Question

If \( f(x) = (2x-1)(3x+2)(4x-3) \) is a real-valued function defined on \(\left[\frac{1}{2}, \frac{3}{4}\right]\), then the value(s) of 'c' as defined in the statement of Rolle's theorem is:

• A. Does not exist
• B. \( \frac{7 + \sqrt{247}}{36} \)
• C. \( \frac{7 - \sqrt{247}}{36} \)
• D. \( \frac{7 + \sqrt{247}}{36} \)

Hint:

Apply calculus tools such as derivatives carefully, and always verify the conditions of theorems before using their results.

Answer 1

The Correct Option is D

To determine the value(s) of \( c \) as defined in Rolle's theorem for the function \( f(x) = (2x-1)(3x+2)(4x-3) \) on the interval \(\left[\frac{1}{2}, \frac{3}{4}\right]\), follow these steps: 1. Verify the Conditions of Rolle's Theorem: Continuity: The function \( f(x) \) is a polynomial, and polynomials are continuous everywhere, including on \(\left[\frac{1}{2}, \frac{3}{4}\right]\). Differentiability: Polynomials are also differentiable everywhere, so \( f(x) \) is differentiable on \(\left(\frac{1}{2}, \frac{3}{4}\right)\). Equal Endpoints: Calculate \( f\left(\frac{1}{2}\right) \) and \( f\left(\frac{3}{4}\right) \): \[ f\left(\frac{1}{2}\right) = (2 \cdot \frac{1}{2} - 1)(3 \cdot \frac{1}{2} + 2)(4 \cdot \frac{1}{2} - 3) = (0)(\frac{7}{2})(-1) = 0 \] \[ f\left(\frac{3}{4}\right) = (2 \cdot \frac{3}{4} - 1)(3 \cdot \frac{3}{4} + 2)(4 \cdot \frac{3}{4} - 3) = (\frac{1}{2})(\frac{17}{4})(0) = 0 \] Since \( f\left(\frac{1}{2}\right) = f\left(\frac{3}{4}\right) = 0 \), the conditions of Rolle's theorem are satisfied. 2. Find the Derivative \( f'(x) \): First, expand \( f(x) \): \[ f(x) = (2x - 1)(3x + 2)(4x - 3) \] Let’s expand step by step: \[ (2x - 1)(3x + 2) = 6x^2 + 4x - 3x - 2 = 6x^2 + x - 2 \] Now multiply by \( (4x - 3) \): \[ f(x) = (6x^2 + x - 2)(4x - 3) = 24x^3 - 18x^2 + 4x^2 - 3x - 8x + 6 = 24x^3 - 14x^2 - 11x + 6 \] Now, differentiate \( f(x) \): \[ f'(x) = \frac{d}{dx}(24x^3 - 14x^2 - 11x + 6) = 72x^2 - 28x - 11 \] 3. Set the Derivative Equal to Zero to Find \( c \): \[ 72x^2 - 28x - 11 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 72 \cdot (-11)}}{2 \cdot 72} = \frac{28 \pm \sqrt{784 + 3168}}{144} = \frac{28 \pm \sqrt{3952}}{144} \] Simplify \( \sqrt{3952} \): \[ \sqrt{3952} = \sqrt{16 \times 247} = 4\sqrt{247} \] Thus: \[ x = \frac{28 \pm 4\sqrt{247}}{144} = \frac{7 \pm \sqrt{247}}{36} \] 4. Determine Which Root Lies in the Interval \(\left(\frac{1}{2}, \frac{3}{4}\right)\):
Calculate the approximate values: \[ \frac{7 + \sqrt{247}}{36} \approx \frac{7 + 15.72}{36} \approx \frac{22.72}{36} \approx 0.631 \] \[ \frac{7 - \sqrt{247}}{36} \approx \frac{7 - 15.72}{36} \approx \frac{-8.72}{36} \approx -0.242 \] Only \( \frac{7 + \sqrt{247}}{36} \approx 0.631 \) lies within the interval \(\left(\frac{1}{2}, \frac{3}{4}\right)\). Therefore, the correct value of \( c \) is: \[ \boxed{\frac{7 + \sqrt{247}}{36}} \]