Verify the truth of Rolle's Theorem for the following functions :
`f(x)=(x-2)(x-3)(x-4)` in the interval `2lexle4`.

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = (x-2)(x-3)(x-4) \) in the interval \([2, 4]\), we will follow these steps:

### Step 1: Check if the function is continuous on the closed interval \([2, 4]\)

Since \( f(x) \) is a polynomial function, it is continuous everywhere. Therefore, it is continuous on the closed interval \([2, 4]\).

### Step 2: Check if the function is differentiable on the open interval \((2, 4)\)

Again, since \( f(x) \) is a polynomial function, it is differentiable everywhere. Thus, it is differentiable on the open interval \((2, 4)\).

### Step 3: Check if \( f(a) = f(b) \)

We need to evaluate \( f(2) \) and \( f(4) \):

\[ f(2) = (2-2)(2-3)(2-4) = 0 \cdot (-1) \cdot (-2) = 0 \]

\[ f(4) = (4-2)(4-3)(4-4) = 2 \cdot 1 \cdot 0 = 0 \]

Since \( f(2) = f(4) = 0 \), the condition \( f(a) = f(b) \) is satisfied.

### Step 4: Find \( c \) in \((2, 4)\) such that \( f'(c) = 0 \)

First, we need to find the derivative \( f'(x) \). We can use the product rule to differentiate:

\[ f(x) = (x-2)(x-3)(x-4) \]

Using the product rule, we find:

\[ f'(x) = (x-3)(x-4) + (x-2)(x-4) + (x-2)(x-3) \]

Calculating each term:

1. \( (x-3)(x-4) = x^2 - 7x + 12 \)
2. \( (x-2)(x-4) = x^2 - 6x + 8 \)
3. \( (x-2)(x-3) = x^2 - 5x + 6 \)

Now, adding these together:

\[ f'(x) = (x^2 - 7x + 12) + (x^2 - 6x + 8) + (x^2 - 5x + 6) \]

Combining like terms:

\[ f'(x) = 3x^2 - 18x + 26 \]

Now, we set \( f'(c) = 0 \):

\[ 3c^2 - 18c + 26 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3 \), \( b = -18 \), and \( c = 26 \):

\[ c = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 3 \cdot 26}}{2 \cdot 3} \]

Calculating the discriminant:

\[ c = \frac{18 \pm \sqrt{324 - 312}}{6} = \frac{18 \pm \sqrt{12}}{6} = \frac{18 \pm 2\sqrt{3}}{6} = 3 \pm \frac{\sqrt{3}}{3} \]

Calculating the two possible values for \( c \):

1. \( c_1 = 3 + \frac{\sqrt{3}}{3} \)
2. \( c_2 = 3 - \frac{\sqrt{3}}{3} \)

### Step 6: Check if \( c \) is in the interval \((2, 4)\)

Since \( \sqrt{3} \approx 1.732 \):

1. \( c_1 \approx 3 + 0.577 \approx 3.577 \) (which is in \((2, 4)\))
2. \( c_2 \approx 3 - 0.577 \approx 2.423 \) (which is also in \((2, 4)\))

### Conclusion

Both values \( c_1 \) and \( c_2 \) are in the interval \((2, 4)\). Therefore, by Rolle's Theorem, there exists at least one \( c \) in \((2, 4)\) such that \( f'(c) = 0 \).

Thus, we have verified the truth of Rolle's Theorem for the function \( f(x) = (x-2)(x-3)(x-4) \) in the interval \([2, 4]\).

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