1. ## Showing absolute minimum

Let $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $\displaystyle f(x) = 2x^4+x^4\sin{1/x}$ for $\displaystyle x\neq0$ and $\displaystyle f(0)=0$ .
Show that f has an absolute minimum at $\displaystyle x=0$, but that its derivative has both positive and negative values in every neighborhood of 0 .

2. Originally Posted by frenchguy87
Let $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $\displaystyle f(x) = 2x^4+x^4\sin{1/x}$ for $\displaystyle x\neq0$ and $\displaystyle f(0)=0$ .
Show that f has an absolute minimum at $\displaystyle x=0$, but that its derivative has both positive and negative values in every neighborhood of 0 .
$\displaystyle f(x)=2x^4+x^4\sin\left(\tfrac{1}{x}\right)\geqslan t 2x^4-x^4=x^4\geqslant 0$. Thus, $\displaystyle f(x)\geqslant0,\text{ }\forall x\in\mathbb{R}$. The fact that $\displaystyle f(0)=0$ let's you draw the proper conclusion.

Now, $\displaystyle f'(x)=\begin{cases} 8x^3+4x^3\sin\left(\tfrac{1}{x}\right)-x^2\cos\left(\tfrac{1}{x}\right) & \mbox{if} \quad x\ne 0 \\ 0 & \mbox{if}\quad x=0\end{cases}$. Notice that $\displaystyle f'(x)=x^2\left(8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)\right)$ and so it suffices to show the claim for $\displaystyle g(x)=8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)$

Now, choose values of $\displaystyle x$ small enough to fit into any range such that the first two terms are irrelevant.

I leave this to you.