Question

For Lagrange's mean value theorem, the value of 'c' for the function \(f(x) = px^2+qx+r, p\neq 0\) in the interval \([1, b]\) and \(c \in ]1, b[\), is:

• A. \(b/2\)
• B. \(b/2+1\)
• C. \((b+1)/4\)
• D. \((b+1)/2\)

Hint:

For a quadratic function \(f(x) = Ax^2 + Bx + C\) on an interval \([a, b]\), the value of \(c\) from the Mean Value Theorem is always the midpoint of the interval, \(c = \frac{a+b}{2}\). Here, the interval is \([1, b]\), so \(c = \frac{1+b}{2}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Lagrange's Mean Value Theorem (LMVT) states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \(c\) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b-a} \]
Step 2: Key Formula or Approach:
1. Find the derivative \(f'(x)\).
2. Calculate \(f(b)\) and \(f(a)\).
3. Substitute these values into the LMVT formula and solve for \(c\).

Step 3: Detailed Explanation:
The given function is \(f(x) = px^2 + qx + r\), which is a polynomial and thus continuous and differentiable everywhere. The interval is \([a, b] = [1, b]\).
1. Find the derivative:
\[ f'(x) = 2px + q \] Therefore, at \(x=c\), we have: \[ f'(c) = 2pc + q \] 2. Calculate function values at endpoints:
\[ f(b) = pb^2 + qb + r \] \[ f(1) = p(1)^2 + q(1) + r = p + q + r \] 3. Apply the LMVT formula:
\[ f'(c) = \frac{f(b) - f(1)}{b-1} \] Substitute the expressions we found: \[ 2pc + q = \frac{(pb^2 + qb + r) - (p + q + r)}{b-1} \] \[ 2pc + q = \frac{pb^2 - p + qb - q}{b-1} \] Factor the numerator: \[ 2pc + q = \frac{p(b^2 - 1) + q(b - 1)}{b-1} \] \[ 2pc + q = \frac{p(b-1)(b+1) + q(b-1)}{b-1} \] Cancel the \((b-1)\) term from the numerator and denominator: \[ 2pc + q = p(b+1) + q \] Now, solve for \(c\): \[ 2pc = p(b+1) \] Since we are given \(p \neq 0\), we can divide both sides by \(2p\): \[ c = \frac{p(b+1)}{2p} = \frac{b+1}{2} \]
Step 4: Final Answer:
The value of \(c\) is \(\frac{b+1}{2}\).