Show from the definition that the function f(x) = x^3 is uniformly continuous on the interval [−1, 1].
let $\displaystyle x,y \in [-1,1]$
then note that
$\displaystyle f(x)-f(y)=x^3-y^3=(x-y)(x^2+2xy+y^2)$
So we need to bound the 2nd factor since we are on the closed interval [-1,1] this biggest it can be is 4 if x=y=1.
so set $\displaystyle \delta = \frac{\epsilon}{4}$
So this delta will for for any $\displaystyle x \in [-1,1]$
I hope this clears it up good luck.