The value of c in Lgrange's Mean Value theorem for the function f(x) =x(x-2) in the interval [1,2] is

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. `(3)/(4)`
D. `(5)/(4)`

Answer

To find the value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = x(x - 2) \) in the interval \([1, 2]\), we can follow these steps:

### Step 1: Verify the conditions of Lagrange's Mean Value Theorem
Lagrange's Mean Value Theorem states that if a function \( f(x) \) 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Here, \( f(x) = x(x - 2) \) is a polynomial function, which is continuous and differentiable everywhere. Thus, it satisfies the conditions of the theorem on the interval \([1, 2]\).

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

### Step 3: Compute the average rate of change
Now, we can calculate the average rate of change of \( f \) over the interval \([1, 2]\):
\[ \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(1)}{2 - 1} = \frac{0 - (-1)}{2 - 1} = \frac{1}{1} = 1 \]

### Step 4: Find \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f(x) = x^2 - 2x \]
Using the power rule, we differentiate:
\[ f'(x) = 2x - 2 \]

### Step 5: Set \( f'(c) \) equal to the average rate of change
According to the Mean Value Theorem, we set \( f'(c) \) equal to the average rate of change:
\[ f'(c) = 2c - 2 \]
Setting this equal to 1 (the average rate of change):
\[ 2c - 2 = 1 \]

### Step 6: Solve for \( c \)
Now, we solve for \( c \):
\[ 2c - 2 = 1 \]
\[ 2c = 1 + 2 \]
\[ 2c = 3 \]
\[ c = \frac{3}{2} \]

### Conclusion
Thus, the value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = x(x - 2) \) in the interval \([1, 2]\) is:
\[ \boxed{\frac{3}{2}} \]