Discuss the applicability of Lagrange's Mean Value Theorem to the following :
`f(x)=2x^(2)-10x+29" in "[2,7]`

Answer

To discuss the applicability of Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = 2x^2 - 10x + 29 \) on the interval \([2, 7]\), we will follow these steps:

### Step 1: Check 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

### Step 2: Verify continuity and differentiability
The function \( f(x) = 2x^2 - 10x + 29 \) is a polynomial function. Polynomial functions are continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous on \([2, 7]\) and differentiable on \((2, 7)\).

### Step 3: Calculate \( f(a) \) and \( f(b) \)
Now we will calculate \( f(2) \) and \( f(7) \):
\[ f(2) = 2(2^2) - 10(2) + 29 = 2(4) - 20 + 29 = 8 - 20 + 29 = 17 \]
\[ f(7) = 2(7^2) - 10(7) + 29 = 2(49) - 70 + 29 = 98 - 70 + 29 = 57 \]

### Step 4: Compute the average rate of change
Next, we compute the average rate of change of \( f \) over the interval \([2, 7]\):
\[ \frac{f(7) - f(2)}{7 - 2} = \frac{57 - 17}{7 - 2} = \frac{40}{5} = 8 \]

### Step 5: Find \( f'(x) \)
Now we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^2 - 10x + 29) = 4x - 10 \]

### Step 6: Set \( f'(c) = 8 \) and solve for \( c \)
We need to find \( c \) such that:
\[ f'(c) = 8 \]
Substituting \( f'(c) \):
\[ 4c - 10 = 8 \]
\[ 4c = 18 \quad \Rightarrow \quad c = \frac{18}{4} = \frac{9}{2} = 4.5 \]

### Conclusion
Since \( c = 4.5 \) lies within the interval \((2, 7)\), we conclude that Lagrange's Mean Value Theorem is applicable to the function \( f(x) = 2x^2 - 10x + 29 \) on the interval \([2, 7]\).