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Step-by-Step Solution
Step 1: Understand the Composition
We have two functions:
$f(x) = \sin^{-1}(x)$
$g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6}$
The composition $f \circ g$ is given by $f(g(x)) = \sin^{-1}\bigl(g(x)\bigr)$. For $f(g(x))$ to be defined (i.e., for $\sin^{-1}$ to be real-valued), the expression $g(x)$ must lie within the closed interval $[-1,1]$:
$\displaystyle |g(x)| \leq 1.$
Step 2: Determine the Value of $g(2)$ via the Given Limit
The question states that $g(2)$ is defined by the limit as $x$ approaches 2:
$\displaystyle g(2) \;=\; \lim_{x \to 2} g(x) \;=\; \lim_{x \to 2} \dfrac{(x^2 - x - 2)}{(2x^2 - x - 6)}.$
Factor the numerator and the denominator to see if the factor $(x-2)$ can be canceled:
Numerator: $x^2 - x - 2 = (x-2)(x+1)$.
Denominator: $2x^2 - x - 6 = (2x+3)(x-2)$.
Thus, for $x \neq 2$,
$\displaystyle g(x) = \dfrac{(x-2)(x+1)}{(2x+3)(x-2)} = \dfrac{x+1}{2x+3} \quad (\text{when } x \neq 2).$
Hence,
$\displaystyle g(2) = \lim_{x \to 2} \dfrac{x+1}{2x+3} = \dfrac{2+1}{2\cdot 2 + 3} = \dfrac{3}{7}.$
Step 3: Express $g(x)$ for $x \neq 2$
Since $g(x)$ becomes $\dfrac{x+1}{2x+3}$ for $x \neq 2$, we use this simplified form to find when $|g(x)| \le 1$:
$\displaystyle \left|\dfrac{x+1}{2x+3}\right| \;\le\; 1.$
We will break this into two inequalities:
$\displaystyle \dfrac{x+1}{2x+3} \;\le\; 1,$
$\displaystyle \dfrac{x+1}{2x+3} \;\ge\; -1.$
Step 4: Solve the Inequalities
4.1 Inequality $\dfrac{x+1}{2x+3} \;\le\; 1$
Rewrite:
$\displaystyle \dfrac{x+1}{2x+3} - 1 \;\le\; 0.$
Combine into a single fraction:
$\displaystyle \dfrac{x+1 - (2x+3)}{2x+3} = \dfrac{x+1 - 2x - 3}{2x+3} = \dfrac{-x - 2}{2x+3} \;\le\; 0.$
So we need
$\displaystyle \dfrac{-\,\bigl(x + 2\bigr)}{2x+3} \;\le\; 0.$
This is equivalent to
$\displaystyle \dfrac{x + 2}{2x+3} \;\ge\; 0.$
4.2 Inequality $\dfrac{x+1}{2x+3} \;\ge\; -1$
Rewrite:
$\displaystyle \dfrac{x+1}{2x+3} + 1 \;\ge\; 0.$
Combine into a single fraction:
$\displaystyle \dfrac{x+1 + (2x+3)}{2x+3} = \dfrac{x+1 + 2x + 3}{2x+3} = \dfrac{3x + 4}{2x+3} \;\ge\; 0.$
So we need
$\displaystyle \dfrac{3x + 4}{2x+3} \;\ge\; 0.$
Step 5: Combine Both Inequalities
We must have both:
$\displaystyle \dfrac{x+2}{2x+3} \;\ge\; 0 \quad \text{and} \quad \dfrac{3x+4}{2x+3} \;\ge\; 0.$
Solve these conditions carefully, noting where the numerator and denominator each become zero or undefined. The combined solution set (after analyzing the signs on each interval) is:
$\displaystyle (-\infty,\, -2] \;\cup\; \bigl[-\tfrac{4}{3},\, \infty\bigr).$
Additionally, we must exclude points that make the original denominator $2x^2 - x - 6 = 0$ or cause any conflict for $x=2$ if not properly handled. However, the continuous analysis and the limit definition at $x=2$ confirm the final combined domain is consistent with
$\displaystyle (-\infty,\, -2] \;\cup\; \Bigl[-\tfrac{4}{3},\, \infty\Bigr).$
Step 6: Final Domain for $f \circ g$
Thus, the domain of $f(g(x)) = \sin^{-1}\bigl(g(x)\bigr)$ is:
$\displaystyle \boxed{(-\infty,\, -2] \;\cup\;\Bigl[-\tfrac{4}{3},\, \infty\Bigr).}$
