Let g(x) be a function defined on [-1,1]. If the area of the equilateral triangle with two of its vertices at (0,0) and `(x,g(x))` is `(sqrt(3))/(4)` , then the function g(x) , is

Options

A. `pm sqrt(1-x^(2))`
B. `-sqrt(1-x^(2)) or sqrt(1-x^(2))`
C. `sqrt(1-x^(2))` only
D. `sqrt(1+x^(2))`

Answer

To find the function \( g(x) \) defined on the interval \([-1, 1]\) such that the area of the equilateral triangle with vertices at \((0,0)\) and \((x, g(x))\) is \(\frac{\sqrt{3}}{4}\), we can follow these steps:

### Step 1: Understand the Area of an Equilateral Triangle
The area \( A \) of an equilateral triangle can be expressed using the formula:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
where \( a \) is the length of a side of the triangle.

### Step 2: Set the Area Equal to Given Value
We are given that the area of the triangle is \(\frac{\sqrt{3}}{4}\). Therefore, we can set up the equation:
\[ \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \]

### Step 3: Simplify the Equation
To simplify, we can multiply both sides by \( 4 \) to eliminate the fraction:
\[ \sqrt{3} a^2 = \sqrt{3} \]
Next, divide both sides by \(\sqrt{3}\):
\[ a^2 = 1 \]

### Step 4: Solve for \( a \)
Taking the square root of both sides gives us:
\[ a = 1 \quad \text{or} \quad a = -1 \]
Since \( a \) represents a length, we only consider the positive value:
\[ a = 1 \]

### Step 5: Find the Length of the Side \( a \)
The length \( a \) is the distance between the points \((0,0)\) and \((x, g(x))\). We can use the distance formula:
\[ a = \sqrt{(x - 0)^2 + (g(x) - 0)^2} = \sqrt{x^2 + g(x)^2} \]
Since we found \( a = 1 \), we can set up the equation:
\[ \sqrt{x^2 + g(x)^2} = 1 \]

### Step 6: Square Both Sides
To eliminate the square root, we square both sides:
\[ x^2 + g(x)^2 = 1 \]

### Step 7: Solve for \( g(x) \)
Rearranging the equation gives:
\[ g(x)^2 = 1 - x^2 \]
Taking the square root of both sides, we find:
\[ g(x) = \sqrt{1 - x^2} \quad \text{or} \quad g(x) = -\sqrt{1 - x^2} \]

### Conclusion
Thus, the function \( g(x) \) can be expressed as:
\[ g(x) = \pm \sqrt{1 - x^2} \]
This indicates that \( g(x) \) could take on both positive and negative values depending on the context.