24 сент. 2013 г. · It is continuous. However, it is not uniformly continuous. Suppose it were; then for every ϵ, there exists a δ for which |x−y|<δ⟹|x2−y2|<ϵ. Proof that $x^2$ is not uniformly continuous on $\mathbb R Show that $f(x) = x^2$ is not uniformly continuous on $[0,\infty) Proof that $x^2$ is not uniformly continuous on $\mathbb{R} Другие результаты с сайта math.stackexchange.com

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Опубликовано: 29 февр. 2024 г.

30 дек. 2023 г. · It is uniformly continuous on the closed interval [a,b] [ a , b ] where 0<a<b<1 0 < a < b < 1 because a continuous function on a closed interval ...

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Опубликовано: 6 мар. 2021 г.

25 июл. 2021 г. · We can do it two ways: EASY WAY: f(x)=x2 f ( x ) = x 2 is a continuous function, and [0,4] [ 0 , 4 ] is a compact.

If E is not bounded, then take for example E = [0,∞) and f(x) = x. Clearly f is uniformly continuous, but f(E) = E is not bounded.

3 дек. 2020 г. · The function f(x) = x2 is not uniformly continuous on R. Proof. We need to choose an ε > 0 and show that for every δ > 0, there are two real ...

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Example 10. The function f(x) = x2 is continuous but not uniformly con- tinuous on the interval S = (0,∞).

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Опубликовано: 11 февр. 2021 г.