If $f(x) = |x|^3$, show $f'''(0)$ does not exist

Question

If $f(x) = |x|^3$, show $f'''(0)$ does not exist.

So I've computed the first, second and third derivatives, so

$f'(x) = 3x|x|$

$f''(x) = \begin{cases} 0, & \text{if x = 0} \\ \frac{6x^2}{|x|}, & \text{if x $\ne$ 0} \end{cases}$

$f'''(x) = \frac{6x^3}{|x|^3}$.

I know if I were to simply plug in $0$ for $x$, I would end up with $\frac{0}{0} = 0$, but that doesn't bring me to the $DNE$ conclusion. Is this someplace I should be using L'Hopital's Rule? The question didn't use the word "limit", but it's the only option I can come up with...

Answer 1

Most straightforward is to say:

$f(x) = \begin{cases} -x^3 & x < 0\\x^3 & x\ge 0 \end{cases}$

The derivatives of $f$ will have the same piece-wise structure.

$f'''(x)$ will appear to have a jump discontinuity.

The derivative is defined as a limit, if the left hand limit does not equal the right hand limit, the limit does not exist. The derivative does not exist at the point of this jump.

Answer 2

By L'Hopital rule we have that $$lim_{x\rightarrow 0^+}\frac{ f{''}(x)}{x}=12$$ and $$lim_{x\rightarrow 0^-}\frac{ f{''}(x)}{x}=-12$$

So the third derivative at $0$ does not exist.

Answer 3

You can also think of $f(x)=|x|^3$ as composite function $f(x)=(g\circ h)(x)$ with $g(x)=x^3$ and $h(x)=|x|$. By chain rule if at a point $x=c$ the function $h$ is differentiable and $g$ is differentiable at $h(c)$ then $f'(c)=g'(h(c))h'(c)$. But at $c=0$ we have that $h$ is not differentiable since $h'(0^-)=-1\neq 1=h'(0^+)$. As consequence $f$ is not differentiable at $c=0$. Surely then the third derivative doesn't exist there for it requires the first derivative to exist in the first place.

Answer 4

Let us first establish a lemma:

Given any function $g : (-\epsilon,\epsilon) \to \mathbb{R}$. If $g'(0)$ exists , then$\color{blue}{{}^{[1]}}$ $$\mathcal{J}(g) \stackrel{def}{=} \lim_{h\to 0+} \frac{g(2h)-g(h) + g(-2h) - g(-h)}{h} = 0$$

Let's say $g'(0)$ exists and equals to $L$, we have

$$\lim_{h\to 0} \frac{g(2h)-g(h)}{h} = \lim_{h\to 0} \left[2\left(\frac{g(2h)-g(0)}{2h}\right) - \frac{g(h)-g(0)}{h} \right] = 2L - L = L $$ This implies $$\mathcal{J}(g) = \lim_{h\to 0+} \left[\frac{g(2h)-g(h)}{h} - \frac{g(-2h)-g(-h)}{-h}\right] = L - L = 0$$ and justify the lemma.

Back to original problem. Since $f(x)$ is smooth for $x \ne 0$, it is easy to deduce $f''(x) = 6|x|$ for $x \ne 0$. By direct computation, we have

$$\begin{align}\mathcal{J}(f'') &= \lim_{h\to 0+} \frac{f''(2h)-f''(h) + f''(-2h) - f''(-h)}{h}\\ &= \lim_{h\to 0+} \frac{6|2h| - 6|h| + 6|-2h| - 6|-h|}{h}\\ &= 12 \end{align}$$ Since $\mathcal{J}(f'') \ne 0$, by above lemma, $f'''(0)$ does not exist.

Notes

• $\color{blue}{[1]}$ - $\mathcal{J}(g)$ is the size of jump discontinuity of $g'(x)$ at $x = 0$. The whole point of this complicated mess is to have a formula for the jump in $f''(x)$ at $x = 0$ without explicit referral to values of $f(0)$, $f'(0)$, $f''(0)$ and $f'''(0)$ which are clumsy to describe and justify.

Answer 5

$$f(x)=\begin{cases} -x^3 & x < 0\\x^3 & x\ge 0 \end{cases}\Rightarrow f'(x)=\begin{cases} -3x^2=3x(-x) & x < 0\\3x^2=3x(x) & x\ge 0 \end{cases}\iff f'(x)=3x|x|$$ $$f^{''}(x)=\begin{cases} -6x & x < 0\\6x & x\ge 0 \end{cases}\iff f^{''}(x)=6|x|$$ It follows that the tangents are constantly $-6$ at the left of $0$ and constantly $6$ at the right of $0$ so $f^{''}(x)$ is not differentiable at $0$.

NOTE.-You can verify, for example, drawing the graphics that $6|x|=\begin{cases} 0, & \text{if x = 0} \\ \dfrac{6x^2}{|x|}, & \text{if x $\ne$ 0} \end{cases}$.

I mean your expression and mine are equivalent.