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Answer :
To solve the inequality
[tex]$$\frac{x+4}{2x-1} < 0,$$[/tex]
we first identify the critical points by setting the numerator and the denominator equal to zero.
1. The numerator is zero when
[tex]$$x + 4 = 0 \quad \Longrightarrow \quad x = -4.$$[/tex]
2. The denominator is zero when
[tex]$$2x - 1 = 0 \quad \Longrightarrow \quad x = \frac{1}{2}.$$[/tex]
These critical points divide the number line into three intervals:
- Interval I: [tex]\(x < -4\)[/tex]
- Interval II: [tex]\(-4 < x < \frac{1}{2}\)[/tex]
- Interval III: [tex]\(x > \frac{1}{2}\)[/tex]
Next, we test a point in each interval to determine the sign of the expression.
Interval I: [tex]\(x < -4\)[/tex]
Choose [tex]\(x = -5\)[/tex]. Then:
[tex]\[
x+4 = -5+4 = -1, \quad 2x-1 = 2(-5)-1 = -10-1 = -11.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{-1}{-11} = \frac{1}{11} > 0.
\][/tex]
The expression is positive in this interval.
Interval II: [tex]\(-4 < x < \frac{1}{2}\)[/tex]
Choose [tex]\(x = 0\)[/tex]. Then:
[tex]\[
x+4 = 0+4 = 4, \quad 2x-1 = 2(0)-1 = -1.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{4}{-1} = -4 < 0.
\][/tex]
The expression is negative in this interval.
Interval III: [tex]\(x > \frac{1}{2}\)[/tex]
Choose [tex]\(x = 1\)[/tex]. Then:
[tex]\[
x+4 = 1+4 = 5, \quad 2x-1 = 2(1)-1 = 1.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{5}{1} = 5 > 0.
\][/tex]
The expression is positive in this interval.
Since we are looking for the part where
[tex]$$\frac{x+4}{2x-1} < 0,$$[/tex]
only the interval [tex]\(-4 < x < \frac{1}{2}\)[/tex] satisfies the inequality.
Finally, notice that the inequality is strict (using “[tex]\(<\)[/tex]”), which means the endpoints [tex]\(x = -4\)[/tex] and [tex]\(x = \frac{1}{2}\)[/tex] are not included (at [tex]\(x = -4\)[/tex], the expression equals 0; at [tex]\(x = \frac{1}{2}\)[/tex], the expression is undefined).
Therefore, the solution to the inequality is:
[tex]$$-4 < x < \frac{1}{2}.$$[/tex]
This corresponds to option 4.
[tex]$$\frac{x+4}{2x-1} < 0,$$[/tex]
we first identify the critical points by setting the numerator and the denominator equal to zero.
1. The numerator is zero when
[tex]$$x + 4 = 0 \quad \Longrightarrow \quad x = -4.$$[/tex]
2. The denominator is zero when
[tex]$$2x - 1 = 0 \quad \Longrightarrow \quad x = \frac{1}{2}.$$[/tex]
These critical points divide the number line into three intervals:
- Interval I: [tex]\(x < -4\)[/tex]
- Interval II: [tex]\(-4 < x < \frac{1}{2}\)[/tex]
- Interval III: [tex]\(x > \frac{1}{2}\)[/tex]
Next, we test a point in each interval to determine the sign of the expression.
Interval I: [tex]\(x < -4\)[/tex]
Choose [tex]\(x = -5\)[/tex]. Then:
[tex]\[
x+4 = -5+4 = -1, \quad 2x-1 = 2(-5)-1 = -10-1 = -11.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{-1}{-11} = \frac{1}{11} > 0.
\][/tex]
The expression is positive in this interval.
Interval II: [tex]\(-4 < x < \frac{1}{2}\)[/tex]
Choose [tex]\(x = 0\)[/tex]. Then:
[tex]\[
x+4 = 0+4 = 4, \quad 2x-1 = 2(0)-1 = -1.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{4}{-1} = -4 < 0.
\][/tex]
The expression is negative in this interval.
Interval III: [tex]\(x > \frac{1}{2}\)[/tex]
Choose [tex]\(x = 1\)[/tex]. Then:
[tex]\[
x+4 = 1+4 = 5, \quad 2x-1 = 2(1)-1 = 1.
\][/tex]
Thus,
[tex]\[
\frac{x+4}{2x-1} = \frac{5}{1} = 5 > 0.
\][/tex]
The expression is positive in this interval.
Since we are looking for the part where
[tex]$$\frac{x+4}{2x-1} < 0,$$[/tex]
only the interval [tex]\(-4 < x < \frac{1}{2}\)[/tex] satisfies the inequality.
Finally, notice that the inequality is strict (using “[tex]\(<\)[/tex]”), which means the endpoints [tex]\(x = -4\)[/tex] and [tex]\(x = \frac{1}{2}\)[/tex] are not included (at [tex]\(x = -4\)[/tex], the expression equals 0; at [tex]\(x = \frac{1}{2}\)[/tex], the expression is undefined).
Therefore, the solution to the inequality is:
[tex]$$-4 < x < \frac{1}{2}.$$[/tex]
This corresponds to option 4.
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