Key Concepts and Formulas

• Domain of $\sin^{-1}(x)$: The domain of the inverse sine function, $\sin^{-1}(x)$, is $[-1, 1]$. This means that the argument of $\sin^{-1}$ must be between -1 and 1, inclusive.
• Domain of a Composite Function (fog)(x): For the composite function $(f \circ g)(x) = f(g(x))$, the domain is determined by two conditions:
  1. $x$ must be in the domain of $g(x)$.
  2. $g(x)$ must be in the domain of $f(x)$.
• Limits of Rational Functions: When evaluating the limit of a rational function at a point where the denominator becomes zero, we often need to factorize the numerator and denominator to simplify the expression or use L'Hôpital's Rule if the indeterminate form $\frac{0}{0}$ arises.

Step-by-Step Solution

Step 1: Analyze the function $f(x) = \sin^{-1}(x)$ The function $f(x) = \sin^{-1}(x)$ is defined for $-1 \le x \le 1$. This means that for the composite function $(f \circ g)(x) = f(g(x))$ to be defined, the output of $g(x)$ must lie within the interval $[-1, 1]$. So, we require $-1 \le g(x) \le 1$.

Step 2: Analyze the function $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$ First, we need to find the domain of $g(x)$. The function $g(x)$ is a rational function, and it is undefined when its denominator is zero. Let's find the roots of the denominator: $2x^2 - x - 6 = 0$ We can factor this quadratic equation. We look for two numbers that multiply to $2 \times -6 = -12$ and add up to -1. These numbers are -4 and 3. $2x^2 - 4x + 3x - 6 = 0$ $2x(x - 2) + 3(x - 2) = 0$ $(2x + 3)(x - 2) = 0$ The roots are $x = 2$ and $x = -\frac{3}{2}$. Therefore, the domain of $g(x)$ is $x \in \mathbb{R} \setminus \{2, -\frac{3}{2}\}$.

The problem states that $g(2) = \lim_{x \to 2} g(x)$. This implies that we need to define $g(2)$ as the limit to make the function continuous at $x=2$ if possible. Let's evaluate the limit: $\mathop {\lim }\limits_{x \to 2} g(x) = \mathop {\lim }\limits_{x \to 2} \frac{x^2 - x - 2}{2x^2 - x - 6}$ The numerator at $x=2$ is $2^2 - 2 - 2 = 4 - 2 - 2 = 0$. The denominator at $x=2$ is $2(2^2) - 2 - 6 = 2(4) - 2 - 6 = 8 - 2 - 6 = 0$. Since we have the indeterminate form $\frac{0}{0}$, we can factorize the numerator and denominator. Numerator: $x^2 - x - 2 = (x-2)(x+1)$ Denominator: $2x^2 - x - 6 = (2x+3)(x-2)$ So, $\mathop {\lim }\limits_{x \to 2} g(x) = \mathop {\lim }\limits_{x \to 2} \frac{(x-2)(x+1)}{(2x+3)(x-2)}$ For $x \ne 2$, we can cancel out the $(x-2)$ term: $\mathop {\lim }\limits_{x \to 2} g(x) = \mathop {\lim }\limits_{x \to 2} \frac{x+1}{2x+3}$ Now, substitute $x=2$: $\frac{2+1}{2(2)+3} = \frac{3}{4+3} = \frac{3}{7}$ So, $g(2) = \frac{3}{7}$. This value is within the domain of $f(x)$ since $-1 \le \frac{3}{7} \le 1$.

Step 3: Determine the condition $-1 \le g(x) \le 1$ We need to find the values of $x$ (excluding $x=2$ and $x=-\frac{3}{2}$ for now, as these are points where $g(x)$ is initially undefined) such that: $-1 \le \frac{x^2 - x - 2}{2x^2 - x - 6} \le 1$ This inequality can be split into two inequalities:

1. $\frac{x^2 - x - 2}{2x^2 - x - 6} \le 1$
2. $\frac{x^2 - x - 2}{2x^2 - x - 6} \ge -1$

