Discussion
If \(f(x) = x\sin(x)\) and \(g(x) = |x|\sin(x)\), then
Answer: D
The correct statement is that \(g(x)\) is differentiable at \(x = 0\).
Near 0, \(\sin x \approx x\), so \(g(x) \approx |x|x\), and
\[ \begin{aligned} g'(0) &= \lim_{h \to 0} \frac{|h|\sin h}{h} \\ &= \lim_{h \to 0} |h|\frac{\sin h}{h} = 0 \end{aligned} \]
from both sides, so the derivative exists and equals 0.
The others fail. \(|f(x)| = |x||\sin x|\), which differs from g wherever \(\sin x \lt 0\). g is odd, the product of the even |x| and the odd sin x. And since \(g = -f\) for \(x \lt 0\), the maxima of g there sit at the minima of f, so the maxima do not coincide.
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