Question

A function \( f \) is defined by \( f(x) = 2 + (x - 1)^{2/3} \) on \( [0, 2] \). Which of the following statements is incorrect?

• A. \( f \) is not derivable in \( (0, 2) \).
• B. \( f \) is continuous in \( [0, 2] \).
• C. \( f(0) = f(2) \).
• D. Rolle's theorem is a
  pplicable on \( [0, 2] \).

Hint:

Remember that a function must be differentiable on the open interval for Rolle's theorem to apply. Points where the derivative is undefined violate this condition.

Answer 1

The Correct Option is D

Step 1: Continuity of \( f(x) \) on \( [0, 2] \).
\n\( g(x) = (x - 1)^{2/3} = (\sqrt[3]{x - 1})^2 \) is continuous on \( [0, 2] \) because it's a composition of continuous functions.
\nTherefore, \( f(x) = 2 + g(x) \) is also continuous on \( [0, 2] \). Statement (B) is confirmed.
\n\n
Step 2: Differentiability of \( f(x) \) on \( (0, 2) \).
\nThe derivative of \( f(x) \) is \( f'(x) = \frac{2}{3}(x - 1)^{-1/3} = \frac{2}{3\sqrt[3]{x - 1}} \). This derivative is undefined at \( x = 1 \), which is in \( (0, 2) \). Hence, \( f(x) \) is not differentiable in \( (0, 2) \). Statement (A) is confirmed.\n\n
Step 3: Check \( f(0) = f(2) \).
\n\( f(0) = 2 + (0 - 1)^{2/3} = 2 + (-1)^{2/3} = 2 + 1 = 3 \)
\n\( f(2) = 2 + (2 - 1)^{2/3} = 2 + (1)^{2/3} = 2 + 1 = 3 \)
\nSince \( f(0) = f(2) = 3 \), statement (C) is confirmed.
\n\n
Step 4: Rolle's theorem on \( [0, 2] \).
\nRolle's theorem has three requirements:
\n1. \( f \) is continuous on \( [0, 2] \). (Met)\n2. \( f \) is differentiable on \( (0, 2) \). (Not met, as \( f \) isn't differentiable at \( x = 1 \))\n3. \( f(0) = f(2) \). (Met)
\n\nSince condition 2 is not satisfied, Rolle's theorem isn't applicable on \( [0, 2] \). Statement (D) is incorrect.