Let f(x) be a real valued function such that the area of an equilateral triangle with two of its vertices at (0, 0) and (x, f(x)) is `(sqrt(3))/(4)` square units. Then
The domain of the function is

Options

A. `[1, oo)`
B. `[-oo, 1)`
C. `(-1, 1)`
D. `[-1, 1]`

Answer

To find the domain of the function \( f(x) \) given that the area of an equilateral triangle with vertices at \( (0, 0) \) and \( (x, f(x)) \) is \( \frac{\sqrt{3}}{4} \) square units, we can follow these steps:

### Step 1: Understand the Area of the Triangle
The area \( A \) of an equilateral triangle can be expressed in terms of its side length \( s \) as:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Given that the area is \( \frac{\sqrt{3}}{4} \), we can set up the equation:
\[ \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} s^2 \]
This implies that:
\[ s^2 = 1 \quad \Rightarrow \quad s = 1 \quad \text{(since side length must be positive)} \]

### Step 2: Find the Side Length
The side length \( s \) of the triangle is the distance between the points \( (0, 0) \) and \( (x, f(x)) \). The distance formula gives us:
\[ s = \sqrt{(x - 0)^2 + (f(x) - 0)^2} = \sqrt{x^2 + f(x)^2} \]
Setting this equal to 1, we have:
\[ \sqrt{x^2 + f(x)^2} = 1 \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides:
\[ x^2 + f(x)^2 = 1 \]

### Step 4: Solve for \( f(x) \)
Rearranging gives us:
\[ f(x)^2 = 1 - x^2 \]
Taking the square root, we find:
\[ f(x) = \sqrt{1 - x^2} \quad \text{or} \quad f(x) = -\sqrt{1 - x^2} \]

### Step 5: Determine the Domain
For \( f(x) \) to be a real-valued function, the expression under the square root must be non-negative:
\[ 1 - x^2 \geq 0 \]
This simplifies to:
\[ x^2 \leq 1 \]
Taking square roots gives:
\[ -1 \leq x \leq 1 \]
Thus, the domain of \( f(x) \) is:
\[ [-1, 1] \]

### Conclusion
The domain of the function \( f(x) \) is \( [-1, 1] \).