Answer to Question #240499 in Real Analysis for saduni

Question #240499

Discuss the continuity/uniform continuity/Lipchitz continuity and differentiability of the functions |đ‘„|, |đ‘„| 3 ,3 âˆšđ‘„ on [−1,1].


1
Expert's answer
2021-09-27T15:01:31-0400

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 |đ‘„|, |đ‘„|3 are not continuous on [−1,1] , and the function 3Â âˆšđ‘„ 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 3âˆšđ‘„ is uniform continuous on [−1,1], and the functions |đ‘„|, |đ‘„|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 x1 and x2

"|f(x_1)-f(x_2)|\\le K|x_1-x_2|"

So, the functions |đ‘„|, |đ‘„|3 , 3Â âˆšđ‘„ are Lipschitz continuous on [−1,1].


A differentiable function is a function whose derivative exists at each point in its domain.

So, the function 3âˆšđ‘„ is differentiable on [−1,1], and the functions |đ‘„|, |đ‘„|3 are not differentiable on [−1,1] (derivative does not exist at x=0).


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