Exploring Local Minimum and Maximum Points in Continuous Functions

[sergey_le]

Homework Statement

Let function ƒ be Continuous in R, x1 local minimum point of f and x2 local maximum point of f.
Existing f(x1)>f(x2).
Prove that there is another local minimum point f (except x1)

Relevant Equations

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Here's what I tried to do:
f Continuous function at R, x1 local minimum point of f, x2 local maximum point of f.
Existing f(x1)>f(x2).
Let's look at the interval [x1,x2]⊆ℝ .
f is continuous in R and therefore continuous in its partial segment. Therefore f continuous in [x1,x2].
Therefore, there is an environment of δ>0 ,So that for all x that holds x1<x<x1+δ Happening : f(x1)≤f(fx)
Because given that x1 is a local minimum point of f.
And thus to all x that holds x2-δ<x<x2 Happening : f(x)≤f(x2).
Because given that x2 is a local maximum of f.
Don't know how to proceed

 

[scottdave]

Can you satisfy to yourself that there exists another local min.? Try sketching different scenarios. Think about the derivative of the function on either side of a minimum (and a maximum).

 

[sergey_le]

Can you satisfy to yourself that there exists another local min.? Try sketching different scenarios. Think about the derivative of the function on either side of a minimum (and a maximum).

I have no problem drawing different options of the given function.
But I can't put formal proof.

 

[Orodruin]

Try plotting $f(x) = x(x-1)(x+1)$.

 

[sergey_le]

Try plotting $f(x) = x(x-1)(x+1)$.

Your example doesn't help me unfortunately.
This is a general case.
I know what it looks like and understand why it's true but don't know how to formalize the proof in words.
If it was a specific function it would have been a simpler case.

 

[scottdave]

On either side of a maximum, f(x) decreases. But we are told that this value of f is less than the local minimum, so what will f have to do, to get there? Hopefully maybe you can work this idea into your formalisation.

 

[scottdave]

Maybe you could go the way of contradiction. Say if there are no other minimums, then how would the derivative of f behave. Then see how it needs to behave to satisfy the conditions.

 

[HallsofIvy]

Orodruin's point is that you CAN'T prove this statement because it is not true! The graph of y= x(x- 1)(x+ 1) has a local minimum at $\left(-\frac{\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}\right)$ and a local minimum at $\left(\frac{\sqrt{3}}{3}, -\frac{-2\sqrt{3}}{3}\right)$. There is NO other minimum point.

I suspect that the problem actually said that there are two maxima which then must have a minimum between them, or two minima which then must have a maximum between them.

 

[Orodruin]

Orodruin's point is that you CAN'T prove this statement because it is not true! The graph of y= x(x- 1)(x+ 1) has a local minimum at $\left(-\frac{\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}\right)$ and a local minimum at $\left(\frac{\sqrt{3}}{3}, -\frac{-2\sqrt{3}}{3}\right)$. There is NO other minimum point.

I suspect that the problem actually said that there are two maxima which then must have a minimum between them, or two minima which then must have a maximum between them.

Upon rereading the OP, my example does not satisfy the given constraints. The important information is that the value of the function at the local minimum is larger than that at the local maximum.

 

[Orodruin]

Maybe you could go the way of contradiction. Say if there are no other minimums, then how would the derivative of f behave. Then see how it needs to behave to satisfy the conditions.

$f$ is assumed continuous, not differentiable.

 

[Mark44]

The graph of y= x(x- 1)(x+ 1) has a local minimum at $\left(-\frac{\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}\right)$

Typo -- there's a local maximum at that point.

 

[archaic]

I can notice that $\text{sgn}(f'(x))=-\text{sgn}\left(\frac{f(b)-f(a)}{b-a}\right)$ as $x\to a^+$ or $b^-$ where $a$ and $b$ are interchangeably $x_1$ and $x_2$. Maybe that'd help since $f(x_1)>f(x_2)$?
EDIT: Oh yeah, continuous $\Rightarrow$ differentiable is false.

 

[Orodruin]

I will give another hint: Extreme value theorem.

