# Math Help - function

1. ## function

Show from the definition that the function f(x) = x^3 is uniformly continuous on the interval [−1, 1].

2. Originally Posted by sgarg
Show from the definition that the function f(x) = x^3 is uniformly continuous on the interval [−1, 1].

let $x,y \in [-1,1]$

then note that

$f(x)-f(y)=x^3-y^3=(x-y)(x^2+2xy+y^2)$

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 $\delta = \frac{\epsilon}{4}$

So this delta will for for any $x \in [-1,1]$

I hope this clears it up good luck.