描述一个由两个变换组成的序列，该序列将 $y = \ln\left(\frac{x^2}{x+1}\right)$ 的图像映射到 $y = \ln\left(\frac{2x+1}{4x^2}\right)$ 的图像上。 下图显示了 $y = \mathrm{f}(x)$ 的图像。它在 $A(-1,-1)$ 处有一个极大值点，在 $B(1, \frac{1}{4})$ 处有一个极小值点。该图像的渐近线为 $y = \frac{1}{2}$，$x = 0$ 和 $x = -2$。 在不同的图上分别画出以下函数的图像草图： (i) $y = f(2 - x)$， (ii) $y = \frac{1}{f(x)}$， 需清晰地标明所有渐近线的方程、与两坐标轴的交点坐标以及对应于 A 和 B 的点。 $y = f(2 - x)$ $y = \frac{1}{f(x)}$

1. Express the target function in terms of the original function $f(x)$.

Let the original function be $f(x) = \ln\left(\frac{x^2}{x+1}\right)$.
We manipulate the target equation to relate it to $f(x)$:
$y = \ln\left(\frac{2x+1}{4x^2}\right)$
Using the property of logarithms $\ln(a^{-1}) = -\ln(a)$:
$y = \ln\left(\left(\frac{4x^2}{2x+1}\right)^{-1}\right)$
$y = -\ln\left(\frac{(2x)^2}{2x+1}\right)$
$y = -f(2x)$

2. Identify the sequence of transformations that maps $f(x)$ to $-f(2x)$.

The transformation from $f(x)$ to $f(2x)$ represents a horizontal scaling.
The transformation from $f(2x)$ to $-f(2x)$ represents a vertical reflection.
Thus, the sequence of transformations is:

1. A scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$ (mapping $f(x)$ to $f(2x)$).
2. A reflection in the $x$-axis (mapping $f(2x)$ to $-f(2x)$).

(Note: These two transformations are independent, so their order can be reversed without affecting the final graph.)

1. Relate the coordinates of the new graph to the original graph.

For the graph of $y = \mathrm{f}(2 - x)$, the $x$-coordinates of the new graph ($x'$) are related to the original $x$-coordinates by $x = 2 - x'$, which simplifies to $x' = 2 - x$. The $y$-coordinates remain unchanged.

This corresponds geometrically to a reflection in the $y$-axis followed by a translation of $2$ units in the positive $x$-direction.

2. Determine the new coordinates of the stationary points and intersections with axes.

Applying the transformation $x' = 2 - x$:

• Maximum point $A(-1, -1) \to A'(2 - (-1), -1) = (3, -1)$
• Minimum point $B(1, \frac{1}{4}) \to B'(2 - 1, \frac{1}{4}) = (1, \frac{1}{4})$

Intersections:

• $x$-intercepts: The original graph $y=\mathrm{f}(x)$ has no $x$-intercepts, so the transformed graph also has none.
• $y$-intercept: Set $x = 0$ in $y = \mathrm{f}(2 - x) \implies y = \mathrm{f}(2)$. The intersection is exactly at $(0, \mathrm{f}(2))$.

3. Determine the new equations of the asymptotes.

Applying the mapping $x' = 2 - x$ to the vertical asymptotes:

• Vertical asymptote $x = -2 \to x = 2 - (-2) \implies x = 4$
• Vertical asymptote $x = 0 \to x = 2 - 0 \implies x = 2$

Since the transformation only affects the $x$-values, the horizontal asymptote remains unchanged:

• Horizontal asymptote $y = \frac{1}{2}$

1. Determine the coordinates and nature of the new turning points.

For a point $(x, y)$ on $y = \mathrm{f}(x)$, the corresponding point on $y = \frac{1}{\mathrm{f}(x)}$ is $\left(x, \frac{1}{y}\right)$.

• $A(-1, -1) \to A'(-1, -1)$. Since $A$ is a local maximum where $y < 0$, its reciprocal becomes a local minimum (the magnitude becomes larger as it moves away from the peak).
• $B\left(1, \frac{1}{4}\right) \to B'(1, 4)$. Since $B$ is a local minimum where $y > 0$, its reciprocal becomes a local maximum.

2. Find the new asymptotes and intersections with the axes.

• Horizontal Asymptote: As $x \to \pm\infty$, $\mathrm{f}(x) \to \frac{1}{2}$, so $\frac{1}{\mathrm{f}(x)} \to 2$. Hence, the new horizontal asymptote is $y = 2$.
• Axis Intersections / Roots: The vertical asymptotes $x = -2$ and $x = 0$ mean $\mathrm{f}(x) \to \pm\infty$ as it approaches these lines. Thus, $\frac{1}{\mathrm{f}(x)} \to 0$. The new graph intersects the axes at $(-2, 0)$ and $(0, 0)$.
• Vertical Asymptotes: Since $\mathrm{f}(x)$ has no $x$-intercepts (never equals $0$), the new graph has no vertical asymptotes.

3. Describe the required sketch of $y = \frac{1}{\mathrm{f}(x)}$.

Sketch the continuous curve with the following behavior:

• $x < -2$: Starts below $y = 2$, curving downwards to intersect the $x$-axis at $(-2, 0)$.
• $-2 < x < 0$: Starts from $(-2, 0)$, curves downwards to form a local minimum at $A'(-1, -1)$, then curves upwards to intersect the axes at $(0, 0)$.
• $x > 0$: Starts from $(0, 0)$, curves upwards to form a local maximum at $B'(1, 4)$, then curves downwards to approach the horizontal asymptote $y = 2$ from above.