If S is the set of all real `x` such that `(2x-1)/(2x^3+3x^2+x)` is positive

Options

A. `(-oo,-1)uu(-1/2,1)uu(1/2,oo)`
B. `(-10,-3)uu(-1/2,1)`
C. `(1/2,3)uu(7,oo)`
D. None of these

Answer

To solve the inequality \((2x-1)/(2x^3+3x^2+x) > 0\), we will follow these steps:

### Step 1: Identify the inequality
We need to determine when the expression \(\frac{2x-1}{2x^3 + 3x^2 + x}\) is greater than zero.

### Step 2: Factor the denominator
First, we factor the denominator \(2x^3 + 3x^2 + x\):
\[ 2x^3 + 3x^2 + x = x(2x^2 + 3x + 1) \]
Next, we will factor \(2x^2 + 3x + 1\). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4} \]
This gives us the roots:
\[ x = \frac{-2}{4} = -\frac{1}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \]
Thus, we can write:
\[ 2x^3 + 3x^2 + x = x(2x + 1)(x + 1) \]

### Step 3: Rewrite the inequality
Now we can rewrite our inequality:
\[ \frac{2x-1}{x(2x + 1)(x + 1)} > 0 \]

### Step 4: Find critical points
The critical points occur when the numerator or denominator is zero:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x = 0\)
3. \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
4. \(x + 1 = 0 \Rightarrow x = -1\)

The critical points are \(x = -1, -\frac{1}{2}, 0, \frac{1}{2}\).

### Step 5: Test intervals
We will test the sign of the expression in the intervals defined by these critical points:
- Interval 1: \((-∞, -1)\)
- Interval 2: \((-1, -\frac{1}{2})\)
- Interval 3: \((- \frac{1}{2}, 0)\)
- Interval 4: \((0, \frac{1}{2})\)
- Interval 5: \((\frac{1}{2}, ∞)\)

**Testing each interval:**
1. For \(x = -2\) in \((-∞, -1)\):
\(\frac{2(-2)-1}{(-2)(2(-2)+1)(-2+1)} = \frac{-5}{-2 \cdot -3 \cdot -1} = \frac{-5}{6} < 0\)

2. For \(x = -0.75\) in \((-1, -\frac{1}{2})\):
\(\frac{2(-0.75)-1}{(-0.75)(2(-0.75)+1)(-0.75+1)} = \frac{-2.5}{-0.75 \cdot -0.5 \cdot 0.25} > 0\)

3. For \(x = -0.25\) in \((- \frac{1}{2}, 0)\):
\(\frac{2(-0.25)-1}{(-0.25)(2(-0.25)+1)(-0.25+1)} = \frac{-1.5}{-0.25 \cdot 0.5 \cdot 0.75} > 0\)

4. For \(x = 0.25\) in \((0, \frac{1}{2})\):
\(\frac{2(0.25)-1}{(0.25)(2(0.25)+1)(0.25+1)} = \frac{-0.5}{0.25 \cdot 1.5 \cdot 1.25} < 0\)

5. For \(x = 1\) in \((\frac{1}{2}, ∞)\):
\(\frac{2(1)-1}{(1)(2(1)+1)(1+1)} = \frac{1}{1 \cdot 3 \cdot 2} > 0\)

### Step 6: Combine intervals
The expression is positive in the intervals:
- \((-1, -\frac{1}{2})\)
- \((- \frac{1}{2}, 0)\)
- \((\frac{1}{2}, ∞)\)

### Final Answer
Thus, the solution set \(S\) is:
\[ S = (-1, -\frac{1}{2}) \cup (-\frac{1}{2}, 0) \cup (\frac{1}{2}, ∞) \]