The domain and range of the real function f defined by `f(x)=(1)/(4x^(2)-1)` are

Options

A. Domain `={-(1)/(2),(1)/(2)}`, Range `=-{-oo,-1)uu(0,oo)`
B. Domain `R={-(1)/(2),(1)/(2)}`, Range `=-{-oo,-1)uu(0,oo)`
C. Domain `=[-(1)/(2),(1)/(2)]`, Range `=-{-oo,-1)uu(0,oo)`
D. Domain `=R-[-(1)/(2),(1)/(2)]`, Range `=-{-oo,-1)uu(2,oo)`

Answer

To find the domain and range of the function \( f(x) = \frac{1}{4x^2 - 1} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function consists of all the values of \( x \) for which the function is defined. In this case, the function \( f(x) \) will be undefined when the denominator is equal to zero.

Set the denominator equal to zero:
\[ 4x^2 - 1 = 0 \]

### Step 2: Solve for \( x \)
Rearranging the equation gives:
\[ 4x^2 = 1 \]
\[ x^2 = \frac{1}{4} \]
Taking the square root of both sides, we find:
\[ x = \pm \frac{1}{2} \]

### Step 3: State the Domain
The function is undefined at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \). Therefore, the domain of \( f(x) \) is all real numbers except these two values:
\[ \text{Domain} = \mathbb{R} \setminus \left\{ -\frac{1}{2}, \frac{1}{2} \right\} \]

### Step 4: Determine the Range
Next, we will find the range of the function. The range consists of all possible values of \( f(x) \).

### Step 5: Analyze the Function
The function can be rewritten as:
\[ f(x) = \frac{1}{4x^2 - 1} \]
We know that \( 4x^2 \) is always non-negative, and thus \( 4x^2 - 1 \) can take values less than, equal to, or greater than zero.

1. When \( 4x^2 > 1 \) (i.e., \( |x| > \frac{1}{2} \)), \( f(x) \) will be positive.
2. When \( 4x^2 < 1 \) (i.e., \( |x| < \frac{1}{2} \)), \( f(x) \) will be negative.
3. The function will never equal zero since the numerator is always 1.

### Step 6: State the Range
From the analysis:
- As \( x \) approaches \( \pm \frac{1}{2} \), \( f(x) \) approaches \( \infty \) (positive).
- As \( x \) approaches \( 0 \), \( f(x) \) approaches \( -1 \) (negative).

Thus, the range of \( f(x) \) is:
\[ \text{Range} = (-\infty, 0) \cup (0, \infty) \]

### Final Answer
- **Domain**: \( \mathbb{R} \setminus \left\{ -\frac{1}{2}, \frac{1}{2} \right\} \)
- **Range**: \( (-\infty, 0) \cup (0, \infty) \)