 

[PeroK]

Homework Statement:: Let function ƒ be Continuous in R, x1 local minimum point of f and x2 local maximum point of f.
Existing f(x1)>f(x2).
Prove that there is another local minimum point f (except x1)
Homework Equations:: -

Here's what I tried to do:
f Continuous function at R, x1 local minimum point of f, x2 local maximum point of f.
Existing f(x1)>f(x2).
Let's look at the interval [x1,x2]⊆ℝ .
f is continuous in R and therefore continuous in its partial segment. Therefore f continuous in [x1,x2].
Therefore, there is an environment of δ>0 ,So that for all x that holds x1<x<x1+δ Happening : f(x1)≤f(fx)
Because given that x1 is a local minimum point of f.
And thus to all x that holds x2-δ<x<x2 Happening : f(x)≤f(x2).
Because given that x2 is a local maximum of f.
Don't know how to proceed

Note that this is not true if you define a local minimum, $x_0$, as follows:

$\exists \ \delta > 0 \ s.t. 0 < |x - x_0| < \delta \ \Rightarrow \ f(x) > f(x_0)$

But, it is true if you define a local minimum as:

$\exists \ \delta > 0 \ s.t. 0 < |x - x_0| < \delta \ \Rightarrow \ f(x) \ge f(x_0)$

 

[sergey_le]

Yes, but I wrote f(x)≥f(x0)

Note that this is not true if you define a local minimum, $x_0$, as follows:

$\exists \ \delta > 0 \ s.t. 0 < |x - x_0| < \delta \ \Rightarrow \ f(x) > f(x_0)$

But, it is true if you define a local minimum as:

$\exists \ \delta > 0 \ s.t. 0 < |x - x_0| < \delta \ \Rightarrow \ f(x) \ge f(x_0)$

 

[sergey_le]

I will give another hint: Extreme value theorem.

I've tried it now, but I'm not sure about my formal wording.
I know that according to the extreme value theorem there is a minimum point, and it is clear to me that because there is an environment of δ>0 So for every x that is found x2-δ<x<x2 Happening f(x1)>f(x2)≥f(x)
Therefore there is at least one value of x so that it is another minimum point

 

[PeroK]

I've tried it now, but I'm not sure about my formal wording.
I know that according to the extreme value theorem there is a minimum point, and it is clear to me that because there is an environment of δ>0 So for every x that is found x2-δ<x<x2 Happening f(x1)>f(x2)≥f(x)
Therefore there is at least one value of x so that it is another minimum point

You need to be careful here. You need to find a point that is not $x_1$ but is a local minimum.

By the extreme value theorem, there is a point - let's call it $x_0$ - where $f$ takes the minimum value on the interval $[x_1, x_2]$ or $[x_2, x_1]$ (depending on whether $x_1 <x_2$).

These details are important.

Now you have to show that $x_0 \ne x_1$.

There is also a nasty technical point to make sure that $x_0 \ne x_2$.

 

[sergey_le]

You need to be careful here. You need to find a point that is not $x_1$ but is a local minimum.

By the extreme value theorem, there is a point - let's call it $x_0$ - where $f$ takes the minimum value on the interval $[x_1, x_2]$ or $[x_2, x_1]$ (depending on whether $x_1 <x_2$).

These details are important.

Now you have to show that $x_0 \ne x_1$.

There is also a nasty technical point to make sure that $x_0 \ne x_2$.

It should be shown that x0 is different from x1 and x2.
The problem I don't know how to do it

 

[PeroK]

It should be shown that x0 is different from x1 and x2.
The problem I don't know how to do it

Try showing that $x_0 \ne x_1$. You were very close in post #16.

Showing that you can choose $x_0 \ne x_2$ is slightly trickier. Do you see why you need to show this?

Hint: why did I write post #14? Do you understand the significance of the $\le$ in the definition?

 

[sergey_le]

Try showing that $x_0 \ne x_1$. You were very close in post #16.

Showing that you can choose $x_0 \ne x_2$ is slightly trickier. Do you see why you need to show this?

Hint: why did I write post #14? Do you understand the significance of the $\le$ in the definition?

Yes I know what the definition of ≤ .
I think I understand what you want to say.
If f(x1)≠f(x0) then f(x2)≠f(x0) Right?

 

[PeroK]

If f(x1)≠f(x0) then f(x2)≠f(x0) Right?

No. $f(x_0) \ne f(x_1)$. From that you know that $x_0 \ne x_1$.

But, $f(x_2)$ might be the minimum for $f$ on $[x_1, x_2]$. As well as a local maximum for $f$. That's the nasty technical point.

 

[sergey_le]

No. $f(x_0) \ne f(x_1)$. From that you know that $x_0 \ne x_1$.

But, $f(x_2)$ might be the minimum for $f$ on $[x_1, x_2]$. As well as a local maximum for $f$. That's the nasty technical point.

Ok that's what I thought from the beginning I just try to understand your clues.
If I understand correctly I should use the Extreme value theorem and show that x0 exists so that
f(x2)>f(x0).
Otherwise x2 is the minimum point of f in [x1,x2] .
But I don't know why this is true.
(If that's true)

 

[Orodruin]

What is the definition of a local maximum such as $x_2$?

