Question

The value(s) of $ c \in (1, 2) $, where the conclusion of Lagrange’s M.V.T. is satisfied for the function $ f(x) = x^2 + 3x + 2 $ in $[1,2]$, is/are:

• A. \( \left( -\frac{3}{2}, \frac{1}{2} \right) \)
• B. \( \left( \frac{1}{2}, \frac{3}{2} \right) \)
• C. \( \left( -\frac{1}{2} \right) \)
• D. \( \left( \frac{3}{2} \right) \)

Hint:

When using Lagrange's Mean Value Theorem, always check that the function is continuous and differentiable over the interval and find the derivative to equate it to the average rate of change.

Answer 1

The Correct Option is A

Step 1: Apply Lagrange's Mean Value Theorem (M.V.T.):
The Lagrange's Mean Value Theorem states that for a function \( f(x) \) continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), there exists some \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] In this problem, we are given the function \( f(x) = x^2 + 3x + 2 \) on the interval \([1, 2]\). We need to find \( c \in (1, 2) \) where the conclusion of the M.V.T. holds.
Step 2: Calculate \( f(2) \) and \( f(1) \):
First, we find the values of the function at the endpoints of the interval: \[ f(2) = 2^2 + 3(2) + 2 = 4 + 6 + 2 = 12, \] \[ f(1) = 1^2 + 3(1) + 2 = 1 + 3 + 2 = 6. \] 
Step 3: Find the average rate of change:
The average rate of change of the function over the interval \([1, 2]\) is: \[ \frac{f(2) - f(1)}{2 - 1} = \frac{12 - 6}{1} = 6. \] 
Step 4: Find the derivative of \( f(x) \):
The derivative of the function \( f(x) = x^2 + 3x + 2 \) is: \[ f'(x) = 2x + 3. \] 
Step 5: Set the derivative equal to the average rate of change:
According to the M.V.T., there exists some \( c \in (1, 2) \) such that: \[ f'(c) = 6. \] Substituting the expression for \( f'(x) \): \[ 2c + 3 = 6. \] 
Step 6: Solve for \( c \):
Now, solve for \( c \): \[ 2c = 6 - 3 = 3 \quad \Rightarrow \quad c = \frac{3}{2}. \] 
Step 7: Verify that \( c \in (1, 2) \):
Since \( \frac{3}{2} \) lies in the interval \((1, 2)\), it is a valid solution. 
Conclusion:
Thus, the value of \( c \) that satisfies the 
conclusion of Lagrange’s M.V.T. is \( \frac{3}{2} \).