To find the domain and range of the function \( f(x) = \frac{x^2}{1 - x^2} \), we need to determine the values of \( x \) for which the function is defined and the set of possible values that the function can take.

### Domain

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

For \( f(x) = \frac{x^2}{1 - x^2} \), the function is undefined when the denominator is zero. Therefore, we need to find the values of \( x \) that make the denominator zero:

\[ 1 - x^2 = 0 \]

Solving for \( x \):

\[ x^2 = 1 \]
\[ x = \pm 1 \]

Thus, the function is undefined at \( x = 1 \) and \( x = -1 \). Therefore, the domain of \( f(x) \) is all real numbers except \( x = 1 \) and \( x = -1 \):

\[ \text{Domain} = \{ x \in \mathbb{R} \mid x \neq 1, x \neq -1 \} \]

### Range

The range of a function is the set of all possible output values (y-values) that the function can take.

To find the range of \( f(x) \), we analyze the behavior of the function as \( x \) approaches certain critical points:

1. **As \( x \to \pm \infty \)**:
   \[
   \lim_{x \to \pm \infty} \frac{x^2}{1 - x^2} = \lim_{x \to \pm \infty} \frac{1}{\frac{1}{x^2} - 1} = \lim_{x \to \pm \infty} \frac{1}{0 - 1} = -1
   \]
   So, as \( x \) approaches \(\pm \infty\), \( f(x) \) approaches -1.

2. **At the vertical asymptotes \( x = 1 \) and \( x = -1 \)**:
   \[
   \lim_{x \to 1^-} \frac{x^2}{1 - x^2} = \infty \quad \text{and} \quad \lim_{x \to 1^+} \frac{x^2}{1 - x^2} = -\infty
   \]
   \[
   \lim_{x \to -1^-} \frac{x^2}{1 - x^2} = \infty \quad \text{and} \quad \lim_{x \to -1^+} \frac{x^2}{1 - x^2} = -\infty
   \]
   So, near the vertical asymptotes \( x = 1 \) and \( x = -1 \), the function takes on all values from \( -\infty \) to \(\infty\).

3. **Behavior between and outside the critical points \( x = 1 \) and \( x = -1 \)**:
   For \( -1 < x < 1 \):
   \[
   \frac{x^2}{1 - x^2} \text{ is positive and increases from 0 to } \infty
   \]
   For \( x < -1 \) or \( x > 1 \):
   \[
   \frac{x^2}{1 - x^2} \text{ is negative and decreases from } -1 \text{ to } -\infty
   \]

Combining these observations, we see that the range of \( f(x) \) covers all real numbers except for \(-1\):

\[ \text{Range} = \{ y \in \mathbb{R} \mid y \neq -1 \} \]

### Summary

- **Domain**: \( \{ x \in \mathbb{R} \mid x \neq 1, x \neq -1 \} \)
- **Range**: \( \{ y \in \mathbb{R} \mid y \neq -1 \} \)

Would you like further details or have any questions?

Here are five related questions you might find useful:

1. How do you determine the vertical asymptotes of a function?
2. How do you find the horizontal asymptotes of a rational function?
3. What are the steps to find the critical points of a function?
4. How do you determine the intervals of increase and decrease for a function?
5. What is the importance of analyzing the behavior of a function at infinity?

**Tip:** When finding the range of a function, it's often helpful to consider the function's behavior as \( x \) approaches critical values and infinity.

Learn how to determine the domain and range of the function f(x) = x² / (1 - x²). Understand the conditions for the function's definition and explore its possible output values. Suitable for advanced high school students studying functions and calculus.

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