If mean value theorem holds for the function `f(x)=(x-1)(x-2)(x-3), x in [0,4],` then c=

Options

A. `2 pm (4)/(sqrt(3))`
B. `2 pm (2)/(sqrt(3))`
C. `2 pm sqrt(2)`
D. `2 pm sqrt(3)`

Answer

To solve the problem using the Mean Value Theorem (MVT) for the function \( f(x) = (x-1)(x-2)(x-3) \) on the interval \([0, 4]\), we will follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Here, \( f(x) \) is a polynomial function, which is continuous and differentiable everywhere. Therefore, the conditions of the MVT are satisfied.

### Step 2: Calculate \( f(0) \) and \( f(4) \)
First, we calculate the values of the function at the endpoints of the interval:
\[ f(0) = (0-1)(0-2)(0-3) = (-1)(-2)(-3) = -6 \]
\[ f(4) = (4-1)(4-2)(4-3) = (3)(2)(1) = 6 \]

### Step 3: Apply the Mean Value Theorem
Now we can find the average rate of change of the function over the interval \([0, 4]\):
\[ \frac{f(4) - f(0)}{4 - 0} = \frac{6 - (-6)}{4 - 0} = \frac{12}{4} = 3 \]

### Step 4: Find the derivative \( f'(x) \)
Next, we differentiate \( f(x) \):
\[ f(x) = (x-1)(x-2)(x-3) \]
Using the product rule or expanding the function:
\[ f(x) = x^3 - 6x^2 + 11x - 6 \]
Now, we differentiate:
\[ f'(x) = 3x^2 - 12x + 11 \]

### Step 5: Set the derivative equal to the average rate of change
According to the MVT, we need to find \( c \) such that:
\[ f'(c) = 3 \]
Setting the derivative equal to 3:
\[ 3c^2 - 12c + 11 = 3 \]
Simplifying this gives:
\[ 3c^2 - 12c + 8 = 0 \]

### Step 6: Solve the quadratic equation
Now we can solve the quadratic equation:
\[ c^2 - 4c + \frac{8}{3} = 0 \]
Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ c = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot \frac{8}{3}}}{2 \cdot 1} \]
\[ c = \frac{4 \pm \sqrt{16 - \frac{32}{3}}}{2} \]
\[ c = \frac{4 \pm \sqrt{\frac{48}{3} - \frac{32}{3}}}{2} \]
\[ c = \frac{4 \pm \sqrt{\frac{16}{3}}}{2} \]
\[ c = \frac{4 \pm \frac{4}{\sqrt{3}}}{2} \]
\[ c = 2 \pm \frac{2}{\sqrt{3}} \]

### Step 7: Determine the valid values of \( c \)
The two possible values for \( c \) are:
\[ c_1 = 2 + \frac{2}{\sqrt{3}}, \quad c_2 = 2 - \frac{2}{\sqrt{3}} \]
Now we need to check if these values lie within the interval \((0, 4)\).

Calculating \( c_1 \) and \( c_2 \):
- For \( c_1 \):
\[ c_1 \approx 2 + 1.155 \approx 3.155 \quad (\text{valid}) \]
- For \( c_2 \):
\[ c_2 \approx 2 - 1.155 \approx 0.845 \quad (\text{valid}) \]

### Final Answer
Both values of \( c \) are valid:
\[ c = 2 + \frac{2}{\sqrt{3}} \quad \text{and} \quad c = 2 - \frac{2}{\sqrt{3}} \]