Continuity of Function - f(x)=|cos(x)|

[EEristavi]

Homework Statement

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We have a function f(x) = |cos(x)|.

It's written that it is piecewise continuous in its domain.
I see that it's not "smooth" function, but why it is not continuous function - from the definition is should be..

Homework Equations

[/B]
We say that a function f is continuous at an interior point c of its domain if
lim(x->c) f(x) = f(c).

The Attempt at a Solution

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I solved the problem but this piecewise continuity is bothering me

 

[fresh_42]

$f(x) =|\cos x|$ is uniformly continuous, continuous and piecewise continuous. Why the piecewise continuity is emphasized in your context doesn't become clear from your post. Maybe it's needed as a condition in another theorem, where piecewise continuity plays a role.

 

[Ray Vickson]

Homework Statement

[/B]
We have a function f(x) = |cos(x)|.

It's written that it is piecewise continuous in its domain.
I see that it's not "smooth" function, but why it is not continuous function - from the definition is should be..

Homework Equations

[/B]
We say that a function f is continuous at an interior point c of its domain if
lim(x->c) f(x) = f(c).

The Attempt at a Solution

[/B]
I solved the problem but this piecewise continuity is bothering me

You are right: there is a little, easily-proved theorem that if $f$ is continuous, then so is $|f|$. (However, as you point out, it is piecewise smooth, but that is a different issue.)

Unless the book you are using has some unusual definitions, I don't think describing the function as "piecewise continuous" is helpful; it is not really wrong, just useless. Basically, by the same token, any continuous function is also piecewise continuous!

The way I have seen the terms "piecewise continuity" employed is for functions that are continuous on intervals $(a_1, a_2), (a_2, a_3), \ldots$ but with $f(a_2 + 0) \neq f(a_2 - 0),$ etc---that is, has discontinuities of the first kind.

 

[member 587159]

Note that this function is continuous, as composition of 2 continuous functions, at least if the functions don't have some kind of strange domain.

I.e.: for $x \in \mathbb{R}:|\cos(x)| = (|\cdot | \circ \cos) (x)$

For me, it is not entirely clear what the relevance of this question is. Can you specify what the domain of the functions is?

 

[EEristavi]

Thank you all for the quick replies.
I will attach the screenshot from book, just in case.

However, I think the case is closed (Thanks again)

 

[member 587159]

Thank you all for the quick replies.
I will attach the screenshot from book, just in case.

However, I think the case is closed (Thanks again)

Since continuity => piecewise continuity, indeed the case is closed.

Maybe they mean with piecewise here that you have to split up the integral with linearity to get rid off the absolute value signs, but this is just a guess in the wild.

 

[fresh_42]

Thank you all for the quick replies.
I will attach the screenshot from book, just in case.

However, I think the case is closed (Thanks again)

I think in this specific case, it may be meant as a hint where to cut the integral to get the values correct - I'd take it as error avoidance.

 

[EEristavi]

I think in this specific case, it is meant as a hint where to cut the integral to get the values correct - I'd take it as error avoidance.

I also though of this. I just asked here also, just to be sure and don't learn something wrong :)

Thanks for help )