在同一幅图中，画出方程为 $y = \frac{|x + 1|}{|x - 2|}$ 和 $y = \ln\left(1 - \frac{x}{2}\right) + \tfrac{1}{2}$ 的曲线，指出所有渐近线的方程以及与坐标轴的交点坐标。清晰地标记这两条曲线。 由此求解不等式 $\frac{|x + 1|}{|x - 2|} \ge \ln\left(1 - \frac{x}{2}\right) + \tfrac{1}{2}$。

1. Find the intercepts and asymptotes of the first curve $y = \frac{|x + 1|}{|x - 2|}$.

For $C_1: y = \frac{|x + 1|}{|x - 2|}$:

• $y$-intercept: Let $x = 0$, $y = \frac{|1|}{|-2|} = \frac{1}{2}$. Point is $\left(0, \frac{1}{2}\right)$.
• $x$-intercept: Let $y = 0$, $|x + 1| = 0 \implies x = -1$. Point is $(-1, 0)$.
• Vertical asymptote: The denominator is zero when $x - 2 = 0 \implies x = 2$.
• Horizontal asymptote: $\lim_{x \to \pm\infty} \frac{|x + 1|}{|x - 2|} = \lim_{x \to \pm\infty} \left| \frac{x+1}{x-2} \right| = 1$. Equation is $y = 1$.

2. Find the intercepts and asymptotes of the second curve $y = \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$.

For $C_2: y = \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$:

• $y$-intercept: Let $x = 0$, $y = \ln(1) + \frac{1}{2} = \frac{1}{2}$. Point is $\left(0, \frac{1}{2}\right)$.
• $x$-intercept: Let $y = 0$,
  $\ln\left(1 - \frac{x}{2}\right) + \frac{1}{2} = 0$
  $\ln\left(1 - \frac{x}{2}\right) = -\frac{1}{2}$
  $1 - \frac{x}{2} = e^{-1/2} \implies x = 2 - 2e^{-1/2}$
  Point is $\left(2 - 2e^{-1/2}, 0\right)$.
• Vertical asymptote: The argument of the natural logarithm is zero when $1 - \frac{x}{2} = 0 \implies x = 2$.

3. Sketch both curves on the same diagram, labelling all required features.

Using a graphing calculator as an aid or reasoning graphically:

1. Draw the $x$ and $y$ axes.
2. Draw the vertical line $x = 2$ and horizontal line $y = 1$ as dashed lines.
3. Plot the axial intercepts $(-1, 0)$, $\left(2 - 2e^{-1/2}, 0\right)$, and the common $y$-intercept $\left(0, \frac{1}{2}\right)$.
4. Sketch $C_1$, noting the sharp "V-shape" minimum at $(-1, 0)$ and its asymptotic behaviour towards $x = 2$ and $y = 1$. The curve is entirely in the region $y \ge 0$.
5. Sketch $C_2$ for its domain $x < 2$. The curve intersects the $y$-axis at $\left(0, \frac{1}{2}\right)$ with a negative gradient, passes through its $x$-intercept, and approaches the asymptote $x = 2$ as $y \to -\infty$.
6. Label the respective curves and their key features clearly.

1. Relate the inequality to the graphs from part (a) and state the point of intersection

Let $y_1 = \frac{|x + 1|}{|x - 2|}$ and $y_2 = \ln\left(1 - \frac{x}{2}\right) + \tfrac{1}{2}$.

The inequality $y_1 \ge y_2$ is satisfied where the curve of $y_1$ lies above or on the curve of $y_2$.
From the intercepts found in part (a), both curves have the same $y$-intercept at $(0, \frac{1}{2})$. Thus, they intersect at $x = 0$.

2. Determine the solution set using the graphs and domain restrictions

From the sketched graphs, the curve of $y_1$ is above $y_2$ for $x > 0$.
However, both functions share a vertical asymptote at $x = 2$, and the logarithmic function $y_2$ is only defined for $1 - \frac{x}{2} > 0 \implies x < 2$.
Therefore, the inequality holds for $0 \le x < 2$.