Question

What are the values of \( c \) for which Rolle’s theorem for the function \( f(x) = x^3 - 3x^2 + 2x \) in the interval \( [0,2] \) is verified?

• A. \( c = \pm 1 \)
• B. \( c = 1 \pm \frac{1}{\sqrt{3}} \)
• C. \( c = \pm 2 \)
• D. None of these

Hint:

After verifying Rolle’s conditions, always solve \( f'(x)=0 \) to find required points.

Answer 1

The Correct Option is B

To verify Rolle's Theorem for the function \( f(x) = x^3 - 3x^2 + 2x \) in the interval \([0, 2]\), we must follow these steps:

1. **Understanding Rolle's Theorem:**
   • The theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there is at least one number \( c \) in the interval \((a, b)\) such that \( f'(c) = 0 \).
2. **Check the conditions of Rolle's Theorem:**
   • The function \( f(x) = x^3 - 3x^2 + 2x \) is a polynomial, which is continuous and differentiable everywhere, hence on \([0, 2]\) as well.
   • Calculate \( f(0) \) and \( f(2) \) to ensure \( f(a) = f(b) \).
   • \( f(0) = (0)^3 - 3(0)^2 + 2(0) = 0 \).
   • \( f(2) = (2)^3 - 3(2)^2 + 2(2) = 8 - 12 + 4 = 0 \).
   • Since \( f(0) = f(2) \), the conditions for Rolle's Theorem are satisfied.
3. **Find the derivative \( f'(x) \):**
   • \( f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2x) = 3x^2 - 6x + 2 \).
4. **Solve \( f'(c) = 0 \) to find \( c \):**
   • Set the derivative equal to zero: \( 3c^2 - 6c + 2 = 0 \).
   • This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -6 \), and \( c = 2 \).
   • Use the quadratic formula: \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
   • \( b^2 - 4ac = (-6)^2 - 4 \times 3 \times 2 = 36 - 24 = 12 \).
   • \( c = \frac{-(-6) \pm \sqrt{12}}{2 \times 3} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{1}{\sqrt{3}} \).

Thus, the values of \( c \) for which Rolle's theorem is verified are \( c = 1 \pm \frac{1}{\sqrt{3}} \).