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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?

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    $\begingroup$ Does $f^3(0)$ exist? $\endgroup$ – who Oct 25 at 4:44
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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.

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  • $\begingroup$ How can I show it is differentiable at $f^{1}$? $\endgroup$ – K. Gibson Oct 25 at 5:24
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    $\begingroup$ You wrote "$f''$ is not differentiable in $x_0=0$". However, I believe you mean $f'''$ instead. $\endgroup$ – John Omielan Oct 25 at 5:30
  • $\begingroup$ 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. $\endgroup$ – Luca Oct 26 at 16:00
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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?

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