Does f(x) being continuous affect |f(x)| being continuous and vice versa?

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Homework Statement



Prove or give a counterexample for each of the following statements:

a) If f(x) is continuous, then the function |f(x)| is continuous.

b) If |f(x)| is continuous, then the function f(x) is continuous.


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The Attempt at a Solution



I am sure I have read the solution somewhere, but cannot think of it for the life of me. I am pretty sure if f(x) is continuous so is |f(x)|, but not the other way around... I just don't know how to prove it!
 
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if f is continuous, and g is continuous, f o g is continuous. Use a delta-epsilon argument for this and set f's epsilon to g's delta.. et c

b is false. try f ( x ) = 1 when x >= 1 and -1 when x < 1
 
A couple of questions:

1) Does that mean for (a) I have to prove f(|f(x)|) is continuous?

2) For (b) I have no idea where to go from here...
 
It says in my textbook f o g is only continuous if g is continuous at c and f is continuous at g(c)...
 
LilTaru said:
2) For (b) I have no idea where to go from here...

Think of f(x)=signum(x)
 
LilTaru said:
1) Does that mean for (a) I have to prove f(|f(x)|) is continuous?

In f\circ g from the theorem set: g(x)=\mbox{your }f(x), and f(y)=|y|

If in the problem statement its says "continuous" it probably means "continuous everywhere".
 
LilTaru said:
A couple of questions:

1) Does that mean for (a) I have to prove f(|f(x)|) is continuous?
Of course not! This has nothing at all to do with f(|f(x)|)! You have to show that if f(x) is continuous then |f(x)| is continuous. That is what it says and that is what it means!

2) For (b) I have no idea where to go from here...
Consider f(x)= 1 if x\le 0, f(x)= -1 if x< 0.
(That's the same as arkajad's suggestion earlier.)
 
I solved part (b)... but part (a) is still giving me problems... I don't know how to prove if f(x) is continuous |f(x)| is continuous. I know that if f(x) is continuous everywhere and |f(x)| is the values of f(x) >= 0, then |f(x)| is continuous, but how do I prove this?
 
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