The value of c in (0, 2) satisfying the mean value theorem for the function `f(x)=x (x-1)^(2), x in [0,2]` is equa to

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To find the value of \( c \) in the interval \( (0, 2) \) that satisfies the Mean Value Theorem (MVT) for the function \( f(x) = x(x-1)^2 \), 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 the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
In our case, \( f(x) = x(x-1)^2 \) is a polynomial function, which is continuous and differentiable everywhere. Therefore, it satisfies the conditions of the MVT on the interval \([0, 2]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 0 \) and \( b = 2 \).
- Calculate \( f(0) \):
\[ f(0) = 0(0-1)^2 = 0 \]
- Calculate \( f(2) \):
\[ f(2) = 2(2-1)^2 = 2(1)^2 = 2 \]

### Step 3: Apply the Mean Value Theorem
Now, we can find the average rate of change from \( a \) to \( b \):
\[ \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(0)}{2 - 0} = \frac{2 - 0}{2 - 0} = \frac{2}{2} = 1 \]

### Step 4: Find the derivative \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f(x) = x(x-1)^2 \]
Using the product rule:
\[ f'(x) = (x)'(x-1)^2 + x((x-1)^2)' \]
Calculating the derivative:
\[ f'(x) = (1)(x-1)^2 + x(2(x-1)(1)) = (x-1)^2 + 2x(x-1) \]
Now simplify:
\[ f'(x) = (x-1)^2 + 2x^2 - 2x = x^2 - 2x + 1 + 2x^2 - 2x = 3x^2 - 4x + 1 \]

### Step 5: Set \( f'(c) \) equal to the average rate of change
We set \( f'(c) = 1 \):
\[ 3c^2 - 4c + 1 = 1 \]
Subtracting 1 from both sides:
\[ 3c^2 - 4c + 1 - 1 = 0 \implies 3c^2 - 4c = 0 \]
Factoring out \( c \):
\[ c(3c - 4) = 0 \]

### Step 6: Solve for \( c \)
This gives us two solutions:
1. \( c = 0 \)
2. \( 3c - 4 = 0 \implies c = \frac{4}{3} \)

### Step 7: Determine which solution is in the interval \( (0, 2) \)
The solution \( c = 0 \) is not in the interval \( (0, 2) \), but \( c = \frac{4}{3} \) is in the interval \( (0, 2) \).

Thus, the value of \( c \) that satisfies the Mean Value Theorem for the function \( f(x) = x(x-1)^2 \) in the interval \( (0, 2) \) is:
\[ \boxed{\frac{4}{3}} \]