0
$\begingroup$

I'm confused about using extreme value theorem here

proof from

https://mathcenter.oxford.emory.edu/site/math111/proofs/rollesTheorem/

Consider the two cases that could occur:

Case 1: $f(x) = 0$ for all $x$ in $[a,b]$. In this case, any value between $a$ and $b$ can serve as the $c$ guaranteed by the theorem, as the function is constant on $[a,b]$ and the derivatives of constant functions are zero.

Case 2: $f(x) \ne 0$ for some $x$ in $(a,b)$. We know by the Extreme Value Theorem, that $f$ attains both its absolute maximum and absolute minimum values somewhere on $[a,b]$.

Recall by our hypothesis, $f(a)=f(b)=0$, and that in this case, $f(x)$ is not zero for some $x$ in $(a,b)$. Thus, $f$ will have either a positive absolute maximum value at some $c_{max}$ in $(a,b)$ or a negative absolute minimum value at some $c_{min}$ in $(a,b)$ or both. Take $c$ to be either $c_{min}$ or $c_{max}$, depending on which you have. Note then, the open interval $(a,b)$ contains $c$, and either:

$f(c) \ge f(x)$ for all $x$ in $(a,b)$; or $f(c) \le f(x)$ for all $x$ in $(a,b)$.

Either way, this means $f$ has a local extremum at $c$. As $f$ is also differentiable at $c$, Fermat's Theorem applies, and concludes that $f'(c)=0$.

Why does the proof of Rolle's Mean Value Theorem require the use of the Extreme Value Theorem or involve extreme values, and what is the starting point?

$\endgroup$
4
  • 1
    $\begingroup$ What is exactly your concern? Do you think that the result is "obvious", i.e. no proof needed? $\endgroup$ Commented Nov 19 at 13:05
  • $\begingroup$ @MauroALLEGRANZA Because I just can't understand why I need M and N(i.e.max and min) $\endgroup$ Commented Nov 19 at 13:11
  • 1
    $\begingroup$ What exactly do you mean by asking "why"? A sufficient enough reason is you cannot prove if yo do not use it. $\endgroup$ Commented Nov 19 at 13:47
  • $\begingroup$ See Tao's Analysis I page 259. $\endgroup$ Commented Nov 19 at 14:01

3 Answers 3

Reset to default
1
$\begingroup$

The simple reason the extreme value theorem is used is because it is necessary. Suppose you are in a situation where the extreme value theorem does not apply, say differentiable functions $\mathbb R\setminus \{0\}\to\mathbb R$. Then Rolle's theorem would also not hold; a counterexample is the function $f(x)=x-1/x$, which has zeroes at $x=\pm1$ but no turning point in between.

$\endgroup$
0
$\begingroup$

i would answer it as i understood.

you want to say something about existence of stationary points in in your domain [a,b]. hence extreme value theorem gives you the existence of atleast one of them in the interior (if both the extreme were at boundaries then you would have f = 0 on your domain, which is ruled out by our hypothesis). then we know that if you have an local extremum (not at boundary!) then its derivative at that point is zero

so to summarize we used the extreme value theorem to get the existence of extremums from which we then deduce the existence such c.

$\endgroup$
0
$\begingroup$

I understood now that by applying Extreme Value Theorem I can show that there is at least a maximum or minimum in the interval $(a,b)$.

And this means there is always such a c that is a maximum or minimum.

Since c is a maximum or a minimum,then by Extreme Value Theorem I can show that $f'(c) = 0$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.