$\displaystyle f(x) = xsinx, \ \ \ \ g(x)=e^{-x^{4}} $

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- May 1st 2009, 03:42 PMsilversandhow to show if these are uniformly continuous
$\displaystyle f(x) = xsinx, \ \ \ \ g(x)=e^{-x^{4}} $

- May 1st 2009, 04:33 PMTheEmptySet

The first one is not uniformly continous( I don't think)

Here is number two

Since g(x) is differentable $\displaystyle g'(x)=-4x^3e^{-x^4}$ and

the derivative is bounded on all of $\displaystyle \mathbb{R}$.

so $\displaystyle f'(x) < M $ for some $\displaystyle M \in \mathbb{R}$

let $\displaystyle \epsilon > 0$ set $\displaystyle \delta=\frac{\epsilon}{M}$

So now if $\displaystyle |x-y|< \delta$

We need to show that $\displaystyle |f(x)-f(y)|< \epsilon$

Now by the Mean Value theorem on $\displaystyle [x,y]$

$\displaystyle f(x)-f(y)=f'(c)(x-y)$

$\displaystyle |f(x)-f(y)|=|f'(c)(x-y)|=|f'(c)||x-y|=f'(c)\cdot \delta =f'(c) \frac{\epsilon}{M}< \epsilon$ - May 1st 2009, 06:43 PMGaloisTheory1
the first one is not uniformly cont on $\displaystyle \mathbb{R}$ b/c the derivative is $\displaystyle xcos(x)+sin(x)(1)$. so $\displaystyle f' \rightarrow \infty$ as $\displaystyle x \rightarrow \infty$ contradicting the fact that the derivative must be bounded.