Proving functions are uniformly continuous help! a) Prove that $f(x)=x^{1/4}$ is uniformly continuous on $[0,\infty)$. Show that this method can be extended inductively to any $f(x)=x^{1/p}$ for any $p=2^n$
b) Prove directly from the definition that $f(x)=(1+x^2)^{1/2}$ is uniformly continuous on
$\mathbb{R}$.
For part a I know how to prove it if $x^{1/2}$ but I'm unsure how to apply it for $x^{1/4}$ or any other p. 
I'm not sure how to start b. any help would be very helpful!

For part a) , show that for $f(x)=x^{1/4}$, $f'(x)$ is bounded in $(0, \infty)$, e.g., by finding $f'(x)$ at $x=1$, say, and then generalize to $f(x)=x^{1/2^n}$ . The mean value theorem should come in handy too.
EDIT: As Jochen pointed out, f'(x) is not bounded in $(0, \infty)$, tho it is bounded in $[c, \infty) ; c>0$. Then, as pointed out, we can use the fact that f is continuous in $[0,c]$ and then, by compactness of the interval, it is uniformly-continuous there, and by the rest of the argument, $f(x)$ is uniformly-continuous in $[0, \infty)$
Start by setting up an arbitrary $\epsilon$. Now you need to show that there is a single $\delta$ so that , for all $x,y$ , you have $$|f(x)-f(y)| < \epsilon $$ when $$|x-y|< \delta$$ . Can you see how to use the mean-value theorem to connect these two?
