Showing absolute minimum

**frenchguy87** · April 6th 2010, 04:56 PM

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x) = 2x^4+x^4\sin{1/x}$ for $x\neq0$ 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 .

**Drexel28** · April 6th 2010, 05:08 PM

Originally Posted by frenchguy87

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x) = 2x^4+x^4\sin{1/x}$ for $x\neq0$ 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 .

$f(x)=2x^4+x^4\sin\left(\tfrac{1}{x}\right)\geqslan t 2x^4-x^4=x^4\geqslant 0$ . Thus, $f(x)\geqslant0,\text{ }\forall x\in\mathbb{R}$ . The fact that $f(0)=0$ let's you draw the proper conclusion.

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

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

I leave this to you.

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