Connect with a global community of knowledgeable individuals on IDNLearn.com. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.
Sagot :
To solve the inequality [tex]\(\frac{5 - x}{x^2 - 2x - 4} \geq 0\)[/tex], let's perform a step-by-step analysis.
### Step 1: Identify Critical Points
First, we need to find the critical points where either the numerator or the denominator of the function is zero.
#### Numerator:
The numerator of the inequality is [tex]\(5 - x\)[/tex].
Set the numerator equal to zero to find the critical points:
[tex]\[ 5 - x = 0 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the numerator is zero at [tex]\(x = 5\)[/tex].
#### Denominator:
The denominator of the inequality is [tex]\(x^2 - 2x - 4\)[/tex].
Set the denominator equal to zero to find the critical points:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(c = -4\)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{20}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 2\sqrt{5}}{2} \][/tex]
[tex]\[ x = 1 \pm \sqrt{5} \][/tex]
So, the denominator is zero at [tex]\(x = 1 - \sqrt{5}\)[/tex] and [tex]\(x = 1 + \sqrt{5}\)[/tex].
### Step 2: Critical Points Summary
The critical points where either the numerator or the denominator is zero are:
[tex]\[ x = 5, \quad x = 1 - \sqrt{5}, \quad x = 1 + \sqrt{5} \][/tex]
### Step 3: Determine the Intervals and Sign Analysis
We need to analyze the sign of the expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex] in the intervals determined by the critical points. These intervals are:
1. [tex]\((-\infty, 1 - \sqrt{5})\)[/tex]
2. [tex]\((1 - \sqrt{5}, 1 + \sqrt{5})\)[/tex]
3. [tex]\((1 + \sqrt{5}, 5)\)[/tex]
4. [tex]\((5, \infty)\)[/tex]
For each interval, we choose a test point and determine the sign of the expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex].
### Step 4: Test the Intervals
1. Interval [tex]\((-\infty, 1 - \sqrt{5})\)[/tex]: Choose [tex]\(x = -10\)[/tex]:
[tex]\[ \frac{5 - (-10)}{(-10)^2 - 2(-10) - 4} = \frac{15}{100 + 20 - 4} = \frac{15}{116} > 0 \][/tex]
2. Interval [tex]\((1 - \sqrt{5}, 1 + \sqrt{5})\)[/tex]: Choose [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{5 - 0}{0^2 - 2(0) - 4} = \frac{5}{-4} = -\frac{5}{4} < 0 \][/tex]
3. Interval [tex]\((1 + \sqrt{5}, 5)\)[/tex]: Choose [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{5 - 3}{3^2 - 2(3) - 4} = \frac{2}{9 - 6 - 4} = \frac{2}{-1} = -2 < 0 \][/tex]
4. Interval [tex]\((5, \infty)\)[/tex]: Choose [tex]\(x = 10\)[/tex]:
[tex]\[ \frac{5 - 10}{10^2 - 2(10) - 4} = \frac{-5}{100 - 20 - 4} = \frac{-5}{76} < 0 \][/tex]
### Step 5: Combine the Results
The expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex] is non-negative (greater than or equal to zero) in the intervals where [tex]\((5 - x)\)[/tex] and [tex]\((x^2 - 2x - 4)\)[/tex] have the same sign or the numerator is zero (on the boundary points).
From our test points:
- The expression is positive in the interval [tex]\((-\infty, 1 - \sqrt{5})\)[/tex] and zero at [tex]\(x = 5\)[/tex].
- It is zero at [tex]\(x = 5\)[/tex].
So, the solution to the inequality [tex]\(\frac{5 - x}{x^2 - 2x - 4} \geq 0\)[/tex] is:
[tex]\[ x \in (-\infty, 1 - \sqrt{5}) \cup \{5\} \][/tex]
### Step 1: Identify Critical Points
First, we need to find the critical points where either the numerator or the denominator of the function is zero.
#### Numerator:
The numerator of the inequality is [tex]\(5 - x\)[/tex].
Set the numerator equal to zero to find the critical points:
[tex]\[ 5 - x = 0 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the numerator is zero at [tex]\(x = 5\)[/tex].
#### Denominator:
The denominator of the inequality is [tex]\(x^2 - 2x - 4\)[/tex].
Set the denominator equal to zero to find the critical points:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(c = -4\)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{20}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 2\sqrt{5}}{2} \][/tex]
[tex]\[ x = 1 \pm \sqrt{5} \][/tex]
So, the denominator is zero at [tex]\(x = 1 - \sqrt{5}\)[/tex] and [tex]\(x = 1 + \sqrt{5}\)[/tex].
### Step 2: Critical Points Summary
The critical points where either the numerator or the denominator is zero are:
[tex]\[ x = 5, \quad x = 1 - \sqrt{5}, \quad x = 1 + \sqrt{5} \][/tex]
### Step 3: Determine the Intervals and Sign Analysis
We need to analyze the sign of the expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex] in the intervals determined by the critical points. These intervals are:
1. [tex]\((-\infty, 1 - \sqrt{5})\)[/tex]
2. [tex]\((1 - \sqrt{5}, 1 + \sqrt{5})\)[/tex]
3. [tex]\((1 + \sqrt{5}, 5)\)[/tex]
4. [tex]\((5, \infty)\)[/tex]
For each interval, we choose a test point and determine the sign of the expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex].
### Step 4: Test the Intervals
1. Interval [tex]\((-\infty, 1 - \sqrt{5})\)[/tex]: Choose [tex]\(x = -10\)[/tex]:
[tex]\[ \frac{5 - (-10)}{(-10)^2 - 2(-10) - 4} = \frac{15}{100 + 20 - 4} = \frac{15}{116} > 0 \][/tex]
2. Interval [tex]\((1 - \sqrt{5}, 1 + \sqrt{5})\)[/tex]: Choose [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{5 - 0}{0^2 - 2(0) - 4} = \frac{5}{-4} = -\frac{5}{4} < 0 \][/tex]
3. Interval [tex]\((1 + \sqrt{5}, 5)\)[/tex]: Choose [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{5 - 3}{3^2 - 2(3) - 4} = \frac{2}{9 - 6 - 4} = \frac{2}{-1} = -2 < 0 \][/tex]
4. Interval [tex]\((5, \infty)\)[/tex]: Choose [tex]\(x = 10\)[/tex]:
[tex]\[ \frac{5 - 10}{10^2 - 2(10) - 4} = \frac{-5}{100 - 20 - 4} = \frac{-5}{76} < 0 \][/tex]
### Step 5: Combine the Results
The expression [tex]\(\frac{5 - x}{x^2 - 2x - 4}\)[/tex] is non-negative (greater than or equal to zero) in the intervals where [tex]\((5 - x)\)[/tex] and [tex]\((x^2 - 2x - 4)\)[/tex] have the same sign or the numerator is zero (on the boundary points).
From our test points:
- The expression is positive in the interval [tex]\((-\infty, 1 - \sqrt{5})\)[/tex] and zero at [tex]\(x = 5\)[/tex].
- It is zero at [tex]\(x = 5\)[/tex].
So, the solution to the inequality [tex]\(\frac{5 - x}{x^2 - 2x - 4} \geq 0\)[/tex] is:
[tex]\[ x \in (-\infty, 1 - \sqrt{5}) \cup \{5\} \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
