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To solve the inequality [tex]\(\frac{x-1}{x+4} \leq 0\)[/tex], we need to analyze the critical points where the numerator and the denominator are zero or undefined, and determine where the expression is less than or equal to zero.
1. Find the Critical Points:
- Set the numerator [tex]\(x - 1 = 0\)[/tex], so [tex]\(x = 1\)[/tex].
- Set the denominator [tex]\(x + 4 = 0\)[/tex], so [tex]\(x = -4\)[/tex].
2. Determine Intervals:
The critical points divide the real number line into three intervals:
[tex]\[ (-\infty, -4), \quad (-4, 1), \quad \text{and} \quad (1, \infty) \][/tex]
3. Test Each Interval:
We need to determine the sign of the expression within each interval.
- For [tex]\(x \in (-\infty, -4)\)[/tex]:
Choose a test point, such as [tex]\(x = -5\)[/tex]:
[tex]\[ \frac{-5 - 1}{-5 + 4} = \frac{-6}{-1} = 6 > 0 \][/tex]
The expression is positive in this interval.
- For [tex]\(x \in (-4, 1)\)[/tex]:
Choose a test point, such as [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{0 - 1}{0 + 4} = \frac{-1}{4} = -0.25 < 0 \][/tex]
The expression is negative in this interval.
- For [tex]\(x \in (1, \infty)\)[/tex]:
Choose a test point, such as [tex]\(x = 2\)[/tex]:
[tex]\[ \frac{2 - 1}{2 + 4} = \frac{1}{6} > 0 \][/tex]
The expression is positive in this interval.
4. Include Boundary Points:
- At [tex]\(x = 1\)[/tex], the expression [tex]\(\frac{1 - 1}{1 + 4} = \frac{0}{5} = 0\)[/tex], which satisfies [tex]\(\leq 0\)[/tex].
- At [tex]\(x = -4\)[/tex], the expression is undefined because the denominator is zero, thus [tex]\(x = -4\)[/tex] is not included in the solution.
5. Combine Results:
The inequality [tex]\(\frac{x - 1}{x + 4} \leq 0\)[/tex] holds in the interval where the expression is negative or zero. This interval is:
[tex]\[ (-4, 1] \][/tex]
Thus, the solution in interval notation is:
[tex]\[ (-4, 1] \][/tex]
1. Find the Critical Points:
- Set the numerator [tex]\(x - 1 = 0\)[/tex], so [tex]\(x = 1\)[/tex].
- Set the denominator [tex]\(x + 4 = 0\)[/tex], so [tex]\(x = -4\)[/tex].
2. Determine Intervals:
The critical points divide the real number line into three intervals:
[tex]\[ (-\infty, -4), \quad (-4, 1), \quad \text{and} \quad (1, \infty) \][/tex]
3. Test Each Interval:
We need to determine the sign of the expression within each interval.
- For [tex]\(x \in (-\infty, -4)\)[/tex]:
Choose a test point, such as [tex]\(x = -5\)[/tex]:
[tex]\[ \frac{-5 - 1}{-5 + 4} = \frac{-6}{-1} = 6 > 0 \][/tex]
The expression is positive in this interval.
- For [tex]\(x \in (-4, 1)\)[/tex]:
Choose a test point, such as [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{0 - 1}{0 + 4} = \frac{-1}{4} = -0.25 < 0 \][/tex]
The expression is negative in this interval.
- For [tex]\(x \in (1, \infty)\)[/tex]:
Choose a test point, such as [tex]\(x = 2\)[/tex]:
[tex]\[ \frac{2 - 1}{2 + 4} = \frac{1}{6} > 0 \][/tex]
The expression is positive in this interval.
4. Include Boundary Points:
- At [tex]\(x = 1\)[/tex], the expression [tex]\(\frac{1 - 1}{1 + 4} = \frac{0}{5} = 0\)[/tex], which satisfies [tex]\(\leq 0\)[/tex].
- At [tex]\(x = -4\)[/tex], the expression is undefined because the denominator is zero, thus [tex]\(x = -4\)[/tex] is not included in the solution.
5. Combine Results:
The inequality [tex]\(\frac{x - 1}{x + 4} \leq 0\)[/tex] holds in the interval where the expression is negative or zero. This interval is:
[tex]\[ (-4, 1] \][/tex]
Thus, the solution in interval notation is:
[tex]\[ (-4, 1] \][/tex]
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