图中显示了方程为 $y = \mathrm{f}(x)$ 的曲线。该曲线在 $x = 1$ 和 $x = 2.5$ 处穿过 $x$ 轴，在 $y = 1$ 处穿过 $y$ 轴，且其极大值点为 $(4, 6)$。渐近线方程为 $x = 2$ 和 $y = 3$。草拟 $y = \mathrm{f}'(x)$ 的图像，尽可能给出渐近线方程、转折点坐标以及与坐标轴的交点。 Lucas 加热了其中一根金属棒。加热时，金属棒均匀膨胀，使其始终保持原有形状。在时间为 $t$ 秒时，六边形每条边的长度为 $x$ cm，金属棒的长度为 $l$ cm，金属棒的体积为 $V\text{ cm}^3$。

1. Determine the horizontal and vertical asymptotes of the derivative graph.

Since the graph of $y = \mathrm{f}(x)$ has a horizontal asymptote $y = 3$ as $x \to \pm\infty$, the curve flattens out, meaning its gradient approaches $0$. Thus, the graph of $y = \mathrm{f}'(x)$ has a horizontal asymptote $y = 0$.

Near the vertical asymptote $x = 2$, the graph of $y = \mathrm{f}(x)$ becomes increasingly steep. As $x \to 2^-$, the curve goes downwards to $-\infty$, meaning the gradient is increasingly negative ($\mathrm{f}'(x) \to -\infty$). As $x \to 2^+$, the curve comes upwards from $-\infty$, meaning the gradient is extremely positive ($\mathrm{f}'(x) \to +\infty$). Thus, $x = 2$ is a vertical asymptote for $y = \mathrm{f}'(x)$.

2. Determine the axial intercepts.

The turning point of $y = \mathrm{f}(x)$ occurs at the maximum point $(4, 6)$. At this maximum, the gradient is zero, so $\mathrm{f}'(4) = 0$. This gives an $x$-intercept at $(4, 0)$ for the derivative graph.

The specific values of the gradient at $x=1$, $x=2.5$ and $x=0$ cannot be evaluated without the explicit function equation, so no other axial intercepts are possible to determine.

3. Analyse the sign of the derivative to establish the full shape of the graph.

• For $x < 2$: The curve $y = \mathrm{f}(x)$ is strictly decreasing, so $\mathrm{f}'(x) < 0$. The graph of $y = \mathrm{f}'(x)$ lies entirely below the $x$-axis, starting near $y = 0$ as $x \to -\infty$ and plunging to $-\infty$ as $x \to 2^-$.
• For $2 < x < 4$: The curve $y = \mathrm{f}(x)$ is strictly increasing, so $\mathrm{f}'(x) > 0$. The derivative graph comes from $+\infty$ near $x = 2^+$ and decreases downwards to the intercept $(4, 0)$.
• For $x > 4$: The curve $y = \mathrm{f}(x)$ is strictly decreasing towards $y = 3$, so $\mathrm{f}'(x) < 0$. After crossing the $x$-axis at $(4,0)$, the graph of $y = \mathrm{f}'(x)$ must go negative, reach a local minimum, and then turn upwards to approach the horizontal asymptote $y = 0$ from below as $x \to \infty$. (The coordinates of this turning point are not possible to compute).