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 under 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: Changing Signs
When we multiply or divide by a negative number, we must reverse the inequality sign:

\[ 16x^2 \leq 4 \]

### Step 4: Simplifying the Inequality
Dividing both sides by 16 gives:

\[ x^2 \leq \frac{4}{16} \]

This simplifies to:

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

### Step 5: Finding the Values of x
Taking the square root of both sides, we find:

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

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

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

### Step 6: Determine the Range
Next, we need to find the range of the function. The minimum value of \( f(x) \) occurs when \( 4 - 16x^2 = 0 \):

\[ f(x) = 0 \quad \text{when} \quad 4 - 16x^2 = 0 \]

This occurs at \( x = \pm \frac{1}{2} \).

### Step 7: Finding the Maximum Value
The maximum value of \( f(x) \) occurs when \( x = 0 \):

\[ f(0) = \sqrt{4 - 16(0)^2} = \sqrt{4} = 2 \]

### Step 8: Conclusion on the Range
Thus, the range of \( f(x) \) is:

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

### Final Answer
- **Domain**: \([-1/2, 1/2]\)
- **Range**: \([0, 2]\)

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