下图显示了 $y = \text{f}(x)$ 的图像。该曲线与 $y$ 轴相交于点 $\left(0, -\frac{11}{2}\right)$，其极大值点为 $(a,b)$，极小值点为 $(c,d)$。渐近线方程为 $y = 3$、$x = -1$ 和 $x = 2$。 在不同的图上，分别画出下列函数的图像，并在必要时标出对应的驻点、坐标轴交点以及渐近线。

1. Identify the geometrical transformation for $y = \mathrm{f}(|x|)$.

The graph of $y = \mathrm{f}(|x|)$ is obtained by retaining the part of the curve $y = \mathrm{f}(x)$ for $x \ge 0$, and discarding the part for $x < 0$. The retained portion is then reflected across the $y$-axis to produce the graph for $x < 0$.

2. Determine the new features (asymptotes, intercepts, stationary points) and sketch the graph.

• Asymptotes: The portion for $x \ge 0$ has a vertical asymptote $x = 2$ and a horizontal asymptote $y = 3$. Reflecting $x = 2$ across the $y$-axis gives an additional vertical asymptote $x = -2$. The horizontal asymptote $y = 3$ extends to both $x \to \infty$ and $x \to -\infty$.
• Stationary points: The existing stationary points for $x > 0$ are the maximum $(a, b)$ and minimum $(c, d)$. Reflecting them across the $y$-axis gives additional stationary points at $(-a, b)$ and $(-c, d)$.
• Axial intercepts: The $y$-intercept is at $(0, -\frac{11}{2})$. Because it lies on the axis of reflection, it remains unchanged and acts as a sharp minimum (kink) connecting the two symmetric halves. There are no $x$-intercepts.

The resulting sketch is drawn with bilateral symmetry across the $y$-axis.

1. Determine the new asymptotes and $x$-intercepts for $y = \frac{1}{\text{f}(x)}$.

As $x \to \pm\infty$, $\text{f}(x) \to 3$, so $y = \frac{1}{\text{f}(x)} \to \frac{1}{3}$. The horizontal asymptote is $y = \frac{1}{3}$.
As $x \to -1$ and $x \to 2$, $\text{f}(x) \to \pm\infty$, so $\frac{1}{\text{f}(x)} \to 0$. This gives $x$-intercepts at $(-1, 0)$ and $(2, 0)$.

2. Determine the $y$-intercept and map the stationary points.

The $y$-intercept maps from $(0, -\frac{11}{2})$ to $(0, -\frac{2}{11})$.
For stationary points, the nature is inverted. The maximum point $(a, b)$ where $b < 0$ maps to a local minimum point $(a, \frac{1}{b})$.
The minimum point $(c, d)$ where $d > 0$ maps to a local maximum point $(c, \frac{1}{d})$.

3. Determine the overall shape and sketch the graph.

For $x < -1$: The curve approaches $y = \frac{1}{3}$ from above as $x \to -\infty$ and decreases smoothly to $(-1, 0)$.
For $-1 < x < 2$: The curve starts at $(-1, 0)$, drops through $(0, -\frac{2}{11})$, reaches a minimum at $(a, \frac{1}{b})$, and then rises to $(2, 0)$.
For $x > 2$: The curve rises from $(2, 0)$ to a maximum at $(c, \frac{1}{d})$. Since $\text{f}(x)$ approaches $y = 3$ from below as $x \to \infty$, we know $d < 3$, meaning $\frac{1}{d} > \frac{1}{3}$. The curve then decreases from this maximum, approaching $y = \frac{1}{3}$ from above.

1. Determine the asymptotes of the derivative graph

As $x \to \pm\infty$, the curve $y = \text{f}(x)$ approaches the horizontal line $y = 3$. This means the gradient approaches $0$. Hence, the horizontal asymptote for $y = \text{f}'(x)$ is $y = 0$.

At $x = -1$ and $x = 2$, $\text{f}(x)$ has vertical asymptotes and the curve becomes infinitely steep. Therefore, $x = -1$ and $x = 2$ remain vertical asymptotes for $y = \text{f}'(x)$.

2. Determine the $x$-intercepts corresponding to the stationary points

The function $\text{f}(x)$ has a maximum point at $x = a$ and a minimum point at $x = c$. At these stationary points, the gradient is zero.

Thus, $\text{f}'(a) = 0$ and $\text{f}'(c) = 0$. The $x$-intercepts of $y = \text{f}'(x)$ are $(a, 0)$ and $(c, 0)$.

3. Analyze the sign of the gradient in each region to sketch the shape

• For $x < -1$: $\text{f}(x)$ increases from $y=3$ to $\infty$. The gradient is positive ($\text{f}'(x) > 0$), approaching $0$ as $x \to -\infty$ and $+\infty$ as $x \to -1^-$.
• For $-1 < x < 2$: $\text{f}(x)$ increases from $-\infty$ to the maximum at $a$, then decreases to $-\infty$. The gradient is positive for $x < a$, zero at $x = a$, and negative for $x > a$. The curve goes from $+\infty$, crosses $(a,0)$, and goes to $-\infty$.
• For $x > 2$: $\text{f}(x)$ decreases from $\infty$ to the minimum at $c$, then increases to approach $y = 3$. The gradient is negative for $x < c$, zero at $x = c$, and positive for $x > c$. Since it approaches a constant $y=3$ as $x \to \infty$, the gradient must peak and then return to $0$.