Prove that $f(x) = a_0+a_1|x|+a_2|x|^2+a_3|x|^3 + \ldots $ is differentiable if and only if $a_1=0$.

Question

Prove or disprove that $f(x) = a_0+a_1|x|+a_2|x|^2+a_3|x|^3 + ... $ is differentiable if and only if $a_1=0$.

I highly suspect this is true (so is to be proven not disproven), I checked for n=3(even n is obvious) and the limits both approach 0 at x=0, which would be the only place where something could go wrong.I assume there is a trick to proving $|x|^n, n\neq1$ is differentiable, and by showing $|x|$ is not differentiable at x=0, the proof is done, however I can't see how I would prove this.

Answer 1

Hint:
$|x|^n= \{ x^n$ for $x\ge 0$ and $(-x)^n$ for $x\lt 0 \}$
Now find the derivative at $x=0$

Answer 2

Let$$g(x)=a_0+a_2\lvert x\rvert^2+a_3\lvert x\rvert^3+\cdots$$Clearly, $g(0)=a_0$. Besides, $g$ is differentiable at $0$, since$$\left\lvert\frac{g(x)-g(0)}x\right\rvert\leqslant\lvert a_2\rvert\lvert x\rvert+\lvert a_3\rvert\lvert x\rvert^2+\cdots$$and $\lim_{x\to0}\lvert a_2\rvert\lvert x\rvert+\lvert a_3\rvert\lvert x\rvert^2+\cdots=0$. So, $f$ is the sum of $x\mapsto a_1\lvert x\rvert$ with $g$. Therefore,\begin{align}f\text{ is differentiable at $0$}&\iff x\mapsto a_1\lvert x\rvert\text{ is differentiable at $0$}\\&\iff a_1=0.\end{align}