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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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    Thanks for the hints. I'm currently having difficulty finding such an n.2012-03-31

1 Answers 1

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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 andf'(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.