For the function `f(x)=(x-1)(x-2)` defined on `[0,1/2]`, the value of 'c' satisfying Lagrange's mean value theorem is

Options

A. `1/3`
B. `1/5`
C. `1/7`
D. `1/4`

Answer

To find the value of 'c' satisfying Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = (x-1)(x-2) \) on the interval \([0, \frac{1}{2}]\), we will follow these steps:

### Step 1: Verify the conditions of LMVT
Lagrange's Mean Value Theorem states that if a function \( f \) 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, \( a = 0 \) and \( b = \frac{1}{2} \).

### Step 2: Calculate \( f(0) \) and \( f\left(\frac{1}{2}\right) \)
First, we need to evaluate the function at the endpoints of the interval:

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

\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right) = \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \frac{3}{4} \]

### Step 3: Calculate \( \frac{f(b) - f(a)}{b - a} \)
Now we can compute the slope of the secant line:

\[ \frac{f\left(\frac{1}{2}\right) - f(0)}{\frac{1}{2} - 0} = \frac{\frac{3}{4} - 2}{\frac{1}{2}} = \frac{\frac{3}{4} - \frac{8}{4}}{\frac{1}{2}} = \frac{-\frac{5}{4}}{\frac{1}{2}} = -\frac{5}{4} \cdot 2 = -\frac{5}{2} \]

### Step 4: Find \( f'(x) \)
Next, we need to find the derivative of \( f(x) \):

\[ f(x) = (x-1)(x-2) = x^2 - 3x + 2 \]

Using the power rule, we differentiate:

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

### Step 5: Set \( f'(c) \) equal to the slope
We need to find \( c \) such that:

\[ f'(c) = -\frac{5}{2} \]

Substituting \( f'(c) \):

\[ 2c - 3 = -\frac{5}{2} \]

### Step 6: Solve for \( c \)
Now we solve for \( c \):

\[ 2c = -\frac{5}{2} + 3 \]

Convert 3 to a fraction:

\[ 3 = \frac{6}{2} \]

Thus:

\[ 2c = -\frac{5}{2} + \frac{6}{2} = \frac{1}{2} \]

Now divide by 2:

\[ c = \frac{1}{4} \]

### Conclusion
The value of \( c \) that satisfies Lagrange's Mean Value Theorem for the function \( f(x) = (x-1)(x-2) \) on the interval \([0, \frac{1}{2}]\) is:

\[ \boxed{\frac{1}{4}} \]