Let $f(x) = |x| \sin(x)$ for all $x \in \mathbb{R}$. Which of the following statements regarding $f(x)$ at $x = 0$ are TRUE?
Options:
(A) $f(x)$ is continuous at $x = 0$
(B) $f(x)$ is differentiable at $x = 0$
(C) $f'(0) = 0$
(D) $f''(0)$ does not exist
Answer Key:
A, B, C, D
Step-by-Step Solution & Explanation
1. $\lim_{x \to 0} f(x) = 0 = f(0)$, so $f(x)$ is continuous at $x = 0$. (Statement A is True)
2. By definition of derivative at $x = 0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{|h| \sin h}{h} = \lim_{h \to 0} |h| \frac{\sin h}{h} = 0 \times 1 = 0$
Since the limit exists and equals 0, $f(x)$ is differentiable at $x = 0$ with $f'(0) = 0$. (Statements B and C are True)
3. For $x > 0$, $f'(x) = \sin x + x \cos x$, and for $x < 0$, $f'(x) = -\sin x - x \cos x$.
Right-hand second derivative: $\lim_{h \to 0^+} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0^+} \frac{\sin h + h \cos h}{h} = 1 + 1 = 2$.
Left-hand second derivative: $\lim_{h \to 0^-} \frac{f'(h) - f'(0)}{h} = -2$.
Since left and right limits disagree, $f''(0)$ does not exist. (Statement D is True)