Step 4: Solve the first inequality: $\frac{x^2 - x - 2}{2x^2 - x - 6} \le 1$ $\frac{x^2 - x - 2}{2x^2 - x - 6} - 1 \le 0$ $\frac{x^2 - x - 2 - (2x^2 - x - 6)}{2x^2 - x - 6} \le 0$ $\frac{x^2 - x - 2 - 2x^2 + x + 6}{2x^2 - x - 6} \le 0$ $\frac{-x^2 + 4}{2x^2 - x - 6} \le 0$ Multiply by -1 and reverse the inequality sign: $\frac{x^2 - 4}{2x^2 - x - 6} \ge 0$ Factorize the numerator and denominator: $\frac{(x-2)(x+2)}{(2x+3)(x-2)} \ge 0$ For $x \ne 2$, we can simplify this to: $\frac{x+2}{2x+3} \ge 0$ The critical points are $x = -2$ and $x = -\frac{3}{2}$. We analyze the sign of $\frac{x+2}{2x+3}$ in the intervals determined by these critical points.

• If $x < -2$: Let $x = -3$. $\frac{-3+2}{2(-3)+3} = \frac{-1}{-6+3} = \frac{-1}{-3} = \frac{1}{3} > 0$. So, $(-\infty, -2]$ is part of the solution.
• If $-2 < x < -\frac{3}{2}$: Let $x = -1.7$. $\frac{-1.7+2}{2(-1.7)+3} = \frac{0.3}{-3.4+3} = \frac{0.3}{-0.4} < 0$.
• If $x > -\frac{3}{2}$: Let $x = 0$. $\frac{0+2}{2(0)+3} = \frac{2}{3} > 0$. So, $(-\frac{3}{2}, \infty)$ is part of the solution.

Considering the critical points: The expression $\frac{x+2}{2x+3}$ is zero at $x=-2$. So $x=-2$ is included. The expression is undefined at $x=-\frac{3}{2}$. So $x=-\frac{3}{2}$ is excluded. So, the solution to $\frac{x+2}{2x+3} \ge 0$ is $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$.

Now we must consider the original inequality $\frac{(x-2)(x+2)}{(2x+3)(x-2)} \ge 0$. The critical points are $x=2, x=-2, x=-\frac{3}{2}$. Let's analyze the sign of $\frac{(x-2)(x+2)}{(2x+3)(x-2)}$:

• If $x < -2$: e.g., $x=-3$. $\frac{(-5)(-1)}{(-3)(-5)} = \frac{5}{15} = \frac{1}{3} > 0$.
• If $-2 < x < -\frac{3}{2}$: e.g., $x=-1.7$. $\frac{(-3.7)(0.3)}{(-0.4)(-3.7)} < 0$.
• If $-\frac{3}{2} < x < 2$: e.g., $x=0$. $\frac{(-2)(2)}{(3)(-2)} = \frac{-4}{-6} = \frac{2}{3} > 0$.
• If $x > 2$: e.g., $x=3$. $\frac{(1)(5)}{(9)(1)} = \frac{5}{9} > 0$.

The expression is equal to 0 when $x=-2$ (numerator is 0, denominator is non-zero). The expression is undefined when $x=2$ or $x=-\frac{3}{2}$. So, the solution to $\frac{(x-2)(x+2)}{(2x+3)(x-2)} \ge 0$ is $(-\infty, -2] \cup (-\frac{3}{2}, 2) \cup (2, \infty)$.

Step 5: Solve the second inequality: $\frac{x^2 - x - 2}{2x^2 - x - 6} \ge -1$ $\frac{x^2 - x - 2}{2x^2 - x - 6} + 1 \ge 0$ $\frac{x^2 - x - 2 + (2x^2 - x - 6)}{2x^2 - x - 6} \ge 0$ $\frac{3x^2 - 2x - 8}{2x^2 - x - 6} \ge 0$ Factorize the numerator $3x^2 - 2x - 8$. We look for two numbers that multiply to $3 \times -8 = -24$ and add up to -2. These numbers are -6 and 4. $3x^2 - 6x + 4x - 8 = 0$ $3x(x - 2) + 4(x - 2) = 0$ $(3x + 4)(x - 2) = 0$ The roots are $x = 2$ and $x = -\frac{4}{3}$. The denominator is $(2x+3)(x-2)$. So the inequality becomes: $\frac{(3x+4)(x-2)}{(2x+3)(x-2)} \ge 0$ For $x \ne 2$, we can simplify this to: $\frac{3x+4}{2x+3} \ge 0$ The critical points are $x = -\frac{4}{3}$ and $x = -\frac{3}{2}$. We analyze the sign of $\frac{3x+4}{2x+3}$ in the intervals determined by these critical points.

