Find the domain and range of `f(x)=sqrt(4-16x^(2))`.

Explanation

To find the domain and range of the function \( f(x) = \sqrt{4 - 16x^2} \), we will follow these steps:

### Step 1: Determine the Domain
The expression inside the square root must be non-negative for the function to be defined. Therefore, we need to solve the inequality:

\[ 4 - 16x^2 \geq 0 \]

### Step 2: Rearranging the Inequality
Rearranging the inequality gives:

\[ -16x^2 \geq -4 \]

### Step 3: Dividing by -16
When dividing by a negative number, we must reverse the inequality sign:

\[ x^2 \leq \frac{1}{4} \]

### Step 4: Taking the Square Root
Taking the square root of both sides gives:

\[ |x| \leq \frac{1}{2} \]

This leads to the interval:

\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

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

\[ \text{Domain} = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Step 5: Determine the Range
Next, we need to find the range of the function. We will evaluate \( f(x) \) at the endpoints of the domain and at \( x = 0 \):

1. **At \( x = -\frac{1}{2} \)**:
\[ f\left(-\frac{1}{2}\right) = \sqrt{4 - 16\left(-\frac{1}{2}\right)^2} = \sqrt{4 - 16 \cdot \frac{1}{4}} = \sqrt{4 - 4} = \sqrt{0} = 0 \]

2. **At \( x = \frac{1}{2} \)**:
\[ f\left(\frac{1}{2}\right) = \sqrt{4 - 16\left(\frac{1}{2}\right)^2} = \sqrt{4 - 16 \cdot \frac{1}{4}} = \sqrt{4 - 4} = \sqrt{0} = 0 \]

3. **At \( x = 0 \)**:
\[ f(0) = \sqrt{4 - 16(0)^2} = \sqrt{4} = 2 \]

### Step 6: Determine the Range
From the evaluations, the minimum value of \( f(x) \) is \( 0 \) and the maximum value is \( 2 \). Therefore, the range of \( f(x) \) is:

\[ \text{Range} = [0, 2] \]

### Final Answer
- **Domain**: \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
- **Range**: \([0, 2]\)