For $n\ge \space 2$ , is a fixed positive integer and $f(x)=x^n\mid x \mid, \space x \space \in \mathbb{R} $

Question

Options given are:

A. $f$ is differentiable everywhere only when $n$ is even.

B. $f$ is differentiable everywhere except at $0$ if $n$ is odd

C.$f$ is differentiable everywhere

D.None of the above.

I cannot understand what has n being even or odd to do anything with $f$ being differentiable?

Please help.

Answer 1

First, $x|x|=x^2$. Second, $x^2|x|=|x^3|$.

So we see that, if $n$ is odd then $x^n|x|=x^{n+1}$ and $n+1$ is even, that is, $x^{n+1}$ is differentiable everywhere.

If, however, $n$ is even the $x^n|x|=|x^{n+1}$|. That is, the question is: if $m$ is odd then is $|x^m|$ differentiable?

Now, $x^m$ is differentiable from the right, no question about that. The derivative is zero at zero. Also, $-x^m$ is differentiable from the left, and the derivative is zero at zero again.

So, $|x^m|$ is differentiable for all $x$ and for all $n>1$.

Answer 2

Let note

• $x\ge 0 \implies f(x)=x^{n+1}\implies f'(x)=(n+1)x^n$
• $x< 0 \implies f(x)=-x^{n+1}\implies f'(x)=-(n+1)x^n$

and by definition of derivative at $x=0$

$$f'(0) = \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{|h|h^{n}}{h}=\lim_{h\to 0} (\operatorname{sign}(h)\cdot h^{n-1})=0$$

and then $f$ is differentiable everywhere.