Question

The set of all points of differentiability of the function $f(x)=e^{-|x|}$ is

• A. $(0, \infty)$
• B. $[0, \infty)$
• C. $(-\infty, \infty)$
• D. $(-\infty, \infty) - \{0\}$

Hint:

Functions involving $|x|$ are typically non-differentiable at $x=0$ because of a sharp corner in the graph.

Answer 1

The Correct Option is D

Step 1: Concept Check for the "sharp turn" or cusp where the absolute value function $|x|$ changes behavior.

Step 2: Meaning $f(x) = e^x$ for $x < 0$ and $f(x) = e^{-x}$ for $x \geq 0$.

Step 3: Analysis Derivative for $x < 0$ is $e^x$; at $0^-$ it is $e^0=1$. Derivative for $x > 0$ is $-e^{-x}$; at $0^+$ it is $-e^0 = -1$.

Step 4: Conclusion Since LHD ($1$) $\neq$ RHD ($-1$) at $x=0$, the function is not differentiable at $x=0$ but is differentiable everywhere else. Final Answer: (D)