Answer to Question #123126 in Real Analysis for Wachira Ann Wangari

Question

Let f(x)=|x|^3. Show that f'''(0) does not exist

Answer 1

$f(x)=|x|^3=\left\{\begin{array}{ll}x^3 & \text { if } x \geq 0 \\ -x^3 & \text { if } x<0\end{array}\right.\\[1 em] So, \lim _{x \rightarrow 0^{+}}|x|^3=\lim _{x \rightarrow 0^{+}} x^3=0 ~~\text{and} ~~\lim _{x \rightarrow 0^{-}}|x|^3=\lim _{x \rightarrow 0^{-}}(-x)^3=0\\[1 em] \therefore ~f(x)=|x|^3~~\text{is continuous at x=0}\\[1 em] \text{To show that} f(x)=|x|^3 ~\text{is not differentiable} \\[1 em] f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\\[1 em] \begin{array}{l} \lim _{h \rightarrow 0} \frac{|0+h|^3-|0|^3}{h}=\lim _{h \rightarrow 0} \frac{|h|^3}{h} =\left\{\begin{array}{ll} 0 & \text { if } h>0 \\ 0 & \text { if } h<0 \end{array}\right. \end{array}\\[1 em] \therefore ~f^{\prime}(x) \text{ exist at }x=0\\[1 em] f^{{\prime }}(x)=\left\{\begin{array}{ll}3x^2 & \text { if } x \geq 0 \\ -3x^2 & \text { if } x<0\end{array}\right.\\[1 em] \text{In the same way, we find that the second derivative is exist at}~x=0\\[1 em] f^{{\prime }{\prime }}(x)=\left\{\begin{array}{ll}6x & \text { if } x \geq 0 \\ -6x & \text { if } x<0\end{array}\right.\\[1 em] \text{Now we are checking the third derivative at}~x=0\\[1 em] f^{{\prime }{\prime }{\prime }}(0)^{+}=\lim _{x \rightarrow 0^{+}}f^{{\prime }{\prime }}(x)= \frac{6(0+h)-f(0)}{h}=6 \\[1 em] f^{{\prime }{\prime }{\prime }}(0)^{-}=\lim _{x \rightarrow 0^{-}}f^{{\prime }{\prime }}(x)= \frac{-6(0+h)-f(0)}{h}=-6 \\[1 em] \therefore~f^{{\prime }{\prime }{\prime }}(0) \text{does not exist}$