# When is the derivative of $x^n\mathbf 1_{[0,\infty)}$ continuous/differentiable?

I am having difficulty proving the following: Let $n \in \mathbb N$ and let $f\colon \mathbb R\to \mathbb R$ be defined by $f(x)=x^n$ for $x\ge0$ and $f(x)=0$ for $x<0$. For which values of $n$ is $f'$ continuous at $0$? For which values of $n$ is $f'$ differentiable at $0$? I'm not sure how to solve the problem but I know the definition of a derivative will probably be utilized somehow for the case when $x<0$.

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Can you find f'? – Salech Alhasov Mar 31 '12 at 22:41
for x>=0 f'(x)=n*x^n-1, for x<=0 f'(0)=0. How should I proceed with finding where it is continuous and differentiable at 0? – Quaternary Mar 31 '12 at 22:51
@Quaternary : I suspect you read "differentiable" but wrote "differential". – Michael Hardy Mar 31 '12 at 23:03
Great! Now find $n$ such that $\lim_{+0}f'=\lim_{-0}f'$ – Salech Alhasov Mar 31 '12 at 23:03
Thanks for the hints. I'm currently having difficulty finding such an n. – Quaternary Mar 31 '12 at 23:20

I'll assume that $\Bbb N$ is the set of positive integers.

First, we need to find when $f$ is differentiable at $0$.

Things are nice for $x\ne0$. We have $$f'(x)=0\quad\text{for}\quad x<0$$ and$$f'(x)=nx^{n-1}\quad\text{for}\quad x>0.$$
From this, we see that the derivative of $f$ from the left at $x=0$ is $0$. Thus, in order for $f'(0)$ to be defined, the derivative of $f$ from the right at $x=0$ must be $0$. The derivative from the right at $x=0$ is given by the limit $$\lim\limits_{h\rightarrow 0^+} {h^{n } \over h} = \lim\limits_{h\rightarrow 0^+}\, {h^{n-1 } };$$ which is $0$ if and only if $n\ge 2$. (Note we could have computed $\lim\limits_{x\rightarrow0^+} nx^{n-1}$ here instead.)

So, for $n\ge 2$, $f$ is differentiable everywhere and, in this case: $$f'(x)=\cases{0, &x\le 0\cr nx^{n-1}, &x>0}.$$

One easily verifies that $f'$ is continuous at $x=0$ (in fact everywhere) for $n\ge2$.

Now for the differentiability of $f'$:

We have $$f''(x)=0\quad\text{for}\quad x< 0$$ and $$f''(x)=n(n-1)x^{n-2}\quad\text{for}\quad x>0.$$ As before, $f''$ exists at $x=0$ if and only if $$\lim\limits_{h\rightarrow 0} {nh^{n-1}\over h}=\lim\limits_{h\rightarrow 0} {nh^{n-2} } =0.$$ This occurs if and only if $n\ge 3$.

So $f'$ is differentiable at $x=0$ if and only if $n\ge3$.

Pictorially, not much is going on here. $f$ is differentiable at $0$ (and the derivative is continuous at $x=0$) if and only if the graph of $f$ on the positive $x$-axis is not a straight line with non-zero slope. $f'$ is differentiable at $x=0$ if and only if its graph on the positive $x$-axis is not a straight line with non-zero slope, which happens if and only if the graph of $f$ on the positive $x$-axis is not a parabola.

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