• If $x < -\frac{3}{2}$: Let $x = -2$. $\frac{3(-2)+4}{2(-2)+3} = \frac{-6+4}{-4+3} = \frac{-2}{-1} = 2 > 0$. So, $(-\infty, -\frac{3}{2})$ is part of the solution.
• If $-\frac{3}{2} < x < -\frac{4}{3}$: Let $x = -1.4$. $\frac{3(-1.4)+4}{2(-1.4)+3} = \frac{-4.2+4}{-2.8+3} = \frac{-0.2}{0.2} = -1 < 0$.
• If $x > -\frac{4}{3}$: Let $x = 0$. $\frac{3(0)+4}{2(0)+3} = \frac{4}{3} > 0$. So, $[-\frac{4}{3}, \infty)$ is part of the solution.

The expression $\frac{3x+4}{2x+3}$ is zero at $x=-\frac{4}{3}$. So $x=-\frac{4}{3}$ is included. The expression is undefined at $x=-\frac{3}{2}$. So $x=-\frac{3}{2}$ is excluded. So, the solution to $\frac{3x+4}{2x+3} \ge 0$ is $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$.

Now we must consider the original inequality $\frac{(3x+4)(x-2)}{(2x+3)(x-2)} \ge 0$. The critical points are $x=2, x=-\frac{4}{3}, x=-\frac{3}{2}$. Let's analyze the sign of $\frac{(3x+4)(x-2)}{(2x+3)(x-2)}$:

• If $x < -\frac{3}{2}$: e.g., $x=-2$. $\frac{(-2)(-4)}{(-1)(-4)} = \frac{8}{4} = 2 > 0$.
• If $-\frac{3}{2} < x < -\frac{4}{3}$: e.g., $x=-1.4$. $\frac{(-1.2)(-3.4)}{(0.2)(-3.4)} < 0$.
• If $-\frac{4}{3} < x < 2$: e.g., $x=0$. $\frac{(4)(-2)}{(3)(-2)} = \frac{-8}{-6} = \frac{4}{3} > 0$.
• If $x > 2$: e.g., $x=3$. $\frac{(13)(1)}{(9)(1)} = \frac{13}{9} > 0$.

The expression is equal to 0 when $x=-\frac{4}{3}$ (numerator is 0, denominator is non-zero). The expression is undefined when $x=2$ or $x=-\frac{3}{2}$. So, the solution to $\frac{(3x+4)(x-2)}{(2x+3)(x-2)} \ge 0$ is $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Step 6: Find the intersection of the solutions from Step 4 and Step 5 We need to find the values of $x$ that satisfy both: Solution 1: $(-\infty, -2] \cup (-\frac{3}{2}, 2) \cup (2, \infty)$ Solution 2: $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, 2) \cup (2, \infty)$

Let's consider the intervals:

• $(-\infty, -2]$: This interval is present in Solution 1. It is also present in Solution 2. So, $(-\infty, -2]$ is part of the intersection.
• $(-2, -\frac{3}{2})$: This interval is NOT in Solution 1. So it is not in the intersection.
• $(-\frac{3}{2}, -\frac{4}{3})$: This interval is in Solution 1. It is NOT in Solution 2. So it is not in the intersection.
• $[-\frac{4}{3}, 2)$: This interval is in Solution 1. It is also in Solution 2. So, $[-\frac{4}{3}, 2)$ is part of the intersection.
• $(2, \infty)$: This interval is in Solution 1. It is also in Solution 2. So, $(2, \infty)$ is part of the intersection.

The intersection is $(-\infty, -2] \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Step 7: Consider the domain of $g(x)$ and the point $x=2$. The original domain of $g(x)$ is $\mathbb{R} \setminus \{2, -\frac{3}{2}\}$. The condition $-1 \le g(x) \le 1$ gives us the set $(-\infty, -2] \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

We need to ensure that $x$ is in the domain of $g(x)$ AND $g(x)$ is in the domain of $f(x)$. The domain of $g(x)$ excludes $x=2$ and $x=-\frac{3}{2}$. The set we found from the inequalities excludes $x=-\frac{3}{2}$ (since it's a boundary in the denominator). We also need to check if $x=2$ is included in our derived set for the domain of $(f \circ g)$. The interval $[-\frac{4}{3}, 2)$ excludes $x=2$. The interval $(2, \infty)$ excludes $x=2$. So, the current combined set is $(-\infty, -2] \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Now, let's re-examine the condition $g(2) = \lim_{x \to 2} g(x)$. We found $g(2) = \frac{3}{7}$. Since $-1 \le \frac{3}{7} \le 1$, the value $x=2$ is valid for the condition on $g(x)$. However, the function $g(x)$ as originally defined is not defined at $x=2$. The problem states "$g(2) = \lim_{x \to 2} g(x)$", implying that we are considering a version of $g(x)$ that is continuous at $x=2$. This means we should include $x=2$ if it satisfies the condition $-1 \le g(x) \le 1$.