 

[sergey_le]

What is the definition of a local maximum such as $x_2$?

There is an environment of δ>0 So that every x that is found comes f(x)≤f(x2)
I don't see how it helps me because that can be the option of f(x)=f(x2)

 

[Orodruin]

There is an environment of δ>0 So that every x that is found comes f(x)≤f(x2)

What do you mean by "an environment of $\delta > 0$"?

 

[sergey_le]

What do you mean by "an environment of $\delta > 0$"?

(x2-δ,x2+δ)
Or in our case( x2-δ,x2]

 

[Orodruin]

So take a point $x_* < x_2$ in that neighbourhood. What can you say about $f(x_*)$?

 

[sergey_le]

So take a point $x_* < x_2$ in that neighbourhood. What can you say about $f(x_*)$?

I think I understood.
Any x found in that neighbourhood f(x)≤f(x2) So there is another local minimum point Whether f(x)<f(x2) or f(x)=f(x2) Right?

 

[PeroK]

I think I understood.
Any x found in that neighbourhood f(x)≤f(x2) So there is another local minimum point Whether f(x)<f(x2) or f(x)=f(x2) Right?

You need to make this argument formal and rigorous. You must work on this. For example, note that it does not say in the question that $x_1 < x_2$. For the final proof you need to think about how you manage that technical point.

You are also mixing up the variable $x$ with specifc values $x_0, x_1, x_2$ etc. You should get into the habit of being clear when you are talking about a specific $x$ and when you are talking about any $x$ in an interval.

Let me recap what we have so far:

$f$ is continuous, has a local minimum at $x_1$ and a local maximum at $x_2$ and $f(x_1) > f(x_2)$.

First assume that $x_1 < x_2$, and consider the interval $[x_1, x_2]$. By the extreme value theorem, $f$ attains its minimum value on this interval. I.e. $\exists \ x_0 \in [x_1, x_2] \$ such that $\forall x \in [x_1, x_2], f(x) \ge f(x_0)$.

Note that $x_0 \ne x_1$ as $f(x_1) > f(x_2)$. I.e. $x_1$ cannot be a minimum on this interval.

If $x_0 \ne x_2$ then $x_0$ is not only a minimum on the interval $[x_1, x_2]$ but also a local minimum for $f$.

**you might want to add a line here to justify this. **

If $x_0 = x_2$, then we have a minimum at the end point of the interval. Hence, this $x_0$ may not be a local minimum for $f$.

However, $x_2$ is a local maximum for $f$. Therefore:

** Now you need a rigorous argument to show that you can find another $x'_0 \ne x_2$ which is a local minimum for $f$. Hint: use the $\delta$ definition for maximum and minimum. **

 

[sergey_le]

You need to make this argument formal and rigorous. You must work on this. For example, note that it does not say in the question that $x_1 < x_2$. For the final proof you need to think about how you manage that technical point.

You are also mixing up the variable $x$ with specifc values $x_0, x_1, x_2$ etc. You should get into the habit of being clear when you are talking about a specific $x$ and when you are talking about any $x$ in an interval.

Let me recap what we have so far:

$f$ is continuous, has a local minimum at $x_1$ and a local maximum at $x_2$ and $f(x_1) > f(x_2)$.

First assume that $x_1 < x_2$, and consider the interval $[x_1, x_2]$. By the extreme value theorem, $f$ attains its minimum value on this interval. I.e. $\exists \ x_0 \in [x_1, x_2] \$ such that $\forall x \in [x_1, x_2], f(x) \ge f(x_0)$.

Note that $x_0 \ne x_1$ as $f(x_1) > f(x_2)$. I.e. $x_1$ cannot be a minimum on this interval.

If $x_0 \ne x_2$ then $x_0$ is not only a minimum on the interval $[x_1, x_2]$ but also a local minimum for $f$.

**you might want to add a line here to justify this. **

If $x_0 = x_2$, then we have a minimum at the end point of the interval. Hence, this $x_0$ may not be a local minimum for $f$.

However, $x_2$ is a local maximum for $f$. Therefore:

** Now you need a rigorous argument to show that you can find another $x'_0 \ne x_2$ which is a local minimum for $f$. Hint: use the $\delta$ definition for maximum and minimum. **

Thank you so much for your understanding.

 

[sergey_le]

Thank you all so much you wonderful and patient people, I am glad I found this forum you are really helping me. Blessed