上面给出了 $y = \text{f}(x)$ 的图像。在不同的图上分别画出以下函数的图像： 你的草图应清晰地标出任何驻点、与坐标轴的交点坐标以及任何渐近线的方程。

1. Establish the coordinate mapping for the transformation

The transformation from $y = \text{f}(x)$ to $y = \text{f}(2x + 3)$ maps each point $(x, y)$ on the original graph to a new point $(x', y)$.
The relationship between the $x$-coordinates is given by:
$2x' + 3 = x \implies x' = \frac{x - 3}{2}$

2. Determine the new equations of the asymptotes

• Vertical asymptote $x = -1 \implies x' = \frac{-1 - 3}{2} = -2$
• Vertical asymptote $x = 3 \implies x' = \frac{3 - 3}{2} = 0$
• The horizontal asymptote is unaffected by horizontal transformations, so it remains $y = 2$.

3. Calculate the new coordinates of the stationary point and intercepts

• The $x$-intercept at $(0, 0) \implies x' = \frac{0 - 3}{2} = -1.5$. The new intercept is $(-\frac{3}{2}, 0)$.
• The $x$-intercept at $(2, 0) \implies x' = \frac{2 - 3}{2} = -0.5$. The new intercept is $(-\frac{1}{2}, 0)$.
• The local maximum at $(1, \frac{2}{3}) \implies x' = \frac{1 - 3}{2} = -1$. The new stationary point is $(-1, \frac{2}{3})$.
  Note: Since $x = 0$ is now a vertical asymptote, there is no $y$-intercept.

1. Deduce the equations of the asymptotes for the derivative graph.

As $x \to \pm\infty$, the graph of $y = \text{f}(x)$ approaches the horizontal asymptote $y = 2$, meaning the gradient flattens out and approaches $0$. Hence, the horizontal asymptote for $y = \text{f}'(x)$ is $y = 0$.
As $x \to -1$ and $x \to 3$, the graph of $y = \text{f}(x)$ approaches vertical asymptotes, so the curve becomes infinitely steep. Hence, the vertical asymptotes remain $x = -1$ and $x = 3$.

2. Identify the x-intercept of the derivative graph from the stationary point of f(x).

The graph of $y = \text{f}(x)$ has a local maximum (stationary point) at $(1, \frac{2}{3})$. At this maximum turning point, the gradient of the tangent is $0$.
Hence, $y = \text{f}'(x)$ has an $x$-intercept at $(1, 0)$.

3. Determine the sign and behaviour of the derivative graph in each region to complete the sketch.

Analyse the gradient of $y = \text{f}(x)$ across the three intervals:

• For $x < -1$: $\text{f}(x)$ is increasing, so $\text{f}'(x) > 0$. The gradient increases from $0$ (at $-\infty$) to $+\infty$ (near $x = -1$).
• For $-1 < x < 1$: $\text{f}(x)$ is increasing from $-\infty$, so $\text{f}'(x) > 0$. The gradient decreases from steepest at $+\infty$ (near $x = -1$) to $0$ (at $x = 1$).
• For $1 < x < 3$: $\text{f}(x)$ is decreasing towards $-\infty$, so $\text{f}'(x) < 0$. The gradient becomes increasingly negative, dropping from $0$ to $-\infty$ (near $x = 3$).
• For $x > 3$: $\text{f}(x)$ is decreasing from $+\infty$, so $\text{f}'(x) < 0$. The gradient starts very steep at $-\infty$ (near $x = 3$) and becomes less steep, approaching $0$ (at $\infty$).

The concavity of $y = \text{f}(x)$ does not change sign within any of the three branches, meaning there are no stationary points for $y = \text{f}'(x)$. Sketch the three branches satisfying these gradient changes.

1. Find the vertical asymptotes and $x$-intercepts of the reciprocal graph

The $x$-intercepts of $y = \text{f}(x)$ occur at $x = 0$ and $x = 2$. These become vertical asymptotes for $y = \frac{1}{\text{f}(x)}$, yielding the lines $x = 0$ and $x = 2$.

The vertical asymptotes of $y = \text{f}(x)$ are $x = -1$ and $x = 3$. As $x \to -1$ and $x \to 3$, $|\text{f}(x)| \to \infty$, which means $\frac{1}{\text{f}(x)} \to 0$. Thus, $y = \frac{1}{\text{f}(x)}$ has $x$-intercepts at $(-1, 0)$ and $(3, 0)$.

2. Find the new horizontal asymptote and stationary point

The horizontal asymptote of $y = \text{f}(x)$ is $y = 2$. Therefore, the horizontal asymptote of the reciprocal graph is $y = \frac{1}{2}$.

The local maximum point of $y = \text{f}(x)$ is $(1, \frac{2}{3})$. Since taking the reciprocal of positive values inverses the nature of the stationary point, $y = \frac{1}{\text{f}(x)}$ will have a local minimum at $(1, \frac{3}{2})$.

3. Determine the limits and asymptotic behaviour to shape the branches

For $x < -1$ and $x > 3$, $y = \text{f}(x)$ is positive and greater than $2$ (approaching $2$ as $x \to \pm\infty$). Thus, $\frac{1}{\text{f}(x)}$ is positive, less than $\frac{1}{2}$, and approaches the horizontal asymptote $y = \frac{1}{2}$ from below.

Between the vertical asymptotes $x = -1$ and $x = 3$, $\text{f}(x)$ changes sign as it passes through $x = 0$ and $x = 2$. Consequently:

• As $x \to -1^-$, $\text{f}(x) \to \infty \implies y \to 0^+$
• As $x \to -1^+$, $\text{f}(x) \to -\infty \implies y \to 0^-$
  This shows the curve cuts through $(-1, 0)$ smoothly. A similar crossing from negative to positive occurs at $(3,0)$.