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

1. The expression inside the square root, \( 1 - x^2 \), must be non-negative because the square root of a negative number is not defined in the real number system.

\[ 1 - x^2 \geq 0 \]

2. Rearranging this inequality gives:

\[ x^2 \leq 1 \]

3. Taking the square root of both sides leads to:

\[ -1 \leq x \leq 1 \]

Thus, the domain of \( f(x) \) is:

\[ \text{Domain} = [-1, 1] \]

### Step 2: Determine the Range
Next, we will find the range of the function \( f(x) \).

1. The output of the function is \( f(x) = \cos(\sqrt{1 - x^2}) \). Since the domain restricts \( x \) to the interval \([-1, 1]\), we need to find the values of \( \sqrt{1 - x^2} \) for \( x \) in this interval.

2. When \( x = -1 \) or \( x = 1 \):

\[ \sqrt{1 - (-1)^2} = \sqrt{0} = 0 \]
\[ \sqrt{1 - 1^2} = \sqrt{0} = 0 \]

3. When \( x = 0 \):

\[ \sqrt{1 - 0^2} = \sqrt{1} = 1 \]

4. The values of \( \sqrt{1 - x^2} \) will range from \( 0 \) (when \( x = -1 \) or \( x = 1 \)) to \( 1 \) (when \( x = 0 \)). Therefore, \( \sqrt{1 - x^2} \) takes values in the interval \([0, 1]\).

5. The cosine function \( \cos(t) \) is decreasing in the interval \([0, 1]\), where:

- \( \cos(0) = 1 \)
- \( \cos(1) \) is a value between \( 0 \) and \( 1 \).

Thus, the range of \( f(x) \) is:

\[ \text{Range} = [\cos(1), 1] \]

### Final Result
Combining the findings, we have:

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