The set of all `x` satisfying the inequality `(4x-1)/(3x+1)ge1`

Options

A. `(-oo,-(1)/(3))uu[(1)/(4),oo)`
B. `(-oo,-(2)/(3))uu[(5)/(4),oo)`
C. `(-oo,-(1)/(3))uu[2,oo)`
D. `(-oo,-(2)/(3))uu[4,oo)`

Answer

To solve the inequality \(\frac{4x-1}{3x+1} \ge 1\), we will follow these steps:

### Step 1: Rearranging the Inequality
Start by rearranging the inequality to bring all terms to one side:
\[ \frac{4x-1}{3x+1} - 1 \ge 0 \]
This can be simplified as:
\[ \frac{4x-1 - (3x+1)}{3x+1} \ge 0 \]
Simplifying the numerator:
\[ \frac{4x - 1 - 3x - 1}{3x + 1} \ge 0 \]
This simplifies to:
\[ \frac{x - 2}{3x + 1} \ge 0 \]

### Step 2: Finding Critical Points
Next, we need to find the critical points where the expression is equal to zero or undefined.

1. **Numerator**: Set \(x - 2 = 0\):
\[ x = 2 \]
2. **Denominator**: Set \(3x + 1 = 0\):
\[ 3x = -1 \implies x = -\frac{1}{3} \]

### Step 3: Testing Intervals
The critical points divide the number line into intervals. We will test the sign of the expression \(\frac{x - 2}{3x + 1}\) in the intervals:
- \( (-\infty, -\frac{1}{3}) \)
- \( (-\frac{1}{3}, 2) \)
- \( (2, \infty) \)

1. **Interval \( (-\infty, -\frac{1}{3}) \)**: Choose \(x = -1\):
\[ \frac{-1 - 2}{3(-1) + 1} = \frac{-3}{-2} = \frac{3}{2} > 0 \]
(Positive)

2. **Interval \( (-\frac{1}{3}, 2) \)**: Choose \(x = 0\):
\[ \frac{0 - 2}{3(0) + 1} = \frac{-2}{1} = -2 < 0 \]
(Negative)

3. **Interval \( (2, \infty) \)**: Choose \(x = 3\):
\[ \frac{3 - 2}{3(3) + 1} = \frac{1}{10} > 0 \]
(Positive)

### Step 4: Analyzing the Sign and Including Critical Points
From the tests:
- The expression is positive in the intervals \( (-\infty, -\frac{1}{3}) \) and \( (2, \infty) \).
- At \(x = 2\), the expression equals zero, so it is included in the solution.
- At \(x = -\frac{1}{3}\), the expression is undefined, so it is not included.

### Step 5: Writing the Solution
Combining the intervals where the expression is non-negative:
\[ x \in (-\infty, -\frac{1}{3}) \cup [2, \infty) \]

### Final Answer
Thus, the solution to the inequality \(\frac{4x-1}{3x+1} \ge 1\) is:
\[ x \in (-\infty, -\frac{1}{3}) \cup [2, \infty) \]