-4, 0)$ and goes towards $+\infty$, it must be negative for $x < -4$. As $y = |\mathrm{f}(x)|$ has a horizontal asymptote $y = 1$, $\mathrm{f}(x)$ must approach a negative value, giving the horizontal asymptote $y = -1$ as $x \to -\infty$.
Similarly, for $x > 2$, $\mathrm{f}'(x) < 0$, so $\mathrm{f}(x)$ is strictly decreasing. It passes through $(4, 0)$, coming from $+\infty$, so it is negative for $x > 4$. Thus, it also approaches the horizontal asymptote $y = -1$ as $x \to \infty$.

3. Summarise the features for the final sketch.

The sketch of $y = \mathrm{f}(x)$ requires:

• Vertical asymptotes at $x = -2$ and $x = 2$.
• A horizontal asymptote at $y = -1$.
• $x$-intercepts at $(-4, 0)$ and $(4, 0)$.
• A $y$-intercept and local maximum at $(0, -2)$.