Let `f(x) =(1)/((1-x^(2)) . `Then range (f )=?

Options

A. `(-oo,1]`
B. `(-oo,0) uu [1,oo)`
C. `[-1,1]`
D. none of these

Explanation

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

### Step 1: Set the function equal to \( y \)
Let \( f(x) = y \). Thus, we have:
\[ y = \frac{1}{1 - x^2} \]

### Step 2: Rearrange the equation
We can rearrange this equation to express \( x^2 \) in terms of \( y \):
\[ y(1 - x^2) = 1 \]
\[ y - yx^2 = 1 \]
\[ yx^2 = y - 1 \]
\[ x^2 = \frac{y - 1}{y} \]

### Step 3: Determine the conditions for \( x^2 \)
Since \( x^2 \) must be non-negative (as it represents a square), we need:
\[ \frac{y - 1}{y} \geq 0 \]

### Step 4: Analyze the inequality
This inequality holds true when:
1. \( y - 1 \geq 0 \) and \( y > 0 \) (which means \( y \geq 1 \))
2. \( y - 1 \leq 0 \) and \( y < 0 \) (which is not possible since \( y \) cannot be negative)

Thus, the only valid condition is:
\[ y \geq 1 \]

### Step 5: Identify where \( y \) is undefined
Next, we check when \( f(x) \) is undefined. The function \( f(x) \) is undefined when the denominator is zero:
\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
At these points, \( f(x) \) approaches infinity.

### Step 6: Conclusion on the range
From the analysis, we find that \( y \) can take any value greater than or equal to 1, but it cannot equal 0 or any negative value. Therefore, the range of the function \( f(x) \) is:
\[ \text{Range}(f) = [1, \infty) \]