Key Concepts and Formulas

• Domain of $\sqrt{g(x)}$: The expression inside the square root must be non-negative, i.e., $g(x) \ge 0$.
• Domain of $\frac{1}{h(x)}$: The denominator cannot be zero, i.e., $h(x) \ne 0$.
• Domain of $\log_b(k(x))$: The argument of the logarithm must be strictly positive, i.e., $k(x) > 0$. The base $b$ must be positive and not equal to 1.
• Intersection of Domains: The domain of a composite function is the intersection of the domains of its individual components.

Step-by-Step Solution

Step 1: Analyze the square root term $\sqrt{x^2-25}$ For $\sqrt{x^2-25}$ to be defined, the expression inside the square root must be non-negative. $x^2 - 25 \ge 0$ Factoring the quadratic expression, we get: $(x-5)(x+5) \ge 0$ The roots are $x = 5$ and $x = -5$. Analyzing the sign of the quadratic $(x-5)(x+5)$, we find that it is non-negative when $x \le -5$ or $x \ge 5$. Thus, the domain for this term is $D_1 = (-\infty, -5] \cup [5, \infty)$.

Step 2: Analyze the denominator term $(4-x^2)$ For the fraction $\frac{\sqrt{x^2-25}}{4-x^2}$ to be defined, the denominator cannot be zero. $4 - x^2 \ne 0$ Factoring the difference of squares, we get: $(2-x)(2+x) \ne 0$ This implies $x \ne 2$ and $x \ne -2$. Thus, the restriction from the denominator is $D_2 = \mathbb{R} \setminus \{-2, 2\}$.

Step 3: Analyze the logarithm term $\log_{10}(x^2+2x-15)$ For $\log_{10}(x^2+2x-15)$ to be defined, the argument of the logarithm must be strictly positive. $x^2 + 2x - 15 > 0$ Factoring the quadratic expression, we get: $(x+5)(x-3) > 0$ The roots are $x = -5$ and $x = 3$. Analyzing the sign of the quadratic $(x+5)(x-3)$, we find that it is strictly positive when $x < -5$ or $x > 3$. Thus, the domain for this term is $D_3 = (-\infty, -5) \cup (3, \infty)$.

Step 4: Determine the overall domain of $f(x)$ The domain of $f(x)$ is the intersection of the domains $D_1$, $D_2$, and $D_3$. First, let's find the intersection of $D_1$ and $D_3$: $D_1 = (-\infty, -5] \cup [5, \infty)$ $D_3 = (-\infty, -5) \cup (3, \infty)$

The intersection of $(-\infty, -5]$ and $(-\infty, -5)$ is $(-\infty, -5)$ because $-5$ is excluded from $D_3$. The intersection of $[5, \infty)$ and $(3, \infty)$ is $[5, \infty)$ because $[5, \infty)$ is a subset of $(3, \infty)$. So, $D_1 \cap D_3 = (-\infty, -5) \cup [5, \infty)$.

Now, we intersect this result with $D_2 = \mathbb{R} \setminus \{-2, 2\}$. The set $(-\infty, -5) \cup [5, \infty)$ does not contain the values $-2$ or $2$. Therefore, excluding $-2$ and $2$ does not change the set. The overall domain of $f(x)$ is $D = (-\infty, -5) \cup [5, \infty)$.

Step 5: Identify $\alpha$ and $\beta$ The problem states that the domain of $f(x)$ is $(-\infty, \alpha) \cup [\beta, \infty)$. Comparing this with our derived domain $D = (-\infty, -5) \cup [5, \infty)$, we can conclude that: $\alpha = -5$ $\beta = 5$

Step 6: Calculate $\alpha^2 + \beta^3$ We need to find the value of $\alpha^2 + \beta^3$. $\alpha^2 + \beta^3 = (-5)^2 + (5)^3$ $= 25 + 125$ $= 150$

Common Mistakes & Tips

• Strict vs. Non-strict Inequalities: Be careful to use $\ge 0$ for square roots and $> 0$ for logarithms. This difference is crucial for determining whether endpoints are included in the domain.
• Intersection of Intervals: When intersecting intervals, ensure that all conditions are met simultaneously. Visualizing the intervals on a number line can be very helpful.
• Excluding Points: Remember to exclude points that make a denominator zero or are not allowed in the domain of a logarithm.

Summary To find the domain of the given function, we analyzed the conditions imposed by the square root, the denominator, and the logarithm separately. The square root required $x^2 - 25 \ge 0$, leading to $x \in (-\infty, -5] \cup [5, \infty)$. The denominator $4-x^2$ could not be zero, so $x \ne \pm 2$. The logarithm $x^2+2x-15$ required $x^2+2x-15 > 0$, leading to $x \in (-\infty, -5) \cup (3, \infty)$. The intersection of these conditions yielded the domain $(-\infty, -5) \cup [5, \infty)$. Comparing this to the given form $(-\infty, \alpha) \cup [\beta, \infty)$, we found $\alpha = -5$ and $\beta = 5$. Finally, calculating $\alpha^2 + \beta^3 = (-5)^2 + (5)^3 = 25 + 125 = 150$.

The final answer is $\boxed{150}$.