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To solve the inequality [tex]\(\frac{(x-3)(x+5)}{(x-4)^2} \geq 0\)[/tex], we will proceed through a systematic analysis of the factors involved. Here are the steps:
1. Identify the Critical Points:
The critical points are where the numerator is zero or where the denominator is zero.
- The numerator [tex]\((x-3)(x+5)\)[/tex] is zero when [tex]\(x = 3\)[/tex] or [tex]\(x = -5\)[/tex].
- The denominator [tex]\((x-4)^2\)[/tex] is zero when [tex]\(x = 4\)[/tex].
2. Determine the behavior around each critical point.
We will test intervals around these points to see where the inequality holds. The critical points split the real number line into intervals: [tex]\((-\infty, -5)\)[/tex], [tex]\((-5, 3)\)[/tex], [tex]\((3, 4)\)[/tex], and [tex]\((4, \infty)\)[/tex].
3. Analyze each interval:
Consider the sign of each factor [tex]\((x-3)\)[/tex], [tex]\((x+5)\)[/tex], and [tex]\((x-4)^2\)[/tex] in these intervals:
- Interval [tex]\( (-\infty, -5) \)[/tex]:
- [tex]\( x < -5\)[/tex]
- [tex]\((x-3) < 0\)[/tex]
- [tex]\((x+5) < 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
- Interval [tex]\((-5, 3)\)[/tex]:
- [tex]\(-5 < x < 3\)[/tex]
- [tex]\((x-3) < 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is negative in this interval because the numerator is negative and the denominator is positive.
- Interval [tex]\((3, 4)\)[/tex]:
- [tex]\(3 < x < 4\)[/tex]
- [tex]\((x-3) > 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
- Interval [tex]\((4, \infty)\)[/tex]:
- [tex]\( x > 4\)[/tex]
- [tex]\((x-3) > 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
4. Determine the inclusion of critical points:
- At [tex]\( x = -5 \)[/tex]:
[tex]\(\frac{(x-3)(x+5)}{(x-4)^2} = \frac{(-8)(0)}{(-9)^2} = 0\)[/tex], thus it satisfies [tex]\(\geq 0\)[/tex]. [tex]\( -5 \)[/tex] is included in the solution.
- At [tex]\( x = 3 \)[/tex]:
[tex]\(\frac{(x-3)(x+5)}{(x-4)^2} = \frac{(0)(8)}{1^2} = 0\)[/tex], thus it satisfies [tex]\(\geq 0\)[/tex]. [tex]\( 3 \)[/tex] is included in the solution.
- At [tex]\( x = 4 \)[/tex]:
[tex]\((x-4)^2 = 0\)[/tex] which makes the denominator zero and the expression undefined. [tex]\( 4 \)[/tex] cannot be included in the solution.
5. Combine the results:
- We include [tex]\((-∞, -5]\)[/tex] because the expression is positive and [tex]\( -5 \)[/tex] is included.
- [tex]\((-5, 3)\)[/tex] is excluded because the expression is negative.
- We include [tex]\([3, 4)\)[/tex] because the expression is positive in the interval [tex]\((3, 4)\)[/tex] and [tex]\(3\)[/tex] is included.
- We include [tex]\((4, ∞)\)[/tex] because the expression is positive in this interval and [tex]\(4\)[/tex] is excluded.
Thus, the solution to the inequality is:
[tex]\[ (-∞, -5] ∪ [3, 4) ∪ (4, ∞) \][/tex]
1. Identify the Critical Points:
The critical points are where the numerator is zero or where the denominator is zero.
- The numerator [tex]\((x-3)(x+5)\)[/tex] is zero when [tex]\(x = 3\)[/tex] or [tex]\(x = -5\)[/tex].
- The denominator [tex]\((x-4)^2\)[/tex] is zero when [tex]\(x = 4\)[/tex].
2. Determine the behavior around each critical point.
We will test intervals around these points to see where the inequality holds. The critical points split the real number line into intervals: [tex]\((-\infty, -5)\)[/tex], [tex]\((-5, 3)\)[/tex], [tex]\((3, 4)\)[/tex], and [tex]\((4, \infty)\)[/tex].
3. Analyze each interval:
Consider the sign of each factor [tex]\((x-3)\)[/tex], [tex]\((x+5)\)[/tex], and [tex]\((x-4)^2\)[/tex] in these intervals:
- Interval [tex]\( (-\infty, -5) \)[/tex]:
- [tex]\( x < -5\)[/tex]
- [tex]\((x-3) < 0\)[/tex]
- [tex]\((x+5) < 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
- Interval [tex]\((-5, 3)\)[/tex]:
- [tex]\(-5 < x < 3\)[/tex]
- [tex]\((x-3) < 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is negative in this interval because the numerator is negative and the denominator is positive.
- Interval [tex]\((3, 4)\)[/tex]:
- [tex]\(3 < x < 4\)[/tex]
- [tex]\((x-3) > 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
- Interval [tex]\((4, \infty)\)[/tex]:
- [tex]\( x > 4\)[/tex]
- [tex]\((x-3) > 0\)[/tex]
- [tex]\((x+5) > 0\)[/tex]
- [tex]\((x-4)^2 > 0\)[/tex]
The expression [tex]\(\frac{(x-3)(x+5)}{(x-4)^2}\)[/tex] is positive in this interval because both the numerator and the denominator are positive.
4. Determine the inclusion of critical points:
- At [tex]\( x = -5 \)[/tex]:
[tex]\(\frac{(x-3)(x+5)}{(x-4)^2} = \frac{(-8)(0)}{(-9)^2} = 0\)[/tex], thus it satisfies [tex]\(\geq 0\)[/tex]. [tex]\( -5 \)[/tex] is included in the solution.
- At [tex]\( x = 3 \)[/tex]:
[tex]\(\frac{(x-3)(x+5)}{(x-4)^2} = \frac{(0)(8)}{1^2} = 0\)[/tex], thus it satisfies [tex]\(\geq 0\)[/tex]. [tex]\( 3 \)[/tex] is included in the solution.
- At [tex]\( x = 4 \)[/tex]:
[tex]\((x-4)^2 = 0\)[/tex] which makes the denominator zero and the expression undefined. [tex]\( 4 \)[/tex] cannot be included in the solution.
5. Combine the results:
- We include [tex]\((-∞, -5]\)[/tex] because the expression is positive and [tex]\( -5 \)[/tex] is included.
- [tex]\((-5, 3)\)[/tex] is excluded because the expression is negative.
- We include [tex]\([3, 4)\)[/tex] because the expression is positive in the interval [tex]\((3, 4)\)[/tex] and [tex]\(3\)[/tex] is included.
- We include [tex]\((4, ∞)\)[/tex] because the expression is positive in this interval and [tex]\(4\)[/tex] is excluded.
Thus, the solution to the inequality is:
[tex]\[ (-∞, -5] ∪ [3, 4) ∪ (4, ∞) \][/tex]
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