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

Answer

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

### Step 1: Check the continuity and differentiability of \( f(x) \)

The function \( f(x) = x(2 - x) \) can be rewritten as:
\[ f(x) = 2x - x^2 \]
This is a polynomial function. Polynomial functions are continuous and differentiable everywhere on the real line, including the interval \([0, 1]\).

### Step 2: Evaluate \( f(0) \) and \( f(1) \)

Next, we need to find the values of the function at the endpoints of the interval:
\[ f(0) = 0(2 - 0) = 0 \]
\[ f(1) = 1(2 - 1) = 1 \]

### Step 3: Apply the Mean Value Theorem

According to the Mean Value Theorem, there exists at least one \( c \) in the interval \((0, 1)\) such that:
\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} \]
Substituting the values we found:
\[ f'(c) = \frac{1 - 0}{1 - 0} = 1 \]

### Step 4: Find \( f'(x) \)

Now we need to compute the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x - x^2) = 2 - 2x \]

### Step 5: Set \( f'(c) = 1 \) and solve for \( c \)

We set the derivative equal to 1:
\[ 2 - 2c = 1 \]
Solving for \( c \):
\[ 2c = 2 - 1 \]
\[ 2c = 1 \implies c = \frac{1}{2} \]

### Step 6: Verify that \( c \) is in the interval \((0, 1)\)

The value \( c = \frac{1}{2} \) is indeed within the interval \((0, 1)\).

### Conclusion

Since \( f(x) \) is continuous and differentiable on \([0, 1]\), and we found a \( c \) in \((0, 1)\) such that \( f'(c) = 1 \), we conclude that Lagrange's Mean Value Theorem is applicable to the function \( f(x) = x(2 - x) \) on the interval \([0, 1]\).

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