To find the domains of the functions \(f(x) = \log_4\log_3\log_7 (8-\log_2(x^2+4x+5))\) and \(g(x) = \sin^{-1} \left(\frac{7x + 10}{x-2}\right)\), we need to ensure the arguments of all logarithmic and inverse trigonometric functions are defined.
Step 1: Domain of \(f(x)\)
1. For the innermost function, \(\log_2(x^2+4x+5)\):
- We need \(x^2 + 4x + 5 > 0\). The discriminant \(b^2 - 4ac = 16 - 20 = -4\), which is negative, indicating the quadratic is always positive.
2. The outer function \(8-\log_2(x^2+4x+5)\):
- Requires \(8-\log_2(x^2+4x+5) > 1\), hence \(\log_2(x^2+4x+5) < 8\). Converting to exponential form, \(x^2 + 4x + 5 < 2^8\).
- Simplifying gives \(x^2 + 4x - 251 < 0\), factoring to \((x - 13)(x + 19) < 0\). Using a number line, this holds for \(x \in (-19, 13)\).
3. Finally, for \(\log_4\log_3(y)\) to be defined:
- The argument must be positive: \(\log_3\log_7(8 - \log_2(x^2+4x+5)) > 0\), implying \(8 - \log_2(x^2+4x+5) > 7\), thus \(\log_2(x^2+4x+5) < 1\). Hence, \(x^2 + 4x + 5 < 2\), or (x + 1)(x + 3) < 0.
- The solutions are \(x \in (-3, -1)\).
The domain of \(f(x)\) is \((\alpha, \beta) = (-3, -1)\).
Step 2: Domain of \(g(x)\)
1. For \(\sin^{-1}\left(\frac{7x+10}{x-2}\right)\) to be defined, the argument must be in \([-1, 1]\):
- Hence, \(-1 \leq \frac{7x+10}{x-2} \leq 1\).
- Consider \(\frac{7x+10}{x-2} \geq -1\): This simplifies to \(7x + 10 \geq -x + 2\) or \(8x \geq -8\). So, \(x \geq -1\).
- Consider \(\frac{7x+10}{x-2} \leq 1\): This simplifies to \(7x + 10 \leq x - 2\) or \(6x \leq -12\), giving \(x \leq -2\).
Hence, intersection of both gives \(x \in [-2, -1]\).
The domain of \(g(x)\) is \([\gamma, \delta] = [-2, -1]\).
Final Calculation
Calculate \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\):
- \(\alpha = -3, \beta = -1, \gamma = -2, \delta = -1\)
- \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2 = 9 + 1 + 4 + 1 = 15\)
Thus, the final answer is 15.
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