If `|(12x)/(4x^(2)+9)| le 1`, then

Options

A. `x in R`
B. `x in [-3,3]`
C. `x in [-1,oo)`
D. `x in (-oo,2]`

Explanation

To solve the inequality \( \left| \frac{12x}{4x^2 + 9} \right| \leq 1 \), we will break it down step by step.

### Step 1: Remove the Absolute Value
The inequality \( \left| A \right| \leq B \) can be rewritten as:
\[ -B \leq A \leq B \]
In our case, we have:
\[ -1 \leq \frac{12x}{4x^2 + 9} \leq 1 \]

### Step 2: Split into Two Inequalities
This gives us two separate inequalities to solve:
1. \( \frac{12x}{4x^2 + 9} \leq 1 \)
2. \( \frac{12x}{4x^2 + 9} \geq -1 \)

### Step 3: Solve the First Inequality
For the first inequality:
\[ \frac{12x}{4x^2 + 9} \leq 1 \]
Multiply both sides by \( 4x^2 + 9 \) (which is always positive since \( 4x^2 + 9 > 0 \)):
\[ 12x \leq 4x^2 + 9 \]
Rearranging gives:
\[ 4x^2 - 12x + 9 \geq 0 \]

### Step 4: Factor the Quadratic
Now we will factor the quadratic:
\[ 4x^2 - 12x + 9 = (2x - 3)^2 \]
Thus, we have:
\[ (2x - 3)^2 \geq 0 \]
Since a square is always non-negative, this inequality holds for all \( x \).

### Step 5: Solve the Second Inequality
Now, we solve the second inequality:
\[ \frac{12x}{4x^2 + 9} \geq -1 \]
Again, multiply both sides by \( 4x^2 + 9 \):
\[ 12x \geq - (4x^2 + 9) \]
Rearranging gives:
\[ 4x^2 + 12x + 9 \geq 0 \]

### Step 6: Factor the Quadratic
Factoring this quadratic:
\[ 4x^2 + 12x + 9 = (2x + 3)^2 \]
Thus, we have:
\[ (2x + 3)^2 \geq 0 \]
Again, since a square is always non-negative, this inequality holds for all \( x \).

### Conclusion
Since both inequalities hold for all \( x \), the solution to the original inequality \( \left| \frac{12x}{4x^2 + 9} \right| \leq 1 \) is:
\[ x \in \mathbb{R} \]