The value of `c` in Rolle's theorem for the function `f(x) = x^(3) - 3x` in the interval `[0,sqrt(3)]` is

Options

A. `1`
B. `-1`
C. `3/2`
D. `1/3`

Explanation

To find the value of \( c \) in Rolle's theorem for the function \( f(x) = x^3 - 3x \) in the interval \([0, \sqrt{3}]\), we will follow these steps:

### Step 1: Verify the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Check continuity**: The function \( f(x) = x^3 - 3x \) is a polynomial, and polynomials are continuous everywhere.
2. **Check differentiability**: The function is also differentiable everywhere, including the interval \((0, \sqrt{3})\).
3. **Check endpoints**: Calculate \( f(0) \) and \( f(\sqrt{3}) \):
\[ f(0) = 0^3 - 3(0) = 0 \]
\[ f(\sqrt{3}) = (\sqrt{3})^3 - 3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0 \]
Since \( f(0) = f(\sqrt{3}) = 0 \), the conditions of Rolle's theorem are satisfied.

### Step 2: Find the derivative of the function
Next, we find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3 \]

### Step 3: Set the derivative equal to zero
According to Rolle's theorem, we need to find \( c \) such that \( f'(c) = 0 \):
\[ 3c^2 - 3 = 0 \]

### Step 4: Solve for \( c \)
Now, we solve the equation:
\[ 3c^2 = 3 \implies c^2 = 1 \implies c = \pm 1 \]

### Step 5: Determine the valid \( c \) in the interval
Since we are looking for \( c \) in the interval \((0, \sqrt{3})\), we only consider the positive solution:
\[ c = 1 \]

### Conclusion
Thus, the value of \( c \) in Rolle's theorem for the function \( f(x) = x^3 - 3x \) in the interval \([0, \sqrt{3}]\) is:
\[ \boxed{1} \]