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To solve the inequality [tex]\((x^2 + 9)(1 - x)(x + 1) \leq 0\)[/tex], we need to determine the values of [tex]\(x\)[/tex] for which the expression is less than or equal to zero. We'll break it down step by step:
1. Analyze the quadratic factor [tex]\(x^2 + 9\)[/tex]:
- The quadratic [tex]\(x^2 + 9\)[/tex] is always positive because [tex]\(x^2\)[/tex] is non-negative for all real numbers, and adding 9 ensures it is always greater than or equal to 9.
- Therefore, [tex]\(x^2 + 9 > 0\)[/tex] for all [tex]\(x \in \mathbb{R}\)[/tex].
2. Identify the critical points from the linear factors:
- The critical points arise where the expression [tex]\( (1 - x)(x + 1) \)[/tex] changes sign. These critical points happen where each linear factor equals zero:
- [tex]\(1 - x = 0 \Rightarrow x = 1\)[/tex]
- [tex]\(x + 1 = 0 \Rightarrow x = -1\)[/tex]
3. Test intervals determined by the critical points:
- The critical points divide the number line into intervals. We need to determine the sign of [tex]\((1 - x)(x + 1)\)[/tex] in each interval:
- Intervals: [tex]\((-∞, -1)\)[/tex], [tex]\((-1, 1)\)[/tex], [tex]\((1, ∞)\)[/tex]
4. Determine the sign in each interval:
- For [tex]\(x \in (-∞, -1)\)[/tex]:
- [tex]\(1 - x > 0\)[/tex] because [tex]\(x\)[/tex] is less than [tex]\(-1\)[/tex]
- [tex]\(x + 1 < 0\)[/tex] because [tex]\(x\)[/tex] is less than [tex]\(-1\)[/tex]
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is negative as a product of a positive and a negative number.
- For [tex]\(x \in (-1, 1)\)[/tex]:
- [tex]\(1 - x > 0\)[/tex] because [tex]\(x\)[/tex] is less than 1
- [tex]\(x + 1 > 0\)[/tex] because [tex]\(x\)[/tex] is greater than [tex]\(-1\)[/tex]
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is positive as a product of two positive numbers.
- For [tex]\(x \in (1, ∞)\)[/tex]:
- [tex]\(1 - x < 0\)[/tex] because [tex]\(x\)[/tex] is greater than 1
- [tex]\(x + 1 > 0\)[/tex] because [tex]\(x\)[/tex] is greater than 1
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is negative as a product of a negative and a positive number.
5. Include the endpoints:
- We examine if the endpoints [tex]\(-1\)[/tex] and [tex]\(1\)[/tex] satisfy the inequality:
- At [tex]\(x = -1\)[/tex]: [tex]\((x^2 + 9)(1 - (-1))((-1) + 1) = (10)(2)(0) = 0\)[/tex], so it satisfies [tex]\(\leq 0\)[/tex].
- At [tex]\(x = 1\)[/tex]: [tex]\((x^2 + 9)(1 - 1)(1 + 1) = (10)(0)(2) = 0\)[/tex], so it satisfies [tex]\(\leq 0\)[/tex].
Given the intervals and endpoint analysis, we conclude that the inequality is satisfied in the intervals [tex]\((-∞, -1]\)[/tex] and [tex]\([1, ∞)\)[/tex].
Therefore, the solution to the inequality [tex]\((x^2 + 9)(1 - x)(x + 1) \leq 0\)[/tex] can be written as:
[tex]\[ (-∞, -1] \cup [1, ∞) \][/tex]
1. Analyze the quadratic factor [tex]\(x^2 + 9\)[/tex]:
- The quadratic [tex]\(x^2 + 9\)[/tex] is always positive because [tex]\(x^2\)[/tex] is non-negative for all real numbers, and adding 9 ensures it is always greater than or equal to 9.
- Therefore, [tex]\(x^2 + 9 > 0\)[/tex] for all [tex]\(x \in \mathbb{R}\)[/tex].
2. Identify the critical points from the linear factors:
- The critical points arise where the expression [tex]\( (1 - x)(x + 1) \)[/tex] changes sign. These critical points happen where each linear factor equals zero:
- [tex]\(1 - x = 0 \Rightarrow x = 1\)[/tex]
- [tex]\(x + 1 = 0 \Rightarrow x = -1\)[/tex]
3. Test intervals determined by the critical points:
- The critical points divide the number line into intervals. We need to determine the sign of [tex]\((1 - x)(x + 1)\)[/tex] in each interval:
- Intervals: [tex]\((-∞, -1)\)[/tex], [tex]\((-1, 1)\)[/tex], [tex]\((1, ∞)\)[/tex]
4. Determine the sign in each interval:
- For [tex]\(x \in (-∞, -1)\)[/tex]:
- [tex]\(1 - x > 0\)[/tex] because [tex]\(x\)[/tex] is less than [tex]\(-1\)[/tex]
- [tex]\(x + 1 < 0\)[/tex] because [tex]\(x\)[/tex] is less than [tex]\(-1\)[/tex]
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is negative as a product of a positive and a negative number.
- For [tex]\(x \in (-1, 1)\)[/tex]:
- [tex]\(1 - x > 0\)[/tex] because [tex]\(x\)[/tex] is less than 1
- [tex]\(x + 1 > 0\)[/tex] because [tex]\(x\)[/tex] is greater than [tex]\(-1\)[/tex]
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is positive as a product of two positive numbers.
- For [tex]\(x \in (1, ∞)\)[/tex]:
- [tex]\(1 - x < 0\)[/tex] because [tex]\(x\)[/tex] is greater than 1
- [tex]\(x + 1 > 0\)[/tex] because [tex]\(x\)[/tex] is greater than 1
- Thus, [tex]\((1 - x)(x + 1)\)[/tex] is negative as a product of a negative and a positive number.
5. Include the endpoints:
- We examine if the endpoints [tex]\(-1\)[/tex] and [tex]\(1\)[/tex] satisfy the inequality:
- At [tex]\(x = -1\)[/tex]: [tex]\((x^2 + 9)(1 - (-1))((-1) + 1) = (10)(2)(0) = 0\)[/tex], so it satisfies [tex]\(\leq 0\)[/tex].
- At [tex]\(x = 1\)[/tex]: [tex]\((x^2 + 9)(1 - 1)(1 + 1) = (10)(0)(2) = 0\)[/tex], so it satisfies [tex]\(\leq 0\)[/tex].
Given the intervals and endpoint analysis, we conclude that the inequality is satisfied in the intervals [tex]\((-∞, -1]\)[/tex] and [tex]\([1, ∞)\)[/tex].
Therefore, the solution to the inequality [tex]\((x^2 + 9)(1 - x)(x + 1) \leq 0\)[/tex] can be written as:
[tex]\[ (-∞, -1] \cup [1, ∞) \][/tex]
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