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: Option A, $f(x)=|x-1|^3$.
Write it as $((x-1)^2)^{3/2}$. Near $x=1$ it behaves like $|x-1|^3$, and both one sided slopes are $0$. So it is smooth at $x=1$ and option A is fine.

Step 2: Option B, $f(x)=|x^2-1|$.
Just past $x=1$ the inside is positive, slope $2x=2$. Just before, the inside is negative, slope $-2x=-2$. The two slopes do not match, so there is a corner. Option B fails.

Step 3: Option C, the piecewise function.
For $|x|\le 1$ we have $x^2 e^{-x^2}$, and for $|x|>1$ it is the constant $e^{-1}$. At $x=1$ both pieces give $e^{-1}$, so it is continuous.

Step 4: Match the slopes for C.
The left slope is $\frac{d}{dx}\big(x^2 e^{-x^2}\big)=(2x-2x^3)e^{-x^2}$, which at $x=1$ is $(2-2)e^{-1}=0$. The right slope is $0$ since that side is constant. They agree, so C is differentiable.

Step 5: Option D, $f(x)=\lfloor x\rfloor$.
The floor function jumps at $x=1$, so it is not even continuous there. Option D fails.

Step 6: Conclusion.
Differentiable at $x=1$ in options A and C.
\[ \boxed{A,\ C} \]