Let `f(x)=(2x)/(2x^(2)+5x+2)`, find the interval for which f(x)>0.

Answer

To find the interval for which \( f(x) > 0 \) for the function \( f(x) = \frac{2x}{2x^2 + 5x + 2} \), we will follow these steps:

### Step 1: Identify the function and set up the inequality
We start with the function:
\[ f(x) = \frac{2x}{2x^2 + 5x + 2} \]
We need to find where \( f(x) > 0 \), which translates to:
\[ \frac{2x}{2x^2 + 5x + 2} > 0 \]

### Step 2: Determine when the numerator and denominator are positive or negative
The fraction \( \frac{2x}{2x^2 + 5x + 2} \) is positive when both the numerator and denominator have the same sign (both positive or both negative).

**Numerator:**
The numerator is \( 2x \). It is positive when:
\[ x > 0 \]
And it is negative when:
\[ x < 0 \]

**Denominator:**
Next, we need to analyze the denominator \( 2x^2 + 5x + 2 \). We will find its roots by using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 5, c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]
The roots are:
\[ x = \frac{-5 \pm 3}{4} \]
Calculating the two roots:
1. \( x_1 = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2} \)
2. \( x_2 = \frac{-5 - 3}{4} = \frac{-8}{4} = -2 \)

### Step 3: Analyze the sign of the denominator
The quadratic \( 2x^2 + 5x + 2 \) opens upwards (since \( a = 2 > 0 \)). Thus, it is:
- Positive when \( x < -2 \) and \( x > -\frac{1}{2} \)
- Negative when \( -2 < x < -\frac{1}{2} \)

### Step 4: Combine the intervals
Now we combine the intervals where the numerator and denominator are both positive or both negative:

1. **Both positive:**
- \( x > 0 \) (numerator positive)
- Denominator is positive for \( x > -\frac{1}{2} \)
- Therefore, the interval is \( (0, \infty) \).

2. **Both negative:**
- \( x < 0 \) (numerator negative)
- Denominator is negative for \( -2 < x < -\frac{1}{2} \)
- Therefore, the interval is \( (-2, -\frac{1}{2}) \).

### Step 5: Final intervals
Combining these intervals, we find that \( f(x) > 0 \) in the intervals:
\[ (-2, -\frac{1}{2}) \cup (0, \infty) \]

### Final Answer
Thus, the intervals for which \( f(x) > 0 \) are:
\[ x \in (-2, -\frac{1}{2}) \cup (0, \infty) \]