下图展示了 $y = |\mathrm{f}(x)|$ 和 $y = \mathrm{f}'(x)$ 的图像。 在不同的图上分别画出以下函数的图像： 在每幅图中清晰地标出与坐标轴的交点、极值点的坐标以及渐近线的方程。

1. Identify the geometrical transformations applied to the original graph $y = |\mathrm{f}(x)|$.

To obtain the graph of $y = |\mathrm{f}(2x)| + 1$ from $y = |\mathrm{f}(x)|$, we apply the following sequence of transformations:

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

In terms of coordinates, any point $(x, y)$ on the original graph is mapped to $\left(\frac{x}{2}, y + 1\right)$.

2. Determine the new equations for the asymptotes.

Apply the transformations to the original asymptotes:

• Horizontal asymptote: $y = 2$ is scaled horizontally (which has no effect) and then translated upwards by $1$. The new horizontal asymptote is $y = 2 + 1 \implies y = 3$.
• Vertical asymptote: $x = 1$ is scaled horizontally by a factor of $\frac{1}{2}$ and translated vertically (which has no effect). The new vertical asymptote is $x = 1 \times \frac{1}{2} \implies x = \frac{1}{2}$.

3. Calculate the transformed coordinates of the intersections and turning points.

Apply the mapping $(x, y) \to \left(\frac{x}{2}, y + 1\right)$ to the key points of $y = |\mathrm{f}(x)|$:

• $y$-intercept: $(0, 1) \to (0, 2)$.
• First minimum: $\left(\frac{1}{2}, 0\right) \to \left(\frac{1}{4}, 1\right)$.
• Second minimum: $(2, 0) \to (1, 1)$.
• Local maximum: $(5, 3) \to \left(\frac{5}{2}, 4\right)$.

Since the minimum $y$-value of the new graph is $1$, the graph lies entirely above the $x$-axis, meaning there are no $x$-intercepts.

1. Find the intersection points with the coordinate axes.

To find the $y$-intercept, substitute $x = 0$:
$y = \frac{1}{\mathrm{f}'(0)} = \frac{1}{-3} = -\frac{1}{3}$
So, the $y$-intercept is $(0, -\frac{1}{3})$.

To find the $x$-intercept, set $y = 0$:
$\frac{1}{\mathrm{f}'(x)} = 0 \implies \mathrm{f}'(x) \to \pm\infty$
From the given graph of $\mathrm{f}'(x)$, this occurs at the vertical asymptote $x = 1$. Thus, the curve intersects the $x$-axis at $(1, 0)$.

2. Identify the asymptotes.

Vertical asymptotes occur where the denominator is zero, i.e., $\mathrm{f}'(x) = 0$. From the given graph, the $x$-intercept is at $x = 5$. Hence, there is a vertical asymptote at $x = 5$.

As $x \to \pm\infty$, the graph shows $\mathrm{f}'(x) \to 0$. As the denominator approaches $0$, the value of $y = \frac{1}{\mathrm{f}'(x)} \to \pm\infty$. Thus, there are no horizontal asymptotes.

3. Find the coordinates and nature of the turning points.

Turning points of $y = \frac{1}{\mathrm{f}'(x)}$ correspond to turning points of $\mathrm{f}'(x)$.
The given graph has a local minimum at $(8, -3)$.
The corresponding point on the reciprocal graph is $\left(8, \frac{1}{-3}\right) = \left(8, -\frac{1}{3}\right)$.
Since $\mathrm{f}'(x)$ is negative and has a local minimum at this point, taking its reciprocal produces a local maximum. Thus, there is a local maximum at $\left(8, -\frac{1}{3}\right)$.

4. Deduce the shape of the graph and sketch.

Synthesising the features for the sketch:

• For $x < 1$: $\mathrm{f}'(x)$ is negative and approaches $-\infty$ from $0$. Thus $y$ increases from $-\infty$ to $0$, crossing the $y$-axis at $(0, -1/3)$.
• For $1 < x < 5$: $\mathrm{f}'(x)$ is positive, decreasing from $\infty$ to $0$. Thus $y$ is positive, increasing from $(1, 0)$ up towards $\infty$ at $x=5^-$.
• For $x > 5$: $\mathrm{f}'(x)$ is negative, goes down to $-3$ and back up to $0$. Thus $y$ is negative, comes up from $-\infty$ at $x=5^+$, reaches a local maximum at $(8, -1/3)$, and curves back down towards $-\infty$ as $x \to \infty$.

1. Determine the shape and sign of $f(x)$ for $x < 1$.

From the graph of $y = \mathrm{f}'(x)$, we observe that $\mathrm{f}'(x) < 0$ for $x < 1$. Thus, $\mathrm{f}(x)$ is strictly decreasing on $x < 1$.
The graph of $y = |\mathrm{f}(x)|$ touches the $x$-axis at $x = 0.5$, which implies $\mathrm{f}(0.5) = 0$.
Since $\mathrm{f}(x)$ is strictly decreasing, we deduce that $\mathrm{f}(x) > 0$ for $x < 0.5$ and $\mathrm{f}(x) < 0$ for $0.5 < x < 1$.
Therefore, $\mathrm{f}(x) = |\mathrm{f}(x)|$ for $x < 0.5$, and $\mathrm{f}(x) = -|\mathrm{f}(x)|$ for $0.5 < x < 1$. The curve reflects below the $x$-axis for $0.5 < x < 1$, approaching $-\infty$ as $x \to 1^-$.

2. Determine the shape and sign of $f(x)$ for $x > 1$.

From the graph of $y = \mathrm{f}'(x)$, $\mathrm{f}'(x) > 0$ for $1 < x < 5$ and $\mathrm{f}'(x) < 0$ for $x > 5$. Thus, $\mathrm{f}(x)$ is strictly increasing on $(1, 5)$ and strictly decreasing on $x > 5$.
The graph of $y = |\mathrm{f}(x)|$ touches the $x$-axis at $x = 2$, implying $\mathrm{f}(2) = 0$.
Since $\mathrm{f}(x)$ is increasing on $(1, 5)$, we deduce that $\mathrm{f}(x) < 0$ for $1 < x < 2$ and $\mathrm{f}(x) > 0$ for $x > 2$.
Therefore, $\mathrm{f}(x) = -|\mathrm{f}(x)|$ for $1 < x < 2$ (approaching $-\infty$ as $x \to 1^+$), and $\mathrm{f}(x) = |\mathrm{f}(x)|$ for $x > 2$.
At $x = 5$, $\mathrm{f}(x)$ reaches a local maximum.

3. Identify all key features to be labeled on the sketch.

To sketch $y = \mathrm{f}(x)$, draw the curve identical to $y = |\mathrm{f}(x)|$ for $x < 0.5$ and $x > 2$, and reflect the sections $0.5 < x < 1$ and $1 < x < 2$ across the $x$-axis.
The required features to be clearly labeled on the sketch are:

• $y$-intercept: $(0, 1)$
• $x$-intercepts: $(0.5, 0)$ and $(2, 0)$ (the curve smoothly crosses the $x$-axis at both points without sharp turns)
• Turning point: Local maximum at $(5, 3)$
• Asymptotes: $x = 1$ (vertical) and $y = 2$ (horizontal)