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

Options

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

Answer

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

### Step 1: Verify the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f(x) \) 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. **Continuity**: The function \( f(x) = x^3 - 3x \) is a polynomial, and polynomials are continuous everywhere. Therefore, \( f(x) \) is continuous on \([0, \sqrt{3}]\).

2. **Differentiability**: Since \( f(x) \) is a polynomial, it is also differentiable everywhere, including the open interval \((0, \sqrt{3})\).

3. **Equal values at endpoints**: We need to check if \( f(0) = f(\sqrt{3}) \):
- Calculate \( f(0) \):
\[ f(0) = 0^3 - 3 \cdot 0 = 0 \]
- Calculate \( f(\sqrt{3}) \):
\[ f(\sqrt{3}) = (\sqrt{3})^3 - 3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0 \]
- Since \( f(0) = f(\sqrt{3}) = 0 \), the condition \( f(a) = f(b) \) is satisfied.

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

### Step 3: Set the derivative equal to zero
To find \( c \), we set the derivative equal to zero:
\[ f'(x) = 3x^2 - 3 = 0 \]
Solving for \( x \):
\[ 3x^2 = 3 \implies x^2 = 1 \implies x = \pm 1 \]
Since we are looking for \( c \) in the interval \((0, \sqrt{3})\), we take \( c = 1 \).

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