Question

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

Answer 1

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 x₁ and x₂

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

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).