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

Concept: Rolle’s theorem states that if a function is continuous in \( [a,b] \), differentiable in \( (a,b) \), and \( f(a)=f(b) \), then \( f'(c)=0 \) for some \( c \in (a,b) \).

Step 1: Check Rolle’s conditions. Polynomial $\Rightarrow$ continuous and differentiable everywhere. \[ f(0)=0, \quad f(2)=8 - 12 + 4 = 0 \] So Rolle’s theorem is applicable.

Step 2: Find derivative: \[ f'(x) = 3x^2 - 6x + 2 \]

Step 3: Solve \( f'(x)=0 \): \[ 3x^2 - 6x + 2 = 0 \] \[ x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} \] \[ x = 1 \pm \frac{1}{\sqrt{3}} \] Final Answer: \[ c = 1 \pm \frac{1}{\sqrt{3}} \]