Show an absolute minimum and positive/negative derivative of function Let $f : \mathbb R \to \mathbb R$ be defined by $f(x) := 2x^4+x^4\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$. Show that f has an absolute minimum at x = 0, but that its derivative has both positive and negative values in every neighborhood of 0.
I'm already confused intuitively about this problem
To prove absolute minimum, I need to show that for all $c$ in $\mathbb R$, $f(c) > f(x)$ when $x = 0$ right? or in other words $f'(x) = 0$, but it says that the derivative has positive and negative values in the neighborhood of $0$? What?
 A: Since $f(x)=x^4\left(2+\sin(1/x)\right)$ when $x\ne 0$ we have
$$x^4\le f(x)\le 3x^4,\quad x\in\mathbb{R}\backslash\{0\}$$
Then $f(x)\ge0=f(0)$ for all $x\in\mathbb{R}$, so $f$ has an absolute minimum at $x=0$.
On the other hand, 
\begin{align*}
f'(0)&=\lim_{h\to 0}\frac{h^4\left[2+\sin\left(\frac{1}{h}\right)\right]-0}{h}\\
&=\lim_{h\to 0}\left[h^3\left(2+\sin\frac{1}{h}\right)\right]
\end{align*}
last limit exists, and is $0$, since $1\le2+\sin(1/h)\le 3$ for every $h\ne 0$ and $h^3\to 0$. So $$\color{blue}{f'(0)=0}$$
Thus $$\color{blue}{f'(x)=\begin{cases}0&x=0\\4x^3\left[2+\sin\frac{1}{x}\right]-x^2\cos\frac{1}{x}&x\ne 0\end{cases}}$$
If $U\ni0$ is a neighborhood of $0$ there exist $r>0$ such that $(-r,r)\subset U$, now, by Archimedean property there exist $N\in\mathbb{N}$ such that $N>\frac{1}{r}$, so 
$$x_N=-\frac{1}{2N\pi}\in U\implies f'(x_N)=8x_N^3-x_N^2<0,\quad\text{and}$$
$$y_N=\frac{1}{(2N+1)\pi}\in U,\quad f'(y_N)=8y_N^3+y_N^2>0$$
A: $f'(x)$ is indeed $0$. But this does not contradict the fact that $f'(x)$ achieves positive and negative values alternately. It is clearly shown in the pictures:


*

*$f(x)=2x^4+x^4 \sin (1/x)$

*It's derivative 
