Rolls theorem for the function `f(x)=x^3-6x^2+11 x-6` is applicable in the interval .

Options

A. `[1,4]`
B. `[1,2]`
C. `[1,3]`
D. Cant say

Explanation

To verify Rolle's Theorem for the function \( f(x) = x^3 - 6x^2 + 11x - 6 \) in the interval \([1, 3]\), we will follow these steps:

### Step 1: Verify that \( f(1) = f(3) \)

First, we need to evaluate the function at the endpoints of the interval.

1. **Calculate \( f(1) \)**:
\[ f(1) = 1^3 - 6(1^2) + 11(1) - 6 \]
\[ = 1 - 6 + 11 - 6 = 0 \]

2. **Calculate \( f(3) \)**:
\[ f(3) = 3^3 - 6(3^2) + 11(3) - 6 \]
\[ = 27 - 54 + 33 - 6 = 0 \]

Since \( f(1) = 0 \) and \( f(3) = 0 \), we have \( f(1) = f(3) \).

### Step 2: Check if \( f(x) \) is continuous and differentiable on \([1, 3]\)

The function \( f(x) = x^3 - 6x^2 + 11x - 6 \) is a polynomial function, which is continuous and differentiable everywhere. Therefore, it is continuous on \([1, 3]\) and differentiable on \((1, 3)\).

### Step 3: Find \( f'(x) \) and solve \( f'(x) = 0 \)

Now we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 11x - 6) \]
\[ = 3x^2 - 12x + 11 \]

Next, we set \( f'(x) = 0 \):
\[ 3x^2 - 12x + 11 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3, b = -12, c = 11 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-12)^2 - 4 \cdot 3 \cdot 11 = 144 - 132 = 12 \]

Now substituting into the quadratic formula:
\[ x = \frac{12 \pm \sqrt{12}}{2 \cdot 3} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3} \]

### Step 5: Check if the roots lie within the interval \((1, 3)\)

Calculating the roots:
1. \( x_1 = 2 + \frac{\sqrt{3}}{3} \)
2. \( x_2 = 2 - \frac{\sqrt{3}}{3} \)

Now we need to check if these values lie between 1 and 3.

- For \( x_1 = 2 + \frac{\sqrt{3}}{3} \):
- Since \( \sqrt{3} \approx 1.732 \), we have \( \frac{\sqrt{3}}{3} \approx 0.577 \).
- Thus, \( x_1 \approx 2 + 0.577 = 2.577 \), which is between 1 and 3.

- For \( x_2 = 2 - \frac{\sqrt{3}}{3} \):
- Thus, \( x_2 \approx 2 - 0.577 = 1.423 \), which is also between 1 and 3.

### Conclusion

Since \( f(1) = f(3) \), \( f(x) \) is continuous on \([1, 3]\), differentiable on \((1, 3)\), and we found points where \( f'(x) = 0 \) within the interval, we conclude that Rolle's Theorem is applicable.