The set of $x$ values for which $-1 \le g(x) \le 1$ is $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$. Let's verify this by looking at the simplified inequalities again.

From Step 4: $\frac{x+2}{2x+3} \ge 0$ (after cancelling $(x-2)$). Solution: $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$. From Step 5: $\frac{3x+4}{2x+3} \ge 0$ (after cancelling $(x-2)$). Solution: $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$.

We need $x$ such that $g(x)$ is defined and $-1 \le g(x) \le 1$. The domain of $g(x)$ is $x \in \mathbb{R} \setminus \{2, -\frac{3}{2}\}$.

Let's consider the condition $-1 \le g(x) \le 1$. This is equivalent to solving:

1. $\frac{-x^2 + 4}{2x^2 - x - 6} \le 0 \implies \frac{x^2 - 4}{2x^2 - x - 6} \ge 0 \implies \frac{(x-2)(x+2)}{(2x+3)(x-2)} \ge 0$. The critical points are $2, -2, -\frac{3}{2}$. Sign analysis of $\frac{(x-2)(x+2)}{(2x+3)(x-2)}$: Interval $(-\infty, -2]$: $\frac{(-)(-)}{(-)(-)} = +$. Valid. Interval $(-2, -\frac{3}{2})$: $\frac{(-)(+)}{(-)(-)} = -$. Invalid. Interval $(-\frac{3}{2}, 2)$: $\frac{(-)(+)}{(+)(-)} = +$. Valid. Interval $(2, \infty)$: $\frac{(+)(+)}{(+)(+)} = +$. Valid. So, for this inequality, the solution is $(-\infty, -2] \cup (-\frac{3}{2}, 2) \cup (2, \infty)$.

2. $\frac{3x^2 - 2x - 8}{2x^2 - x - 6} \ge 0 \implies \frac{(3x+4)(x-2)}{(2x+3)(x-2)} \ge 0$. The critical points are $2, -\frac{4}{3}, -\frac{3}{2}$. Sign analysis of $\frac{(3x+4)(x-2)}{(2x+3)(x-2)}$: Interval $(-\infty, -\frac{3}{2})$: $\frac{(-)(-)}{(-)(-)} = +$. Valid. Interval $(-\frac{3}{2}, -\frac{4}{3})$: $\frac{(-)(-)}{(+)(-)} = -$. Invalid. Interval $(-\frac{4}{3}, 2)$: $\frac{(+)(-)}{(+)(-)} = +$. Valid. Interval $(2, \infty)$: $\frac{(+)(+)}{(+)(+)} = +$. Valid. So, for this inequality, the solution is $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

We need the intersection of these two solution sets, AND $x$ must be in the domain of $g(x)$. The domain of $g(x)$ is $\mathbb{R} \setminus \{2, -\frac{3}{2}\}$.

Intersection of the two solution sets: Set 1: $(-\infty, -2] \cup (-\frac{3}{2}, 2) \cup (2, \infty)$ Set 2: $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, 2) \cup (2, \infty)$

Intersection: $(-\infty, -2]$ is common to both. $(-2, -\frac{3}{2})$ is not in Set 2. $(-\frac{3}{2}, -\frac{4}{3})$ is not in Set 2. $[-\frac{4}{3}, 2)$ is common to both. $(2, \infty)$ is common to both.

So, the intersection is $(-\infty, -2] \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Now, consider the domain of $g(x)$, which is $\mathbb{R} \setminus \{2, -\frac{3}{2}\}$. The points $x=2$ and $x=-\frac{3}{2}$ are already excluded from the intervals we found. We need to check if the condition $g(2) = \lim_{x \to 2} g(x)$ affects the inclusion of $x=2$. The question is about the domain of the function $f \circ g$. The domain of $f \circ g$ is $\{x \in \text{Dom}(g) \mid g(x) \in \text{Dom}(f)\}$. Dom($f$) = $[-1, 1]$. Dom($g$) = $\mathbb{R} \setminus \{2, -\frac{3}{2}\}$. So we need $x \in \mathbb{R} \setminus \{2, -\frac{3}{2}\}$ and $-1 \le g(x) \le 1$.

