Differentiability of function

Question

Given $f(x)=a_0+a_1 |x|+a_2 |x|^2+a_3 |x|^3 $ We have to find the range of the constants for which the function is differentiable.

I tried to solve, knowing that sum of differentiable functions is differentiable.

so $a_0+a_2 |x|^2$ is differentiable. I don't know how to proceed further.

I would, with my narrow span of knowledge conclude that $a_3$ and $a_1$ are 0. But my answer is incorrect. Can someone help me with this?

If you know the answer, maybe you could post it here. Anyway, I agree that $a_3$ and $a_1$ are 0: else it is not differentiable at x=0. And it seems that $a_0$ and $a_2$ can take any value, so I would say: $a_3,a_1=0$ and $a_2,a_0\in R$ — Pim, Jun 15 '14 at 09:27
I don't know the answer. I have been given 4 options, none of them matches this. — user141561, Jun 15 '14 at 09:31

Dario · Answer 1 · 2014-06-15T10:00:53.367

0

In general if $x\in\mathbb{R}$ $|x|^n$ is a $C^{n-1}$ for odd $n$ and $C^\infty$ for even $n$. The only summand that "prevent" differentiability is $|x|$, $|x|^3$ is fine (even if $x\in\mathbb{R}^n$). So you should have just to impose $a_1=0$.

To get an idea of how $|x|^3$ is at least $C^1$ you can look at this picture of its graph: enter image description here

edited Jun 15 '14 at 10:00

answered Jun 15 '14 at 09:40

Dario

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score 0 · Answer 2 · answered Jun 15 '14 at 09:40

We have $$f(x) = \begin{cases} a_0 + a_1x + a_2 x^2 + a_3 x^3 & {\text{if $x \geq 0$}};\\ a_0 - a_1x + a_2 x^2 - a_3 x^3 & {\text{if $x \leq 0$}}. \end{cases}$$ Hence $f$ is differentiable everywhere except possibly at $0$. The left- and right-sided derivatives $\partial_- f, \partial_+ f$ of $f$ exist everywhere; $f$ is differentiable exactly where they agree. Since $\partial_\pm f(0) = \pm a_1$, it follows that $f$ is differentiable iff $a_1 = 0$.

Differentiability of function

2 Answers2