To solve the inequality [tex]\(\left|x^2 + 6x + 9\right| + \left|\log \left(x^2 - x - 12\right)\right| \geq 0\)[/tex], we need to first find the critical points where the expressions inside the absolute values and the logarithm might change or be undefined.
1. Analyzing [tex]\(|x^2 + 6x + 9|\)[/tex]:
- The expression inside the absolute value is [tex]\(x^2 + 6x + 9\)[/tex].
- This can be factored as [tex]\((x + 3)^2\)[/tex].
- The critical point for this quadratic is where it equals zero: [tex]\(x + 3 = 0\)[/tex], thus [tex]\(x = -3\)[/tex].
2. Analyzing [tex]\(|\log(x^2 - x - 12)|\)[/tex]:
- The expression inside the logarithm is [tex]\(x^2 - x - 12\)[/tex].
- This can be factored as [tex]\((x - 4)(x + 3)\)[/tex].
- The critical points are where the expression inside the logarithm equals zero: [tex]\(x - 4 = 0\)[/tex] and [tex]\(x + 3 = 0\)[/tex], thus [tex]\(x = 4\)[/tex] and [tex]\(x = -3\)[/tex].
3. Considering the domain of [tex]\(\log(x^2 - x - 12)\)[/tex]:
- The logarithm is defined when [tex]\(x^2 - x - 12 > 0\)[/tex].
- Solving the inequality [tex]\((x - 4)(x + 3) > 0\)[/tex] gives us the intervals [tex]\(x \in (-\infty, -3) \cup (4, \infty)\)[/tex].
4. Combining the Analysis:
- We have identified critical points at [tex]\(x = -3\)[/tex] and [tex]\(x = 4\)[/tex].
- For [tex]\(x\)[/tex] to make the given inequality [tex]\(\left|x^2 + 6x + 9\right| + \left|\log \left(x^2 - x - 12\right)\right| \geq 0\)[/tex] true, it's important that [tex]\(|\log(x^2 - x - 12)|\)[/tex] is defined. Hence, we discard intervals where the logarithm is not defined.
- The resulting valid intervals based on our analysis are [tex]\(x \in (-\infty, -3]\)[/tex] and [tex]\(x \in (4, \infty)\)[/tex].
Therefore, the values of [tex]\(x\)[/tex] that satisfy the inequality are:
[tex]\[
x \in (-\infty, -3] \cup (4, \infty).
\][/tex]
The correct answer is (c).