The values of $x$ for which $-1 \le g(x) \le 1$ are $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$. Let's re-evaluate the inequalities.

Inequality 1: $\frac{x^2 - 4}{2x^2 - x - 6} \ge 0$. Critical points: $-2, 2, -\frac{3}{2}$. Sign analysis: $(-\infty, -2]$: $\frac{(-)(-)}{(-)(-)} = +$. Valid. $x=-2$ is included. $(-2, -\frac{3}{2})$: $\frac{(-)(+)}{(-)(-)} = -$. Invalid. $(-\frac{3}{2}, 2)$: $\frac{(-)(+)}{(+)(-)} = +$. Valid. $x=-\frac{3}{2}$ is excluded. $x=2$ is excluded for now. $(2, \infty)$: $\frac{(+)(+)}{(+)(+)} = +$. Valid. $x=2$ is excluded for now.

So, from inequality 1, we get $(-\infty, -2] \cup (-\frac{3}{2}, 2) \cup (2, \infty)$.

Inequality 2: $\frac{3x^2 - 2x - 8}{2x^2 - x - 6} \ge 0$. Critical points: $-\frac{4}{3}, 2, -\frac{3}{2}$. Sign analysis: $(-\infty, -\frac{3}{2})$: $\frac{(-)(-)}{(-)(-)} = +$. Valid. $x=-\frac{3}{2}$ is excluded. $(-\frac{3}{2}, -\frac{4}{3})$: $\frac{(-)(-)}{(+)(-)} = -$. Invalid. $[-\frac{4}{3}, 2)$: $\frac{(+)(-)}{(+)(-)} = +$. Valid. $x=-\frac{4}{3}$ is included. $x=2$ is excluded for now. $(2, \infty)$: $\frac{(+)(+)}{(+)(+)} = +$. Valid. $x=2$ is excluded for now.

So, from inequality 2, we get $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Intersection of the valid regions for $-1 \le g(x) \le 1$: We need to find the values of $x$ for which both inequalities hold, and $x \ne 2$ and $x \ne -\frac{3}{2}$.

Consider the intervals: $(-\infty, -2]$: Satisfies inequality 1. Satisfies inequality 2. So, $(-\infty, -2]$ is part of the domain. $(-2, -\frac{3}{2})$: Satisfies inequality 1. Does NOT satisfy inequality 2. $(-\frac{3}{2}, -\frac{4}{3})$: Does NOT satisfy inequality 1. Does NOT satisfy inequality 2. $[-\frac{4}{3}, 2)$: Satisfies inequality 1. Satisfies inequality 2. So, $[-\frac{4}{3}, 2)$ is part of the domain. $(2, \infty)$: Satisfies inequality 1. Satisfies inequality 2. So, $(2, \infty)$ is part of the domain.

The set of $x$ for which $-1 \le g(x) \le 1$ and $x \in \text{Dom}(g)$ is $(-\infty, -2] \cup [-\frac{4}{3}, 2) \cup (2, \infty)$.

Let's re-examine the problem statement: "If $g(2) = \lim_{x \to 2} g(x)$". This implies that we are considering a function $g$ which is defined at $x=2$ with the value $\frac{3}{7}$. This modified $g$ function is then considered for the composite function $f \circ g$.

So, we need to find the domain of $f \circ g$ where $g$ is defined as: $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$ for $x \ne 2, x \ne -\frac{3}{2}$. $g(2) = \frac{3}{7}$. The domain of this modified $g$ is $\mathbb{R} \setminus \{-\frac{3}{2}\}$.

Now we need $x$ such that $x \in \text{Dom}(g)$ and $g(x) \in [-1, 1]$. So, $x \ne -\frac{3}{2}$ and $-1 \le g(x) \le 1$.

We found that the condition $-1 \le g(x) \le 1$ holds for $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$. Let's check the critical points of the inequalities again.

