Proof that Arcsin x is continuous ....

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SUMMARY

The function f(x) = Arcsin x is continuous on the interval [-1, 1]. According to Stephen Abbott's "Understanding Analysis," a function is continuous at a point if for every ε > 0, there exists a δ > 0 such that if |x - c| < δ, then |Arcsin x - Arcsin c| < ε. This definition establishes the framework for proving the continuity of the Arcsin function within the specified interval. The discussion emphasizes the importance of understanding the definitions and properties of continuous functions.

PREREQUISITES
  • Understanding of the definition of continuous functions from real analysis.
  • Familiarity with the properties of inverse functions.
  • Knowledge of the function Arcsin and its domain.
  • Basic proficiency in mathematical proofs and ε-δ definitions.
NEXT STEPS
  • Study the ε-δ definition of continuity in more depth.
  • Explore the properties of inverse functions and their continuity.
  • Review Theorem 4.3.2 from Stephen Abbott's "Understanding Analysis."
  • Practice proving the continuity of other functions using similar methods.
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Students of real analysis, mathematicians, and educators looking to deepen their understanding of continuity and inverse functions in calculus.

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Can someone please help me to prove that the function f(x) = Arcsin x is continuous on the interval [-1, 1] ...

Peter
 
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What is the definition of continuous function in general?
 
An invertible function, y= f(x), is continuous at x= x_0 if and only if y= f^{-1}(x) is continuous at x= f(x_0).
 
Cbarker1 said:
What is the definition of continuous function in general?
The definition of continuity in $$\mathbb{R}$$ is given in Stephen Abbott's book: Understanding Analysis, as follows:
View attachment 9348
Alternative characterizations of continuity are given by Abbott in Theorem 4.3.2 as follows:
View attachment 9349So to show (from first principles) that $$\text{Arcsin } x$$ is continuous on $$[-1, 1]$$ we would have to show that given an arbitrary point $$c \in [-1, 1]$$ that for every $$\epsilon \gt 0$$ we can find $$\delta \gt 0$$ such that

$$\mid x - c \mid \lt \delta \ \Longrightarrow \ \mid \text{Arcsin x } - \text{Arcsin } c \mid \lt \epsilon$$ ...But how do we proceed ... ?

Peter

- - - Updated - - -

HallsofIvy said:
An invertible function, y= f(x), is continuous at x= x_0 if and only if y= f^{-1}(x) is continuous at x= f(x_0).
Thanks for the help, HallsofIvy ...

Peter
 

Attachments

  • Abbott - Defintion 4.3.1 ... Continuity ... .png
    Abbott - Defintion 4.3.1 ... Continuity ... .png
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  • Abbott - Theorem 4.3.2 ... Characterizations of Continuity ... .png
    Abbott - Theorem 4.3.2 ... Characterizations of Continuity ... .png
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