Question

If Rolle’s theorem is applied for $ f(x) = x^3 - x $ in $ [-1,1] $, then the values of ‘c’ of this theorem are:

• A. \( 0 \)
• B. \( \pm \frac{1}{\sqrt{2}} \)
• C. \( \pm \frac{1}{\sqrt{3}} \)
• D. \( \frac{1}{2} \pm \frac{1}{\sqrt{3}} \)

Hint:

To apply Rolle's theorem, always check if the function values at the endpoints are equal, and then find the points where the derivative equals zero.

Answer 1

The Correct Option is C

Step 1: Check if the function satisfies the conditions for Rolle's theorem

• The function must be continuous on the closed interval \([-1,1]\).
• The function must be differentiable on the open interval \((-1,1)\).
• The function must satisfy \( f(-1) = f(1) \).

Step 2: Check if the function values at the endpoints are equal

First, compute \( f(-1) \) and \( f(1) \):

\[ f(-1) = (-1)^3 - (-1) = -1 + 1 = 0 \]

\[ f(1) = (1)^3 - (1) = 1 - 1 = 0 \]

Since \( f(-1) = f(1) = 0 \), the function satisfies the condition for Rolle's theorem.

Step 3: Compute the derivative of the function

Next, compute the derivative of \( f(x) = x^3 - x \):

\[ f'(x) = 3x^2 - 1 \]

Step 4: Solve for \( c \) where \( f'(c) = 0 \)

According to Rolle’s theorem, there exists at least one \( c \) in the open interval \((-1, 1)\) such that \( f'(c) = 0 \). Set \( f'(x) = 0 \):

\[ 3x^2 - 1 = 0 \]

\[ 3x^2 = 1 \]

\[ x^2 = \frac{1}{3} \]

\[ x = \pm \frac{1}{\sqrt{3}} \]

Thus, the values of \( c \) are \( \pm \frac{1}{\sqrt{3}} \).