Question

Which of the following functions is/are differentiable at \( x = 1 \)?

• A. \( f(x) = |x - 1|^3 \)
• B. \( f(x) = |x^2 - 1| \)
• C. \( f(x) = \begin{cases} x^2 e^{-x^2}, & \text{if} \ |x| \leq 1 \\ e^{-1}, & \text{if} \ |x| > 1 \end{cases} \)
• D. \( f(x) = \lfloor x \rfloor \)

Hint:

When checking for differentiability at a point, ensure the function is continuous at that point and that the derivatives from both sides match at that point.

Answer 1

The Correct Option is A, C

Step 1: Analyze option (A).
The function \( f(x) = |x - 1|^3 \) is the cube of the absolute value function. The absolute value function \( |x - 1| \) is differentiable at \( x = 1 \), and the cube of a differentiable function is also differentiable.
Therefore, \( f(x) \) is differentiable at \( x = 1 \). Thus, option (A) is true.

Step 2: Analyze option (B).
The function \( f(x) = |x^2 - 1| \) involves the absolute value of \( x^2 - 1 \), which is not differentiable at \( x = 1 \) because the derivative at \( x = 1 \) involves a sharp corner. Therefore, option (B) is false.

Step 3: Analyze option (C).
The function \( f(x) = \begin{cases} x^2 e^{-x^2}, & \text{if} \ |x| \leq 1 \\ e^{-1}, & \text{if} \ |x| > 1 \end{cases} \) is continuous at \( x = 1 \), but we need to check the derivative.
The two pieces of the function must also have matching derivatives at \( x = 1 \) for differentiability. Upon differentiating both sides, we find that the derivatives do not match at \( x = 1 \), so the function is not differentiable at \( x = 1 \). Thus, option (C) is false.

Step 4: Analyze option (D).
The function \( f(x) = \lfloor x \rfloor \) is the floor function, which has jumps at integer values, including at \( x = 1 \).
Since the floor function is not continuous at integer values, it is not differentiable at \( x = 1 \). Thus, option (D) is false.

Step 5: Conclusion.
From the analysis, we conclude that option (A) is the correct one, as \( f(x) = |x - 1|^3 \) is differentiable at \( x = 1 \). Therefore, the correct answer is (A).