The function is given by $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$.

Domain of Inverse Sine Function

The domain of the inverse sine function, $\sin^{-1}(y)$, is $[-1, 1]$. Therefore, we must have:

$ -1 \le \frac{1}{x^2 - 2x - 2} \le 1 $

Let $y = x^2 - 2x - 2$. The inequality becomes $-1 \le \frac{1}{y} \le 1$. This requires $y \neq 0$. We split this into two inequalities:

1. $ \frac{1}{y} \le 1 \implies \frac{1-y}{y} \le 0 \implies \frac{y-1}{y} \ge 0 $. This holds if ($y \ge 1$ and $y>0$) or ($y \le 1$ and $y<0$). Thus, $y \ge 1$ or $y < 0$.
2. $ \frac{1}{y} \ge -1 \implies \frac{1+y}{y} \ge 0 $. This holds if ($y \ge -1$ and $y>0$) or ($y \le -1$ and $y<0$). Thus, $y > 0$ or $y \le -1$.

Solving Inequalities for x

We need to satisfy both conditions simultaneously. Substitute back $y = x^2 - 2x - 2$. The roots of $y=0$ are $x = 1 \pm \sqrt{3}$.

Condition 1 ($y \ge 1$ or $y < 0$):

• $x^2 - 2x - 2 \ge 1 \implies x^2 - 2x - 3 \ge 0 \implies (x-3)(x+1) \ge 0$. So, $x \in (-\infty, -1] \cup [3, \infty)$.
• $x^2 - 2x - 2 < 0$. Since roots are $1 \pm \sqrt{3}$, this holds for $x \in (1-\sqrt{3}, 1+\sqrt{3})$.
• Combining these gives: $ x \in (-\infty, -1] \cup (1-\sqrt{3}, 1+\sqrt{3}) \cup [3, \infty) $.

Condition 2 ($y > 0$ or $y \le -1$):

• $x^2 - 2x - 2 > 0$. This holds for $x \in (-\infty, 1-\sqrt{3}) \cup (1+\sqrt{3}, \infty)$.
• $x^2 - 2x - 2 \le -1 \implies x^2 - 2x - 1 \le 0$. Roots are $1 \pm \sqrt{2}$. This holds for $x \in [1-\sqrt{2}, 1+\sqrt{2}]$.
• Combining these gives: $ x \in (-\infty, 1-\sqrt{3}) \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup (1+\sqrt{3}, \infty) $.

Finding the Intersection for the Domain

We need the intersection of the sets derived from Condition 1 and Condition 2.

Intersection:

• $(-\infty, -1]$ intersects with $(-\infty, 1-\sqrt{3})$ yielding $(-\infty, -1]$.
• $(1-\sqrt{3}, 1+\sqrt{3})$ intersects with $[1-\sqrt{2}, 1+\sqrt{2}]$ yielding $[1-\sqrt{2}, 1+\sqrt{2}]$.
• $[3, \infty)$ intersects with $(1+\sqrt{3}, \infty)$ yielding $[3, \infty)$.

The complete domain is $ (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty) $.

Calculating the Sum

The domain is given in the format $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$. Comparing this with our result:

• $ \alpha = -1 $
• $ \beta = 1-\sqrt{2} $
• $ \gamma = 1+\sqrt{2} $
• $ \delta = 3 $

We need to calculate $ \alpha + \beta + \gamma + \delta $:

$ \alpha + \beta + \gamma + \delta = -1 + (1-\sqrt{2}) + (1+\sqrt{2}) + 3 $

$ = -1 + 1 - \sqrt{2} + 1 + \sqrt{2} + 3 $

$ = (-1 + 1 + 1 + 3) + (-\sqrt{2} + \sqrt{2}) $

$ = 4 + 0 = 4 $

The value of $ \alpha + \beta + \gamma + \delta $ is 4.