The diagram shows the curve with equation $y = \mathrm{f}(x)$. The curve crosses the $x$-axis at $x = 1$ and $x = 2.5$, crosses the $y$-axis at $y = 1$ and has a maximum point at $(4, 6)$. The equations of the asymptotes are $x = 2$ and $y = 3$. Sketch the graph of $y = \mathrm{f}'(x)$, giving the equations of asymptotes, coordinates of turning points and axial intercepts, where possible. Lucas heats up one of these metal rods. When heated, the metal rod expands uniformly such that it always retains its shape. At time $t$ seconds, the length of each side of the hexagon is $x$ cm, the length of the rod is $l$ cm and the volume of the rod is $V\text{ cm}^3$.

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).