Find the domain of the following ` y = cos^(-1) ( x^(2)/( 1 + x^(2)))`

Options

A. `[-1,1]
B. R
C. `[0,1]`
D. `[-1,5]`

Answer

To find the domain of the function \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \), we need to ensure that the expression inside the inverse cosine function is valid. The domain of the inverse cosine function, \( \cos^{-1}(x) \), is defined for \( x \) in the interval \([-1, 1]\).

### Step-by-Step Solution:

1. **Identify the expression inside the inverse cosine:**
\[ z = \frac{x^2}{1 + x^2} \]

2. **Determine the range of \( z \):**
- The numerator \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)).
- The denominator \( 1 + x^2 \) is always positive (i.e., \( 1 + x^2 > 0 \)).
- Therefore, \( z \) is always non-negative:
\[ z \geq 0 \]

3. **Find the maximum value of \( z \):**
- We can analyze the function \( z = \frac{x^2}{1 + x^2} \).
- As \( x \) approaches \( \infty \), \( z \) approaches \( 1 \):
\[ \lim_{x \to \infty} \frac{x^2}{1 + x^2} = 1 \]
- At \( x = 0 \):
\[ z(0) = \frac{0^2}{1 + 0^2} = 0 \]
- Thus, as \( x \) varies from \( 0 \) to \( \infty \), \( z \) increases from \( 0 \) to \( 1 \).

4. **Establish the range of \( z \):**
- The range of \( z \) is \( [0, 1] \).

5. **Set the range of \( z \) within the domain of \( \cos^{-1}(z) \):**
- Since \( z \) must be in the interval \([-1, 1]\) for \( \cos^{-1}(z) \) to be defined, and we already established that \( z \) ranges from \( 0 \) to \( 1 \), we conclude:
\[ z \in [0, 1] \]

6. **Conclusion about the domain of the original function:**
- The function \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \) is defined for all \( x \) such that \( z \) is in the range \( [0, 1] \), which is satisfied for all real numbers \( x \).

### Final Answer:
The domain of the function \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \) is all real numbers \( x \).