Proving Continuity in Functions: A Comparison of Two Statements

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Homework Help Overview

The discussion revolves around proving the continuity of functions based on two statements involving the sine function. The original poster presents two claims regarding the continuity of a function \( f \) defined on the interval \([-1, 1]\) and its relationship with the continuity of compositions involving the sine function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the continuity of compositions of functions, particularly focusing on the second statement. There are attempts to disprove the first statement and to understand the definitions of continuity in relation to the problem.

Discussion Status

Some participants have made attempts to analyze the statements, with one indicating a potential disproof of the first statement. Others are engaged in exploring the second statement, discussing the continuity of the sine function and how to apply definitions of continuity to prove or disprove the claims.

Contextual Notes

There is a reminder for participants to show their attempts at solving the problem to avoid deletion of the thread. The discussion includes references to the continuity of functions and the need for specific definitions to approach the problem effectively.

stukbv
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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?
 
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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...
 

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