Question

Discuss the continuity/uniform continuity/Lipchitz continuity and
differentiability of the functions |𝑥|, |𝑥| 3 , √𝑥 3 on
[−1,1]

Accepted Answer

A continuous function is a function that does not have any abrupt changes in value, known as discontinuities. More precisely, a function is continuous if arbitrarily small changes in its output can be assured by restricting to sufficiently small changes in its input. So, the functions |𝑥|, |𝑥|3 are not continuous on [−1,1] , and the function 3 √𝑥 is continuous on [−1,1]. A function is is uniformly continuous if for every real number varepsilon>0 there exists real delta>0 such that (|x_1-x_2|< delta) implies (|f(x_1)-f(x_2)|< varepsilon) So, the function 3√𝑥 is uniform continuous on [−1,1], and the functions |𝑥|, |𝑥|3 are not uniform continuous on [−1,1]. A function is called Lipschitz continuous if there exists a positive real constant K such that, for all real _x_1 and _x_2 |f(x_1)-f(x_2)| le K|x_1-x_2| So, the functions |𝑥|, |𝑥|3 , 3 √𝑥 are Lipschitz continuous on [−1,1]. A differentiable function is a function whose derivative exists at each point in its domain. So, the function 3√𝑥 is differentiable on [−1,1], and the functions |𝑥|, |𝑥|3 are not differentiable on [−1,1] (derivative does not exist at x=0).

Question #241375

Expert's answer

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