Discuss the applicability of Lagrange's Mean Value Theorem to the following :
`f(x)=x^(2)-2x+3" in "[0,4]`

Answer

To discuss the applicability of Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = x^2 - 2x + 3 \) on the interval \([0, 4]\), we will follow these steps:

### Step 1: Verify the conditions for 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 point \( c \) in \((a, b)\) such that:

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

Here, \( a = 0 \) and \( b = 4 \).

### Step 2: Check continuity
The function \( f(x) = x^2 - 2x + 3 \) is a polynomial function. Polynomial functions are continuous everywhere. Therefore, \( f(x) \) is continuous on the interval \([0, 4]\).

### Step 3: Check differentiability
Similarly, polynomial functions are also differentiable everywhere. Thus, \( f(x) \) is differentiable on the interval \((0, 4)\).

### Step 4: Calculate \( f(a) \) and \( f(b) \)
Now, we will evaluate the function at the endpoints of the interval:

\[ f(0) = 0^2 - 2(0) + 3 = 3 \]
\[ f(4) = 4^2 - 2(4) + 3 = 16 - 8 + 3 = 11 \]

### Step 5: Apply LMVT
Now we can apply the LMVT to find \( c \):

\[ f'(c) = \frac{f(4) - f(0)}{4 - 0} = \frac{11 - 3}{4} = \frac{8}{4} = 2 \]

### Step 6: Find \( f'(x) \)
Next, we need to find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^2 - 2x + 3) = 2x - 2 \]

### Step 7: Set \( f'(c) \) equal to 2
Now we set \( f'(c) \) equal to 2:

\[ 2c - 2 = 2 \]

### Step 8: Solve for \( c \)
Now, we solve for \( c \):

\[ 2c = 4 \implies c = 2 \]

### Conclusion
Since \( c = 2 \) lies in the open interval \((0, 4)\), we conclude that the Lagrange's Mean Value Theorem is applicable for the function \( f(x) = x^2 - 2x + 3 \) on the interval \([0, 4]\).