Answer :
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]
