Question

A value of \(c\) according to the Lagrange's mean value theorem for \(f(x) = (x - 1)(x - 2)(x - 3)\) in \([0,4]\) is:

• A. \(2 + \frac{2}{\sqrt{3}}\)
• B. \(2 - \frac{\sqrt{16}}{\sqrt{3}}\)
• C. \(1 + \frac{\sqrt{5}}{4}\)
• D. \(2 + \frac{\sqrt{8}}{3}\)

Hint:

When applying the Mean Value Theorem, always calculate the derivative of the function and solve for \(c\) where the derivative equals the average rate of change.

Answer 1

The Correct Option is A

Lagrange's Mean Value Theorem (LMVT) states that for a function \( f \) continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one \( c \) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] Here, the function is \(f(x) = (x - 1)(x - 2)(x - 3)\), and the interval is \([0, 4]\). We are to find \(c\) where the Mean Value Theorem holds. 
Step 1: Calculate \( f(0) \) and \( f(4) \). - For \(f(0)\): \[ f(0) = (0 - 1)(0 - 2)(0 - 3) = (-1)(-2)(-3) = -6. \] - For \(f(4)\): \[ f(4) = (4 - 1)(4 - 2)(4 - 3) = (3)(2)(1) = 6. \] Step 2: Find the derivative of \(f(x)\). Using the product rule, the derivative of \(f(x) = (x - 1)(x - 2)(x - 3)\) is: \[ f'(x) = \frac{d}{dx} \left( (x - 1)(x - 2)(x - 3) \right). \] Using the product rule for three functions: \[ f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2). \] Simplifying each term: \[ f'(x) = (x^2 - 5x + 6) + (x^2 - 4x + 3) + (x^2 - 3x + 2). \] Combine like terms: \[ f'(x) = 3x^2 - 12x + 11. \] Step 3: Apply the Mean Value Theorem. The Mean Value Theorem states: \[ f'(c) = \frac{f(4) - f(0)}{4 - 0}. \] Substituting \(f(4) = 6\) and \(f(0) = -6\): \[ f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3. \] Thus, \(f'(c) = 3\). Step 4: Solve for \(c\) by setting \(f'(c) = 3\). From \(f'(x) = 3x^2 - 12x + 11\), set this equal to 3: \[ 3x^2 - 12x + 11 = 3. \] Simplifying: \[ 3x^2 - 12x + 8 = 0. \] Divide through by 3: \[ x^2 - 4x + \frac{8}{3} = 0. \] Use the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)\left(\frac{8}{3}\right)}}{2(1)} = \frac{4 \pm \sqrt{16 - \frac{32}{3}}}{2}. \] Simplifying the discriminant: \[ x = \frac{4 \pm \sqrt{\frac{48}{3} - \frac{32}{3}}}{2} = \frac{4 \pm \sqrt{\frac{16}{3}}}{2}. \] \[ x = \frac{4 \pm \frac{4}{\sqrt{3}}}{2}. \] Thus: \[ x = 2 \pm \frac{2}{\sqrt{3}}. \] So, the value of \(c\) is: \[ c = 2 + \frac{2}{\sqrt{3}}. \]