# Find all $n \in \mathbb{N}$ S.T. $f^{n}$ exists on all of $\mathbb{R}$

Define $$f$$ on $$\mathbb{R}$$ by

$$f(x) := \begin{cases} x^3, & \text{if x \geq 0} \\[2ex] 0, & \text{if x \lt 0} \end{cases}$$

find all $$n \in \mathbb{N}$$ such that $$f^{n}$$ exists on all of $$\mathbb{R}$$

I am not sure how to approach this problem. On one hand I can take derivatives all the way down to a constant, 6. Then I need to show that each derivative is defined everywhere on $$\mathbb{R}$$ i.e. show that each derivative is continuous through out $$\mathbb{R}$$. Theoretically I can show this for $$f^{3}, f^{2},f^{1}$$ right?

• Does $f^3(0)$ exist? – who Oct 25 at 4:44

You have that

$$\mathop {\lim }\limits_{h \to 0^ + } \frac{{f''(0 + h) - f''(0)}} {h} = 6$$ while $$\mathop {\lim }\limits_{h \to 0^ - } \frac{{f''(0 + h) - f''(0)}} {h} = 0$$ $$f''$$ is not differentiable in $$x_0=0$$. Therefore your function does have a derivative on $$\mathbb R$$ only up to order 2.

In order to prove that $$f'(0)$$ does exists we have that

$$\mathop {\lim }\limits_{h \to 0^ + } \frac{{f\left( {0 + h} \right) - f(0)}} {h} = \mathop {\lim }\limits_{h \to 0^ + } \frac{{h^3 }} {h} = \mathop {\lim }\limits_{h \to 0^ + } h^2 = 0$$ $$\mathop {\lim }\limits_{h \to 0^ - } \frac{{f\left( {0 + h} \right) - f(0)}} {h} = \mathop {\lim }\limits_{h \to 0^ - } \frac{0} {h} = \mathop {\lim }\limits_{h \to 0^ - } 0 = 0$$ therefore we have that $$f'(x) = \left\{ \begin{gathered} 3x^2 ,\,\,\,\,\,\,x \geqslant 0 \hfill \\ 0,\,\,\,\,\,x < 0 \hfill \\ \end{gathered} \right.$$ Now we can go further in the same way and prove that $$f''(x) = \left\{ \begin{gathered} 6x,\,\,\,\,\,\,x \geqslant 0 \hfill \\ 0,\,\,\,\,\,x < 0 \hfill \\ \end{gathered} \right.$$ When we investigate about f'''(0) we find the left and right limits above and therefore f'''(0) does not exists.

• How can I show it is differentiable at $f^{1}$? – K. Gibson Oct 25 at 5:24
• You wrote "$f''$ is not differentiable in $x_0=0$". However, I believe you mean $f'''$ instead. – John Omielan Oct 25 at 5:30
• Since the right and the left limits are different it means that f'' does not have a derivative. Therefore it is true that it is not differentiable. In other words f''' does not exists. – Luca Oct 26 at 16:00

It may be helpful to consider what happens at $$x = 0$$, and to treat other values of x separately. Basically, it's obvious that for $$x \neq 0$$ all of the derivatives exist. You may be required to actually write something to explain why, but you should focus most attention on $$x = 0.$$

Do you have any trouble finding all $$n \in \mathbb{N}$$ such that $$f^{n}(0)$$ exists?