Verify the truth of Rolle's Theorem for the following functions :
`f(x)=x^(3)-2x^(2)-3x` in the interval [-1, 3].

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = x^3 - 2x^2 - 3x \) in the interval \([-1, 3]\), we need to check the three conditions of Rolle's Theorem:

1. **Continuity on the closed interval \([a, b]\)**.
2. **Differentiability on the open interval \((a, b)\)**.
3. **Equality of the function values at the endpoints, i.e., \( f(a) = f(b) \)**.

### Step 1: Check Continuity
The function \( f(x) = x^3 - 2x^2 - 3x \) is a polynomial function. Polynomial functions are continuous everywhere. Therefore, \( f(x) \) is continuous on the closed interval \([-1, 3]\).

### Step 2: Check Differentiability
Again, since \( f(x) \) is a polynomial function, it is differentiable everywhere, including on the open interval \((-1, 3)\).

### Step 3: Check Function Values at Endpoints
Now we need to calculate \( f(-1) \) and \( f(3) \):

- Calculate \( f(-1) \):
\[ f(-1) = (-1)^3 - 2(-1)^2 - 3(-1) = -1 - 2 + 3 = 0 \]

- Calculate \( f(3) \):
\[ f(3) = (3)^3 - 2(3)^2 - 3(3) = 27 - 18 - 9 = 0 \]

Now we check if \( f(-1) = f(3) \):
\[ f(-1) = 0 \quad \text{and} \quad f(3) = 0 \quad \Rightarrow \quad f(-1) = f(3) \]

### Conclusion
Since all three conditions of Rolle's Theorem are satisfied:
1. \( f(x) \) is continuous on \([-1, 3]\).
2. \( f(x) \) is differentiable on \((-1, 3)\).
3. \( f(-1) = f(3) = 0 \).

Therefore, by Rolle's Theorem, there exists at least one \( c \) in the interval \((-1, 3)\) such that \( f'(c) = 0 \).