Key Concepts and Formulas

• Domain of Logarithmic Functions: For $\log_b A$ to be defined, we need $A > 0$, $b > 0$, and $b \neq 1$.
• Domain of Inverse Trigonometric Functions: For $\sin^{-1}(y)$ to be defined, we need $-1 \le y \le 1$.
• Quadratic Inequalities: For a quadratic $ax^2+bx+c$, if $a>0$, then $ax^2+bx+c > 0$ implies roots are real and distinct or complex. If $ax^2+bx+c < 0$, then roots must be real and distinct. The expression $x^2+4x+5$ is always positive since its discriminant is negative and the leading coefficient is positive.

Step-by-Step Solution

Finding the Domain of $f(x)$: The function is $f(x)=\log _4 \log _3 \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)$. For $f(x)$ to be defined, the arguments of each logarithm must be strictly positive.

• Step 1: Analyze the innermost term, $x^2+4x+5$. The quadratic $x^2+4x+5$ has a discriminant $\Delta = 4^2 - 4(1)(5) = 16 - 20 = -4$. Since the discriminant is negative and the coefficient of $x^2$ (which is 1) is positive, the quadratic $x^2+4x+5$ is always positive for all real values of $x$. Thus, $\log_2(x^2+4x+5)$ is defined for all $x \in \mathbb{R}$.

• Step 2: Consider the argument of $\log_7$. We need $8-\log _2\left(x^2+4 x+5\right) > 0$. This implies $\log_2\left(x^2+4 x+5\right) < 8$. Since the base of the logarithm is $2 > 1$, we can exponentiate both sides with base 2 without changing the inequality direction: $x^2+4x+5 < 2^8$ $x^2+4x+5 < 256$ $x^2+4x - 251 < 0$.

• Step 3: Solve the quadratic inequality $x^2+4x - 251 < 0$. We find the roots of $x^2+4x - 251 = 0$ using the quadratic formula: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-251)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 1004}}{2} = \frac{-4 \pm \sqrt{1020}}{2}$ $\sqrt{1020} = \sqrt{4 \times 255} = 2\sqrt{255}$. So, the roots are $x = \frac{-4 \pm 2\sqrt{255}}{2} = -2 \pm \sqrt{255}$. Since the parabola $y=x^2+4x-251$ opens upwards, the inequality $x^2+4x-251 < 0$ holds for values of $x$ between the roots. Thus, $-2 - \sqrt{255} < x < -2 + \sqrt{255}$.

• Step 4: Consider the argument of $\log_3$. We need $\log_7\left(8-\log _2\left(x^2+4 x+5\right)\right) > 0$. Since the base of the logarithm is $7 > 1$, we have: $8-\log _2\left(x^2+4 x+5\right) > 7^0$ $8-\log _2\left(x^2+4 x+5\right) > 1$ $7 > \log_2\left(x^2+4 x+5\right)$.

• Step 5: Solve the inequality $7 > \log_2\left(x^2+4 x+5\right)$. Since the base of the logarithm is $2 > 1$, we have: $2^7 > x^2+4x+5$ $128 > x^2+4x+5$ $0 > x^2+4x - 123$. So, $x^2+4x - 123 < 0$.

• Step 6: Solve the quadratic inequality $x^2+4x - 123 < 0$. We find the roots of $x^2+4x - 123 = 0$: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-123)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 492}}{2} = \frac{-4 \pm \sqrt{508}}{2}$ $\sqrt{508} = \sqrt{4 \times 127} = 2\sqrt{127}$. So, the roots are $x = \frac{-4 \pm 2\sqrt{127}}{2} = -2 \pm \sqrt{127}$. Since the parabola $y=x^2+4x-123$ opens upwards, the inequality $x^2+4x-123 < 0$ holds for values of $x$ between the roots. Thus, $-2 - \sqrt{127} < x < -2 + \sqrt{127}$.

