Solutions

a) Notice that

$$f(x)=|x|^3=\begin{cases} x^3, & x\ge 0,\\ -x^3, & \text{otherwise}. \end{cases}$$

Hence,

$$f'(x)=\begin{cases} 3x^2, & x>0,\\ -3x^2, & x<0, \end{cases}$$

and

$$\lim_{h\to 0^-}\frac{-h^3}{h}=\lim_{h\to 0^-}-h^2=\lim_{h\to 0^+}h^2=\lim_{h\to 0^+}\frac{h^3}{h}=0=f'(0).$$

In addition,

$$f''(x)=\begin{cases} 6x, & x>0,\\ -6x, & x<0, \end{cases}$$

and

$$\lim_{h\to 0^-}\frac{-3h^2}{h}=\lim_{h\to 0^-}-3h=\lim_{h\to 0^+}3h=\lim_{h\to 0^+}\frac{3h^2}{h}=0=f''(0).$$

Finally,

$$f'''(x)=\begin{cases} 6, & x>0,\\ -6, & x<0. \end{cases}$$

But

$$\lim_{h\to 0^-}\frac{-6h}{h}=-6\ne 6=\lim_{h\to 0^+}\frac{6h}{h}.$$

Thus $f'''(0)$ does not exist.

b) We have

$$f'(x)=\begin{cases} 4x^3, & x\ge 0,\\ -4x^3, & \text{otherwise}, \end{cases}$$$$f''(x)=\begin{cases} 12x^2, & x\ge 0,\\ -12x^2, & \text{otherwise}, \end{cases}$$

and

$$f'''(x)=\begin{cases} 24x, & x\ge 0,\\ -24x, & \text{otherwise}. \end{cases}$$

But $f^{(4)}(0)$ does not exist.