Proving Continuity in Functions: A Comparison of Two Statements

  • Thread starter Thread starter stukbv
  • Start date Start date
  • Tags Tags
    Continuity
stukbv
Messages
112
Reaction score
0

Homework Statement



1. if f : [-1,1] --> Reals is such that sin(f(x) is continuous on the reals then f is continuous.

2. if f : [-1,1] --> Reals is such that f(sin(x)) is continuous on the reals then f is continuous.

Are these true or false how do i prove / give a counter example?
 
Physics news on Phys.org


Again "If you do not show at least some attempt to do this problem yourself, this thread will be deleted."
 


ok, its just i have no idea where to start really.
I know that the composition of 2 continuous functions is continuous too so I think that the second one is true, since sinx is continuous everywhere, the only way the composition could be discontinuous is is f was discontinuous??
 


Hi stukbv! :smile:

Take your favorite discontinuous function and try it out!
 


are you talking about number 1 or 2? I think i have disproved number 1 so that's okay.. now just 2 :(
 


stukbv said:
are you talking about number 1 or 2? I think i have disproved number 1 so that's okay.. now just 2 :(

Try to prove that one. What definition of continuity would you like to use?
 


e-d i think ,

so for |x-c| < d we have that |f(sin(x))-f(sin(c))| < e
 


So for our epsilon we need to find a delta such that

|x-c|&lt;\delta~\Rightarrow~|f(x)-f(c)|&lt;\varepsilon

Now, the trick is, can you write that x and c as sines of something close together?
 


i don't know what you mean
 
  • #10


Can you write x=sin(y) and c=sin(d)?
 
  • #11


oh ok so we get |sin(y)-sin(d)| < delta => |f(sin(y))-f(sin(d))| < epsilon?
 
  • #12


stukbv said:
oh ok so we get |sin(y)-sin(d)| < delta => |f(sin(y))-f(sin(d))| < epsilon?

Yes, and if you know that |y-d| is small, then |f(sin(y))-f(sin(d))| is also small, by hypothesis...
 
  • #13


so is that it then ?
 
  • #14


Once you've checked that it is indeed possible to take y and d close together, then that's that! :smile:
 
  • #15


arghh I am confused how would i check - sorry to be such a pain~!
 
  • #16


Use the continuity of the inverse sine function...
 
Back
Top