If the domain of the function

$f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^2+\beta^3$ is equal to :

<p>To find the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$ we need to consider the domain conditions for both the square root function and the logarithmic function.</p> <p>The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so $x^2 - 25 \geq 0.$ This inequality is satisfied when $x \leq -5 \quad \text{or} \quad x \geq 5.$</p> <p>The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have $4 - x^2 \neq 0.$ This inequality is violated when $x = \pm2.$</p> <p>Combining these conditions gives us the domain for the rational part of the function: $x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$</p> <p>Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive: $x^2 + 2x - 15 &gt; 0.$ This is a quadratic inequality, which we can factor to find the solution: $(x+5)(x-3) &gt; 0.$ From this, we see that the inequality is satisfied for $x &lt; -5 \quad \text{or} \quad x &gt; 3.$</p> <p>The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us: $x \in (-\infty, -5) \cup (5, \infty),$</p> <p>Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that $\alpha = -5 \quad \text{and} \quad \beta = 5.$</p> <p>We calculate $\alpha^2 + \beta^3$ as follows: $\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$</p> <p>So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is: Option D $150$.</p>