Let:
$f(x)=|x|^\alpha$, where $f:\mathbb{R} \to \mathbb{R}$ and $x\in[a,b]$, $(\alpha \ge 1)$
and
$g(x)=(1+\frac{1}{x})^x$, where $g:\mathbb{R} \to \mathbb{R}$ and $x\in[1,+\infty)$
I have to prove that $f$ and $g$ functions satisfy Lipschitz-Condition.
With the $f$, I went like: $$||x|^\alpha-|y|^\alpha|\le||a|^\alpha-|b|^\alpha|$$ And now I am a bit clueless, since if $a$ and $b$ are both positive I can go like: $$||a|^\alpha-|b|^\alpha|\le||a^\alpha|-|b^\alpha|\le|a^\alpha-b^\alpha|$$ And now since $b\ge a$, we have that $$|a^\alpha-b^\alpha|\le|b^\alpha|=b^\alpha$$ since $b$ positive number higher than $a$. What on the other hand means that for every $M=\lceil b^\alpha\rceil$ this function satisfies Lipschitz condition. But I do not know how to get to it, when one $a<0$ and $b>0$ or $a,b<0$.
With the $g$, I went like:
Assume that $x>y$
$$|(1+\frac{1}{x})^x-(1-\frac{1}{y})^y|\le|(\frac{x+1}{y})^x-(\frac{y+1}{y})^y|\le|\frac{(x+1)^x-(y+1)^y}{y^y})\le|(x+1)^x-(y+1)^y)|$$
And now I am stuck. Any help/clue would be appreciated.