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    function

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