Sketch, on the same diagram, the curves with equations $y = \left|\frac{x + 1}{x - 2}\right|$ and $y = \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$, stating the equations of any asymptotes and the coordinates of any points of intersection with the axes. Label the two curves clearly. Hence solve the inequality $\left|\frac{x + 1}{x - 2}\right| \ge \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$.

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

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

• $x$-intercept: Let $y = 0 \implies \frac{x + 1}{x - 2} = 0 \implies x = -1$. The coordinate is $(-1, 0)$.
• $y$-intercept: Let $x = 0 \implies y = \left|\frac{0 + 1}{0 - 2}\right| = \left|-\frac{1}{2}\right| = \frac{1}{2}$. The coordinate is $(0, \frac{1}{2})$.
• Vertical asymptote: The expression is undefined when the denominator is zero: $x - 2 = 0 \implies x = 2$.
• Horizontal asymptote: Consider the limit as $x \to \pm\infty$: $\lim_{x \to \pm\infty} \left|\frac{x + 1}{x - 2}\right| = |1| = 1$. The equation is $y = 1$.

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

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

• $x$-intercept: Let $y = 0$.
  $\ln\left(1 - \frac{x}{2}\right) = -\frac{1}{2}$
  $1 - \frac{x}{2} = e^{-1/2}$
  $\frac{x}{2} = 1 - e^{-1/2} \implies x = 2 - 2e^{-1/2} \approx 0.787$
  The coordinate is $(2 - 2e^{-1/2}, 0)$.
• $y$-intercept: Let $x = 0 \implies y = \ln(1) + \frac{1}{2} = 0 + \frac{1}{2} = \frac{1}{2}$. The coordinate is $(0, \frac{1}{2})$.
• Vertical asymptote: The argument of the natural logarithm approaches zero, so $1 - \frac{x}{2} = 0 \implies x = 2$.
• Domain requirement: $1 - \frac{x}{2} > 0 \implies x < 2$. The curve exists strictly to the left of the vertical asymptote.

3. Sketch both graphs on the same set of axes.

With the analytical features determined and verified via graphing calculator, draw the sketch on a single coordinate plane:

• Draw and label the shared vertical asymptote $x = 2$ and the horizontal asymptote $y = 1$.
• Plot the common $y$-intercept intersection point at $(0, 0.5)$.
• Sketch the V-shaped curve for $y = \left|\frac{x + 1}{x - 2}\right|$ bouncing off the $x$-axis at $(-1, 0)$ and gradually approaching $y = 1$ from below (for large negative $x$) and from above (for large positive $x$).
• Sketch the logarithmic curve for $y = \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$ rising slowly from the left, crossing the $y$-axis at $(0, 0.5)$, intersecting the $x$-axis at $(2 - 2e^{-1/2}, 0)$, and plunging towards $-\infty$ as it approaches the vertical asymptote $x = 2$.
• Clearly label both curves with their equations.

1. Identify the point of intersection of the two curves.

By checking the intercepts, we can verify that the curves intersect at $x = 0$:
When $x = 0$, for the first curve:
$y = \left|\frac{0 + 1}{0 - 2}\right| = \frac{1}{2}$
For the second curve:
$y = \ln\left(1 - \frac{0}{2}\right) + \frac{1}{2} = \ln(1) + \frac{1}{2} = \frac{1}{2}$
Thus, the graphs intersect at the point $(0, \frac{1}{2})$.

2. Determine the solution to the inequality using the relative positions of the graphs.

The inequality $\left|\frac{x + 1}{x - 2}\right| \ge \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$ is satisfied when the graph of $y = \left|\frac{x + 1}{x - 2}\right|$ is above or intersects the graph of $y = \ln\left(1 - \frac{x}{2}\right) + \frac{1}{2}$.

From the graphs sketched in part (a), the modulus curve is above the logarithmic curve for $x > 0$.
However, the logarithmic function requires its argument to be strictly positive:
$1 - \frac{x}{2} > 0 \implies x < 2$
Both curves share a vertical asymptote at $x = 2$.
Combining this domain restriction with the visual deduction, the inequality holds for $0 \le x < 2$.