• Step 7: Consider the argument of $\log_4$. We need $\log_3 \log_7\left(8-\log _2\left(x^2+4 x+5\right)\right) > 0$. Since the base of the logarithm is $3 > 1$, we have: $\log_7\left(8-\log _2\left(x^2+4 x+5\right)\right) > 3^0$ $\log_7\left(8-\log _2\left(x^2+4 x+5\right)\right) > 1$.

• Step 8: Solve the inequality $\log_7\left(8-\log _2\left(x^2+4 x+5\right)\right) > 1$. Since the base of the logarithm is $7 > 1$, we have: $8-\log _2\left(x^2+4 x+5\right) > 7^1$ $8-\log _2\left(x^2+4 x+5\right) > 7$ $1 > \log_2\left(x^2+4 x+5\right)$.

• Step 9: Solve the inequality $1 > \log_2\left(x^2+4 x+5\right)$. Since the base of the logarithm is $2 > 1$, we have: $2^1 > x^2+4x+5$ $2 > x^2+4x+5$ $0 > x^2+4x+3$. So, $x^2+4x+3 < 0$.

• Step 10: Solve the quadratic inequality $x^2+4x+3 < 0$. We find the roots of $x^2+4x+3 = 0$: $(x+1)(x+3) = 0$. The roots are $x = -1$ and $x = -3$. Since the parabola $y=x^2+4x+3$ opens upwards, the inequality $x^2+4x+3 < 0$ holds for values of $x$ between the roots. Thus, $-3 < x < -1$.

• Step 11: Determine the domain of $f(x)$ by intersecting the conditions. We have the following conditions:

  1. From Step 3: $-2 - \sqrt{255} < x < -2 + \sqrt{255}$.
  2. From Step 6: $-2 - \sqrt{127} < x < -2 + \sqrt{127}$.
  3. From Step 10: $-3 < x < -1$.

  Let's approximate the values: $\sqrt{255}$ is slightly less than $\sqrt{256}=16$. So, $-2 - \sqrt{255} \approx -2 - 16 = -18$ and $-2 + \sqrt{255} \approx -2 + 16 = 14$. $\sqrt{127}$ is slightly more than $\sqrt{121}=11$. So, $-2 - \sqrt{127} \approx -2 - 11 = -13$ and $-2 + \sqrt{127} \approx -2 + 11 = 9$.

  Comparing the intervals: $(-2 - \sqrt{255}, -2 + \sqrt{255})$ $(-2 - \sqrt{127}, -2 + \sqrt{127})$ $(-3, -1)$

  The most restrictive interval is $(-3, -1)$. We need to verify if the other conditions are satisfied within this interval. For $x \in (-3, -1)$: $-3 < x < -1$. Adding $-2$ to all parts: $-5 < x-2 < -3$. Squaring the absolute values: $9 < (x+2)^2 < 25$. Adding $1$: $10 < x^2+4x+5 < 26$.

  Let's check if these values satisfy the inequalities derived in Steps 3 and 6. For Step 3: $x^2+4x - 251 < 0$. If $x \in (-3, -1)$, then $x^2+4x+5$ is between 10 and 26. So $x^2+4x$ is between 5 and 21. Then $x^2+4x-251$ is between $5-251=-246$ and $21-251=-230$. These are indeed less than 0. For Step 6: $x^2+4x - 123 < 0$. If $x \in (-3, -1)$, then $x^2+4x$ is between 5 and 21. Then $x^2+4x-123$ is between $5-123=-118$ and $21-123=-102$. These are indeed less than 0.

  Therefore, the domain of $f(x)$ is $(\alpha, \beta) = (-3, -1)$. So, $\alpha = -3$ and $\beta = -1$.

Finding the Domain of $g(x)$: The function is $g(x)=\sin ^{-1}\left(\frac{7 x+10}{x-2}\right)$. For $g(x)$ to be defined, the argument of $\sin^{-1}$ must be between -1 and 1, inclusive. So, $-1 \le \frac{7 x+10}{x-2} \le 1$.

