Discuss the applicability of Lagrange's Mean Value Theorem to the following :
`f(x)=2x-x^(2)" in "[0,1]`

Answer

To determine the applicability of Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = 2x - x^2 \) on the interval \([0, 1]\), we will follow these steps:

### Step 1: Check Continuity
The first condition for LMVT is that the function must be continuous on the closed interval \([a, b]\).

**Solution:**
The function \( f(x) = 2x - x^2 \) is a polynomial function. Polynomial functions are continuous everywhere, including the interval \([0, 1]\).

### Step 2: Check Differentiability
The second condition for LMVT is that the function must be differentiable on the open interval \((a, b)\).

**Solution:**
Since \( f(x) \) is a polynomial function, it is also differentiable everywhere. Thus, it is differentiable on the open interval \((0, 1)\).

### Step 3: Apply the Mean Value Theorem
Since both conditions of continuity and differentiability are satisfied, we can apply LMVT. According to LMVT, there exists at least one \( c \) in the interval \((0, 1)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

where \( a = 0 \) and \( b = 1 \).

**Solution:**
First, we calculate \( f(0) \) and \( f(1) \):

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

Now, substituting these values into the formula:

\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 0}{1} = 1 \]

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

\[ f'(x) = \frac{d}{dx}(2x - x^2) = 2 - 2x \]

### Step 5: Solve for \( c \)
Now we set the derivative equal to 1 to find \( c \):

\[ 2 - 2c = 1 \]

Rearranging gives:

\[ 2c = 2 - 1 \implies 2c = 1 \implies c = \frac{1}{2} \]

### Conclusion
Thus, we have shown that Lagrange's Mean Value Theorem is applicable to the function \( f(x) = 2x - x^2 \) on the interval \([0, 1]\), and the value of \( c \) that satisfies the theorem is \( c = \frac{1}{2} \).