Extreme value nonexistence proof

[mcastillo356]

TL;DR Summary

Don't know if it is meaningless something said about a proof on extreme value existence

Hi, PF

"It is more difficult to draw the graph of a function whose domain has an endpoint at which the function fails to have an extreme value", states my textbook, "Calculus: A Complete Course"

A function with no max or min at an endpoint Let

$f(x)=\begin{cases}{x\sin{\left(\dfrac{1}{x}\right)}}&\text{si}& x>0\\0 & \text{si}& x=0\end{cases}$

Show that $f$ is continuous on $[0,\infty)$ and differentiable on $(0, \infty)$ but it has neither a local maximum nor a local minimun at the endpoint $x=0$
I think the continuity and the differentiability are meaningless for the proof

 

[FactChecker]

A function with no max or min at an endpoint Let

$f(x)=\begin{cases}{x\sin{\left(\dfrac{1}{x}\right)}}&\text{si}& x>0\\0 & \text{si}& x=0\end{cases}$

Show that $f$ is continuous on $[0,\infty)$ and differentiable on $(0, \infty)$ but it has neither a local maximum nor a local minimun at the endpoint $x=0$
I think the continuity and the differentiability are meaningless for the proof

Continuity and differentiability are not meaningless as a counterexample. It shows that there is even a fairly well-behaved function which is a counterexample.

 

[mcastillo356]

Such as $f(x)=x$, $x\geq{0}$?
What does it mean well-behaved?

 

[Mark44]

Such as $f(x)=x$, $x\geq{0}$?

What is your question about this function? It is continuous on $[0, \infty)$ and differentiable on the open interval $(0, \infty)$. It has an absolute min. point at x = 0.

What does it mean well-behaved?

The example you posted is continuous on $[0, \infty)$ and differentiable on the open interval $(0, \infty)$.

 

[Mark44]

Summary:: Don't know if it is meaningless something said about a proof on extreme value existence

"It is more difficult to draw the graph of a function whose domain has an endpoint at which the function fails to have an extreme value"

The point is that it's very difficult to draw the graph of this function due to its wild oscillations near x = 0. So even though it is continuous for $x \ge 0$ and differentiable for $x > 0$, it fails to have an extreme value; i.e., a maximum or minimum.

BTW, along the lines of what @FactChecker said, what you posted isn't a proof -- it is a counterexample.

 

[mcastillo356]

Continuity and differentiability are not meaningless as a counterexample. It shows that there is even a fairly well-behaved function which is a counterexample.

A counterexample is an example that disproves a statement. What statement does disprove the function I've mentioned, i.e., $f(x)=x$ for $x\geq{0}$.

 

[FactChecker]

What statement does disprove the function I've mentioned, i.e., $f(x)=x$ for $x\geq{0}$.

What do you mean by "disprove the function"?
Perhaps I should not have called the $x\sin(1/x)$ function a "counterexample". It is an example of a well-behaved function (continuous on $[0,\infty)$ and with a derivative in $(0,\infty)$) that does not have a local extreme at the endpoint $x=0$.

 

[mcastillo356]

I should have said "What function does disprove the statement at #1 post?":

There are functions like
$f(x)=\begin{cases}{x\sin{\left(\dfrac{1}{x}\right)}}&\text{if}&x>0\\0&\text{if}&x=0\end{cases}$
that are continuous on $[0,\infty)$, differentiable on $(0,\infty)$, but with neither a local max or local min at a endpoint

I'm not native
Thanks, understood!

 

[Mark44]

I should have said "What function does disprove the statement at #1 post?":

There are functions like
$f(x)=\begin{cases}{x\sin{\left(\dfrac{1}{x}\right)}}&\text{if}&x>0\\0&\text{if}&x=0\end{cases}$
that are continuous on $[0,\infty)$, differentiable on $(0,\infty)$, but with neither a local max or local min at a endpoint

I think that you are missing the point here. The statement in your book is

It is more difficult to draw the graph of a function whose domain has an endpoint at which the function fails to have an extreme value", states my textbook, "Calculus: A Complete Course"

The function above is nothing more than an example that confirms the statement. That is, the author has shown that such a function exists by giving its formula.

You are overthinking this, IMO.

 

[mcastillo356]

The question is: what means "that are continuous on $[0,\infty)$, differentiable on $(0,\infty)$, but ..."?

 

[Mark44]

The question is: what means "that are continuous on $[0,\infty)$, differentiable on $(0,\infty)$, but ..."?

Do you understand what the terms continuous and differentiable mean? These terms have very specific and rigorous mathematical definitions.

The point of the example is that we would (naively) expect that a function that is both continuous on some interval and differentiable at all but the endpoint of that interval would have an extreme value there. The example provides evidence that our naive expectation isn't always the case.

 

[mcastillo356]

Fine, @Mark44 . Perfect now