• Step 12: Solve the inequality $\frac{7 x+10}{x-2} \le 1$. $\frac{7 x+10}{x-2} - 1 \le 0$ $\frac{7 x+10 - (x-2)}{x-2} \le 0$ $\frac{6x+12}{x-2} \le 0$ $\frac{6(x+2)}{x-2} \le 0$. The critical points are $x=-2$ and $x=2$. We analyze the sign of the expression in different intervals:

  • If $x < -2$: Numerator is negative, denominator is negative. Ratio is positive.
  • If $-2 < x < 2$: Numerator is positive, denominator is negative. Ratio is negative.
  • If $x > 2$: Numerator is positive, denominator is positive. Ratio is positive. So, the inequality holds for $-2 \le x < 2$. Note that $x=2$ is excluded because it makes the denominator zero.

• Step 13: Solve the inequality $-1 \le \frac{7 x+10}{x-2}$. $\frac{7 x+10}{x-2} + 1 \ge 0$ $\frac{7 x+10 + (x-2)}{x-2} \ge 0$ $\frac{8 x+8}{x-2} \ge 0$ $\frac{8(x+1)}{x-2} \ge 0$. The critical points are $x=-1$ and $x=2$. We analyze the sign of the expression in different intervals:

  • If $x < -1$: Numerator is negative, denominator is negative. Ratio is positive.
  • If $-1 < x < 2$: Numerator is positive, denominator is negative. Ratio is negative.
  • If $x > 2$: Numerator is positive, denominator is positive. Ratio is positive. So, the inequality holds for $x \le -1$ or $x > 2$. Note that $x=2$ is excluded.

• Step 14: Determine the domain of $g(x)$ by intersecting the conditions. We have:

  1. From Step 12: $-2 \le x < 2$.
  2. From Step 13: $x \le -1$ or $x > 2$.

  The intersection of these two conditions is: $(-2 \le x < 2) \cap (x \le -1 \text{ or } x > 2)$ This gives us $-2 \le x \le -1$. Therefore, the domain of $g(x)$ is $[\gamma, \delta] = [-2, -1]$. So, $\gamma = -2$ and $\delta = -1$.

Calculating $\alpha^2+\beta^2+\gamma^2+\delta^2$: We have $\alpha = -3$, $\beta = -1$, $\gamma = -2$, $\delta = -1$. $\alpha^2 = (-3)^2 = 9$ $\beta^2 = (-1)^2 = 1$ $\gamma^2 = (-2)^2 = 4$ $\delta^2 = (-1)^2 = 1$

$\alpha^2+\beta^2+\gamma^2+\delta^2 = 9 + 1 + 4 + 1 = 15$.

Common Mistakes & Tips

• Logarithm Arguments: Always ensure the argument of a logarithm is strictly positive ($>0$), not just non-negative ($\ge 0$).
• Inequality Direction: Be careful when multiplying or dividing inequalities by negative numbers, as the direction of the inequality must be reversed. When dealing with $\log_b A$, if $0 < b < 1$, the inequality direction also reverses upon exponentiation. In this problem, all bases are greater than 1.
• Domain Intersection: When finding the domain of a composite function or when multiple conditions apply, remember to find the intersection of all valid intervals.

Summary

To find the domain of $f(x)$, we applied the conditions for logarithmic functions at each level of nesting. This involved solving a series of quadratic inequalities. For $g(x)$, we used the condition that the argument of the inverse sine function must lie between -1 and 1, which led to solving rational inequalities. By carefully determining the intervals for $x$ in both functions and then calculating the sum of the squares of the endpoints, we arrived at the final answer.

The domain of $f(x)$ was found to be $(\alpha, \beta) = (-3, -1)$, so $\alpha = -3$ and $\beta = -1$. The domain of $g(x)$ was found to be $[\gamma, \delta] = [-2, -1]$, so $\gamma = -2$ and $\delta = -1$. The required sum is $\alpha^2+\beta^2+\gamma^2+\delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2 = 9 + 1 + 4 + 1 = 15$.

The final answer is \boxed{15}.