Let f(x) = begin{cases} x^p sin left( frac{1} | vedclass

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(C) Let $g(x) = x|x^3|$. Since $g(x) = x^4$ for $x \ge 0$ and $g(x) = -x^4$ for $x < 0$,$g(x)$ is $3$ times differentiable at $x=0$ and $g''(0) = 0$.

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$(4, \infty)$

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(C) Let $g(x) = x|x^3|$. Since $g(x) = x^4$ for $x \ge 0$ and $g(x) = -x^4$ for $x Now consider $h(x) = x^p \sin \left( \frac{1}{x} ight)$ for $x 
eq 0$ and $h(0) = 0$.For $h'(0)$ to exist,$\lim_{h 	o 0} \frac{h^p \sin(1/h)}{h} = 0$,which requires $p > 1$. Then $h'(0) = 0$.For $x 
eq 0$,$h'(x) = p x^{p-1} \sin \left( \frac{1}{x} ight) - x^{p-2} \cos \left( \frac{1}{x} ight)$.For $h''(0)$ to exist,$\lim_{h 	o 0} \frac{h'(h) - h'(0)}{h} = \lim_{h 	o 0} \left( p h^{p-2} \sin \left( \frac{1}{h} ight) - h^{p-3} \cos \left( \frac{1}{h} ight) ight) = 0$,which requires $p > 3$. Then $h''(0) = 0$.For $x 
eq 0$,$h''(x) = p(p-1) x^{p-2} \sin \left( \frac{1}{x} ight) - p x^{p-3} \cos \left( \frac{1}{x} ight) - (p-2) x^{p-3} \cos \left( \frac{1}{x} ight) - x^{p-4} \sin \left( \frac{1}{x} ight) = p(p-1) x^{p-2} \sin \left( \frac{1}{x} ight) - 2(p-1) x^{p-3} \cos \left( \frac{1}{x} ight) - x^{p-4} \sin \left( \frac{1}{x} ight)$.For $h''(x)$ to be continuous at $x=0$,we need $\lim_{x 	o 0} h''(x) = h''(0) = 0$. This requires the power of $x$ in all terms to be positive,specifically $p-4 > 0$,so $p > 4$.

Let $f(x) = \begin{cases} x^p \sin \left( \frac{1}{x} \right) + x|x^3|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then the complete set of values of $p$ for which $f''(x)$ is continuous at $x = 0$ is:

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