For the function `f(x)=x^(3)-6x^(2)+ax+b`, it is given that f(1) = f(3) = 0. Find the values of 'a' and 'b', and hence, verify Rolle's Theorem on [1, 3].

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) for the function \( f(x) = x^3 - 6x^2 + ax + b \) given that \( f(1) = 0 \) and \( f(3) = 0 \). We will then verify Rolle's Theorem on the interval \([1, 3]\).

### Step 1: Set up the equations based on the given conditions

1. **Using \( f(1) = 0 \)**:
\[ f(1) = 1^3 - 6 \cdot 1^2 + a \cdot 1 + b = 0 \]
Simplifying this:
\[ 1 - 6 + a + b = 0 \implies a + b - 5 = 0 \implies a + b = 5 \quad \text{(Equation 1)} \]

2. **Using \( f(3) = 0 \)**:
\[ f(3) = 3^3 - 6 \cdot 3^2 + a \cdot 3 + b = 0 \]
Simplifying this:
\[ 27 - 54 + 3a + b = 0 \implies 3a + b - 27 = 0 \implies 3a + b = 27 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations

Now we have two equations:
1. \( a + b = 5 \) (Equation 1)
2. \( 3a + b = 27 \) (Equation 2)

We can eliminate \( b \) by subtracting Equation 1 from Equation 2:
\[ (3a + b) - (a + b) = 27 - 5 \]
This simplifies to:
\[ 2a = 22 \implies a = 11 \]

### Step 3: Substitute back to find \( b \)

Now substitute \( a = 11 \) into Equation 1:
\[ 11 + b = 5 \implies b = 5 - 11 = -6 \]

Thus, we have:
\[ a = 11, \quad b = -6 \]

### Step 4: Write the function with found values

The function now becomes:
\[ f(x) = x^3 - 6x^2 + 11x - 6 \]

### Step 5: Verify Rolle's Theorem

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

1. **Continuity**: The function \( f(x) \) is a polynomial, and polynomials are continuous everywhere, including on \([1, 3]\).

2. **Differentiability**: The function \( f(x) \) is also differentiable everywhere, including on \((1, 3)\).

3. **Check \( f(1) = f(3) \)**:
\[ f(1) = 0, \quad f(3) = 0 \quad \text{(as given)} \]

4. **Find \( f'(x) \)**:
\[ f'(x) = 3x^2 - 12x + 11 \]

5. **Set \( f'(c) = 0 \)**:
\[ 3c^2 - 12c + 11 = 0 \]
Using the quadratic formula:
\[ c = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 3 \cdot 11}}{2 \cdot 3} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3} \]

The values of \( c \) are:
\[ c_1 = 2 + \frac{\sqrt{3}}{3}, \quad c_2 = 2 - \frac{\sqrt{3}}{3} \]

Both values \( c_1 \) and \( c_2 \) lie in the interval \((1, 3)\).

### Conclusion

Thus, we have verified that Rolle's Theorem holds for the function \( f(x) \) on the interval \([1, 3]\).