IDNLearn.com provides a comprehensive solution for all your question and answer needs. Join our community to receive prompt and reliable responses to your questions from experienced professionals.
Sagot :
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).
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).
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.