Question

At x = 1, the function f(x) = \(\left\{ \begin{aligned}     & x^3-1\ \ \ \ \ \ 1<x<\infin \\     & x-1\ \ \ \ \ -\infin<x≤1 \\      \end{aligned} \right.\) is

• A. continuous and differentiable
• B. continuous and non-differentiable
• C. discontinuous and differentiable
• D. discontinuous and non-differentiable

Answer 1

The Correct Option is B

We need to check the continuity and differentiability of \( f(x) \) at \( x = 1 \).
Step 1: Checking Continuity at \( x = 1 \)}
For continuity, we need to check if the left-hand limit, right-hand limit, and the value of the function at \( x = 1 \) are equal.
Left-hand limit as \( x \to 1^- \): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x - 1) = 0 \]
Right-hand limit as \( x \to 1^+ \): \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^3 - 1) = 0 \]
Value of the function at \( x = 1 \): \[ f(1) = 1 - 1 = 0 \] Since both the left-hand and right-hand limits are equal and \( f(1) = 0 \), the function is continuous at \( x = 1 \).
Step 2: Checking Differentiability at \( x = 1 \)}
To check differentiability, we need to verify if the left-hand and right-hand derivatives at \( x = 1 \) are equal.
Left-hand derivative (for \( x \leq 1 \)): \[ \frac{d}{dx} (x - 1) = 1 \]
Right-hand derivative (for \( x > 1 \)): \[ \frac{d}{dx} (x^3 - 1) = 3x^2 \] At \( x = 1 \), the right-hand derivative is: \[ 3(1)^2 = 3 \] Since the left-hand derivative is \( 1 \) and the right-hand derivative is \( 3 \), they are not equal.
Therefore, the function is not differentiable at \( x = 1 \).

Thus, the function is continuous but not differentiable at \( x = 1 \), so the correct answer is (B) continuous and non-differentiable.