discuss the continuity/uniform continuity/Lipchitz continuity and differentiability of the functions |x| on [−1,1].
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Expert's answer
2021-09-27T14:31:49-0400
ANSWER.Function f(x)=∣x∣ is Lipschitz continuous, therefore, uniformly continuous, hence continuous on [-1,1]. The function is not differentiable at all points of the [-1,1] (At the point x=0 the function is not differentiable).
EXPLANATION.
Function f(x)=∣x∣ is Lipschitz continuous, because for all x1,x2∈[−1,1]:∣f(x1)−f(x2)∣=∣∣x1∣−∣x2∣∣≤∣x1−x2∣ (1).
Since for any ε>0 and δ=ε for any x1,x2∈[−1,1] such that ∣x1−x2∣<δfrom (1) it follows ∣f(x1)−f(x2)∣=∣∣x1∣−∣x2∣∣<ε , then f is uniformly continuous.
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