Inequality 1: $\frac{x^2 - 4}{2x^2 - x - 6} \ge 0$. This is $\frac{(x-2)(x+2)}{(2x+3)(x-2)} \ge 0$. If $x=2$, the expression is $\frac{0}{0}$, which is indeterminate. We need to evaluate the limit. The limit as $x \to 2$ is $\frac{2+2}{2(2)+3} = \frac{4}{7} \ge 0$. So $x=2$ satisfies this part. The values of $x$ satisfying $\frac{x+2}{2x+3} \ge 0$ are $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$. Including $x=2$ (since the limit is valid) and excluding $x=-\frac{3}{2}$ (denominator is zero), we get $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$.

Inequality 2: $\frac{3x^2 - 2x - 8}{2x^2 - x - 6} \ge 0$. This is $\frac{(3x+4)(x-2)}{(2x+3)(x-2)} \ge 0$. If $x=2$, the expression is $\frac{0}{0}$. The limit as $x \to 2$ is $\frac{3(2)+4}{2(2)+3} = \frac{10}{7} \ge 0$. So $x=2$ satisfies this part. The values of $x$ satisfying $\frac{3x+4}{2x+3} \ge 0$ are $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$. Including $x=2$ and excluding $x=-\frac{3}{2}$, we get $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$.

Now we need the intersection of these two sets, and $x \ne -\frac{3}{2}$. Set 1: $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$ Set 2: $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$

Intersection: $(-\infty, -2]$ is common. $(-2, -\frac{3}{2})$ is in Set 1 but not Set 2. $(-\frac{3}{2}, -\frac{4}{3})$ is in Set 1 but not Set 2. $[-\frac{4}{3}, \infty)$ is common to both (since $-\frac{4}{3} > -\frac{3}{2}$).

So, the intersection is $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

We must ensure that $x$ is in the domain of $g$. The domain of the modified $g$ is $\mathbb{R} \setminus \{-\frac{3}{2}\}$. The interval $[-\frac{4}{3}, \infty)$ does not contain $-\frac{3}{2}$ since $-\frac{4}{3} = -1.333...$ and $-\frac{3}{2} = -1.5$. So, the domain of $f \circ g$ is $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

This matches option (A).

Common Mistakes & Tips

• Forgetting the domain of $\sin^{-1}(x)$: Always remember that the argument of $\sin^{-1}$ must be within $[-1, 1]$.
• Handling removable discontinuities: When dealing with limits and continuity, be careful with points where both numerator and denominator are zero. Factorization and cancellation are key. The problem statement about $g(2)$ explicitly asks to consider the continuous extension of $g(x)$ at $x=2$.
• Sign analysis of inequalities: Carefully construct sign tables for rational inequalities, paying attention to where the numerator and denominator are zero or undefined.

Summary

To find the domain of the composite function $(f \circ g)(x) = f(g(x))$, we need to ensure two conditions: (1) $x$ is in the domain of $g(x)$, and (2) $g(x)$ is in the domain of $f(x)$. The domain of $f(x) = \sin^{-1}(x)$ is $[-1, 1]$. The function $g(x)$ is given as a rational function with a removable discontinuity at $x=2$, which is resolved by defining $g(2)$ as its limit. This means we consider $g(x)$ as a function defined for all real numbers except $x = -\frac{3}{2}$. We then establish the inequalities $-1 \le g(x) \le 1$ and solve them. By finding the intersection of the solution sets of these inequalities and ensuring $x$ is in the domain of $g$, we determine the domain of $f \circ g$.

The domain of $f(x) = \sin^{-1}(x)$ is $[-1, 1]$. The function $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$ has a removable discontinuity at $x=2$ and an asymptote at $x=-\frac{3}{2}$. The condition $g(2) = \lim_{x \to 2} g(x)$ implies we consider the continuous extension of $g(x)$ at $x=2$. Thus, the domain of $g(x)$ is $\mathbb{R} \setminus \{-\frac{3}{2}\}$.

We require $-1 \le g(x) \le 1$. Solving $-1 \le \frac{x^2 - x - 2}{2x^2 - x - 6} \le 1$ yields the intervals $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$. We must also ensure $x$ is in the domain of $g$, which is $x \ne -\frac{3}{2}$. The interval $[-\frac{4}{3}, \infty)$ starts at $-\frac{4}{3} \approx -1.33$, which is greater than $-\frac{3}{2} = -1.5$. So, $-\frac{3}{2}$ is not in this interval. Therefore, the domain of $f \circ g$ is $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

The final answer is $\boxed{( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)}$.