Intermediate Value Theorem ....Silva, Theorem 4.2.1 .... ....

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SUMMARY

This discussion focuses on the clarification of Theorem 4.2.1, the Intermediate Value Theorem, as presented in Cesar E. Silva's "Invitation to Real Analysis." Participants emphasize the importance of continuity in proving that for values of \( x \) within the interval \( b > x > \beta \), the function \( f(x) < 0 \). The continuity argument is critical, as it ensures that \( |f(x) - f(\beta)| < \epsilon \) leads to \( f(x) < \epsilon + f(\beta) < 0 \), thereby confirming the theorem's assertion.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem
  • Familiarity with continuity in real analysis
  • Knowledge of epsilon-delta definitions in calculus
  • Basic concepts of function behavior in intervals
NEXT STEPS
  • Study the proof of the Intermediate Value Theorem in detail
  • Explore the sign-preserving property of continuous functions
  • Review epsilon-delta definitions in calculus
  • Examine related theorems in real analysis, such as Apostol's Theorem 3.7
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Students of real analysis, educators teaching calculus, and anyone seeking a deeper understanding of the Intermediate Value Theorem and its implications in continuous functions.

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I am reading Cesar E. Silva's book entitled "Invitation to Real Analysis" ... and am focused on Chapter 4: Continuous Functions ...

I need help to clarify an aspect of the proof of Theorem 4.2.1, the Intermediate Value Theorem ... ...

Theorem 4.2.1 and its related Corollary read as follows:
View attachment 9562
View attachment 9563
In the above proof by Silva, we read the following:

" ... ... So there exists $$x$$ with $$b \gt x \gt \beta$$ and such that $$f(x) \lt 0$$ ... ... "My question is as follows:

How can we be sure that $$f(x) \lt 0$$ given $$x$$ with $$b \gt x \gt \beta$$ ... indeed how do we show rigorously that for $$x$$ such that $$b \gt x \gt \beta$$ we have $$f(x) \lt 0$$ ...Help will be much appreciated ...

Peter
 

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  • Silva - 1 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 1.png
    Silva - 1 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 1.png
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  • Silva - 2 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 2 .png
    Silva - 2 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 2 .png
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Hi Peter,

Since $\beta<b$ and $\delta >0$, there are values of $x$ such that $|x-\beta|<\delta$ and $\beta < x< b$. The key here is to remember that $|x-\beta| <\delta$. By the continuity argument, $f(x)<0$ for all such $x$.
 
GJA said:
Hi Peter,

Since $\beta<b$ and $\delta >0$, there are values of $x$ such that $|x-\beta|<\delta$ and $\beta < x< b$. The key here is to remember that $|x-\beta| <\delta$. By the continuity argument, $f(x)<0$ for all such $x$.
Thanks for the help GJA!

At first I struggled with what you meant by ... " By the continuity argument, $f(x)<0$ for all such $x$ ... "

But then I found Apostol Theorem 3.7 (Calculus Vol. 1, page 143) which reads as follows:View attachment 9565Were you indeed invoking something like what Apostol calls the sign-preserving property of continuous functions?Thanks again for your help ...

Peter
 

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  • Ap[ostol - Calculus - Theorem 3.7 .png
    Ap[ostol - Calculus - Theorem 3.7 .png
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Hi Peter,

Happy to help!

I wasn't quoting that purposely, though it is true. In fact, it's essentially what the author is proving by their choice of epsilon.

What I meant was: $|f(x)-f(\beta)|<\epsilon\,\Longrightarrow\, f(x)<\epsilon + f(\beta)<0.$

Hope this helps clear up the confusion on my earlier post.
 

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