Question

If the domain of the function $f(x) = \frac{\sqrt{x^2 - 25}}{(4 - x^2)} + \log_{10}(x^2 + 2x - 15)$is $(-\infty, \alpha) \cup [\beta, \infty)$, then $\alpha^2 + \beta^3$ is equal to:

• A. 140
• B. 175
• C. 150
• D. 125

Answer 1

Answer: (C)

150

Answer 2

To find the domain of the function, we consider the restrictions imposed by each term separately.

Step 1: Analyzing $\frac{\sqrt{x^2 - 25}}{4 - x^2}$

The term $\sqrt{x^2 - 25}$ requires:
$x^2 - 25 \geq 0 \implies x^2 \geq 25 \implies x \leq -5 \text{ or } x \geq 5$

The term $\frac{1}{4 - x^2}$ requires:
$4 - x^2 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2$

Combining these conditions:
$x \leq -5 \text{ or } x \geq 5$

Step 2: Analyzing $\log_{10}(x^2 + 2x - 15)$
For the logarithmic term to be defined:
x^2 + 2x - 15 $>$ 0

Factoring the quadratic:
(x + 5)(x - 3) $>$ 0

Using the sign chart for this inequality:
$x \in (-\infty, -5) \cup (3, \infty)$

Step 3: Combining the Conditions
The overall domain of $f(x)$ is given by the intersection of the two sets of conditions:
$x \in (-\infty, -5) \cup [5, \infty)$

Thus, $\alpha = -5$ and $\beta = 5$.

Calculating $\alpha^2 + \beta^3$
$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$

Conclusion: $\alpha^2 + \beta^3 = 150$.