Domain and range of the function `f(x)=cos(sqrt(1-x^(2)))` are

Options

A. `[-1,1],[-1,1]`
B. `[-1,1], [0,cos1]`
C. `[0,1],[0,cos1]`
D. `[-1,1],[cos1,1]`

Answer

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

### Step 1: Determine the Domain

1. **Identify the expression inside the square root**: The function includes a square root, \( \sqrt{1 - x^2} \). For the square root to be defined, the expression inside must be non-negative:
\[ 1 - x^2 \geq 0 \]

2. **Solve the inequality**:
\[ 1 \geq x^2 \]
This can be rewritten as:
\[ x^2 \leq 1 \]
Taking the square root of both sides, we find:
\[ -1 \leq x \leq 1 \]

3. **Conclusion about the domain**: The domain of \( f(x) \) is:
\[ \text{Domain} = [-1, 1] \]

### Step 2: Determine the Range

1. **Evaluate the function at the endpoints of the domain**:
- At \( x = -1 \):
\[ f(-1) = \cos(\sqrt{1 - (-1)^2}) = \cos(\sqrt{1 - 1}) = \cos(0) = 1 \]
- At \( x = 1 \):
\[ f(1) = \cos(\sqrt{1 - 1^2}) = \cos(\sqrt{1 - 1}) = \cos(0) = 1 \]
- At \( x = 0 \):
\[ f(0) = \cos(\sqrt{1 - 0^2}) = \cos(\sqrt{1}) = \cos(1) \]

2. **Determine the minimum and maximum values**:
- The maximum value of \( f(x) \) occurs at both endpoints \( x = -1 \) and \( x = 1 \), giving \( f(-1) = f(1) = 1 \).
- The minimum value occurs at \( x = 0 \), giving \( f(0) = \cos(1) \).

3. **Conclusion about the range**: The range of \( f(x) \) is:
\[ \text{Range} = [\cos(1), 1] \]

### Final Result

- **Domain**: \([-1, 1]\)
- **Range**: \([\cos(1), 1]\)