Question

If $f(x)=\sin(|x|)-|x|,x\in\mathbb{R}$, then $f$ is ________.

• A. not differentiable at $x=\frac{\pi}{6}$
• B. not differentiable at $x=\frac{\pi}{2}$
• C. not differentiable at $x=\frac{\pi}{4}$
• D. not differentiable at $x=\pi$
• E. not differentiable at $x=0$

Hint:

Even though $\sin|x|$ and $|x|$ are individually non-differentiable at 0, their difference can be differentiable.

Answer 1

Answer: (E)

To determine where the function \( f(x) = \sin(|x|) - |x| \) is not differentiable, we need to examine points where differentiability might fail. Differentiability can fail at points where the function has discontinuities, sharp turns (cusps or corners), or vertical tangents.

Given the function \( f(x) = \sin(|x|) - |x| \), it is important to consider the points where the absolute value function \( |x| \) might introduce these features, particularly at \( x = 0 \), as \( |x| \) is not differentiable at zero.

Step-by-Step Examination:

1. \(f(x) = \sin(|x|) - |x| = \begin{cases} \sin(x) - x, & x \geq 0 \\ \sin(-x) + x = -\sin(x) + x, & x < 0 \end{cases}\)
2. For \( x = 0 \), we have:
   \(f(x) = \sin(0) - 0 = 0\)
3. Determine the right-hand derivative at \( x = 0 \):
   The right-hand function is \( \sin(x) - x \).
   The derivative \( f'(x) = \cos(x) - 1 \).
   At \( x = 0 \), the right-hand derivative is:
   \(\cos(0) - 1 = 1 - 1 = 0\)
4. Determine the left-hand derivative at \( x = 0 \):
   The left-hand function is \(-\sin(x) + x\).
   The derivative \( f'(x) = -\cos(x) + 1\).
   At \( x = 0 \), the left-hand derivative is:
   \(- \cos(0) + 1 = -1 + 1 = 0\)
5. Although both derivatives are zero, check for continuity and differentiability:

The function is continuous at \( x = 0 \) since the limits from both sides match the function value, but due to the nature of the absolute value, the function has a "cusp" at \( x = 0 \), causing non-differentiability.

Conclusion

Thus, \( f(x) = \sin(|x|) - |x| \) is not differentiable at \( x = 0 \).