Show Existence of $a,b$: Proof for $f'(\xi) \neq 0$

[evinda]

Hello! (Wave)

Let $I$ an interval and $f: I \to \mathbb{R}$ a differentiable (as many times as we want) function.
If $\xi \in I$ with $f''(\xi) \neq 0$, then there are $a,b \in I: a< \xi <b$ and $\frac{f(b)-f(a)}{b-a}=f'(\xi)$.

Hint: First suppose that $f'(\xi)=0$ and show that at $\xi$ we have a local extremum. Then find (for example with the intermediate value theorem) $a,b$ with $a< \xi<b$ and $f(a)=f(b)$, therefore... etc.
For $f'(\xi) \neq 0$, consider the function $g(x)=f(x)-f'(\xi)x$ to get reduced to the previous case.So we suppose that $f'(\xi)=0$.

How do we show that $f$ has a local extremum at $\xi$ ?

I have thought to calculate the Taylor series of second order.

Then, $f(x)=f(\xi)+f'(\xi) (x-\xi)+\frac{f''(\xi)}{2!}(x-\xi)^2=f(\xi)+\frac{f''(\xi)}{2}(x-\xi)^2$.

From this we get that $f(x)-f(\xi) \neq 0$.

But this does not help somehow, does it?

How else can we start the proof? (Thinking)

 

[I like Serena]

Hello!

Let $I$ an interval and $f: I \to \mathbb{R}$ a differentiable (as many times as we want) function.
If $\xi \in I$ with $f''(\xi) \neq 0$, then there are $a,b \in I: a< \xi <b$ and $\frac{f(b)-f(a)}{b-a}=f'(\xi)$.

Hint: First suppose that $f'(\xi)=0$ and show that at $\xi$ we have a local extremum. Then find (for example with the intermediate value theorem) $a,b$ with $a< \xi<b$ and $f(a)=f(b)$, therefore... etc.
For $f'(\xi) \neq 0$, consider the function $g(x)=f(x)-f'(\xi)x$ to get reduced to the previous case.So we suppose that $f'(\xi)=0$.

How do we show that $f$ has a local extremum at $\xi$ ?

Hey evinda!

If $f'(\xi)=0$ then we either have a local extremum or a saddle point don't we?
Can it be a saddle point? (Wondering)

I have thought to calculate the Taylor series of second order.

Then, $f(x)=f(\xi)+f'(\xi) (x-\xi)+\frac{f''(\xi)}{2!}(x-\xi)^2=f(\xi)+\frac{f''(\xi)}{2}(x-\xi)^2$.

From this we get that $f(x)-f(\xi) \neq 0$.

Couldn't we still have that $f(x)-f(\xi) = 0$?

Suppose we pick $I=[-2,2]$, $f(x)=x^2-1$, $x=-1$, and $\xi=+1$.
Don't we have $f(x)-f(\xi)=0$ then? (Worried)

 

[evinda]

Hey evinda!

If $f'(\xi)=0$ then we either have a local extremum or a saddle point don't we?
Can it be a saddle point? (Wondering)

When do we have a saddle point given a single-valued function? (Worried) (Thinking)

Couldn't we still have that $f(x)-f(\xi) = 0$?

Suppose we pick $I=[-2,2]$, $f(x)=x^2-1$, $x=-1$, and $\xi=+1$.
Don't we have $f(x)-f(\xi)=0$ then? (Worried)

Oh yes, right! (Tmi) (Thinking)

 

[I like Serena]

When do we have a saddle point given a single-valued function?

Ah, sorry, it's not called a saddle point for a function of one variable. (Blush)
Instead it's called an inflection point.
Such as is the case for $f(x)=x^3$ at $x=0$.
It has $f'(0)=0$ doesn't it? But it's not a local extremum is it? (Thinking)

 

[evinda]

Ah, sorry, it's not called a saddle point for a function of one variable. (Blush)
Instead it's called an inflection point.
Such as is the case for $f(x)=x^3$ at $x=0$.
It has $f'(0)=0$ doesn't it? But it's not a local extremum is it? (Thinking)

Ah, I see... A necessary condition for $x$ to be an inflection point is that $f''(x)=0$.

So in our case since $f'(\xi)=0$ and $f''(\xi) \neq 0$, we have that $\xi$ is a local extremum.

Then let $a, b \in \mathbb{R}$ with $a<b$ such that $a< \xi<b$ and $[a,b] \subseteq I$. Then $f$ is continuous on $[a,b]$.
From the Intermediate Value Theorem we have that for every $y_0$ between $f(a)$ and $f(b)$, there exists a number $x_0 \in [a,b]$ such that $f(x_0)=y_0$.

First of all can we pick $a,b$ such that $a< \xi<b$ ? Because here $\xi$ is specific.

Secondly, the condition $f(x_0)=y_0$ does not help to conclude that $f(a)=f(b)$, does it? (Thinking)

 

[I like Serena]

Ah, I see... A necessary condition for $x$ to be an inflection point is that $f''(x)=0$.

So in our case since $f'(\xi)=0$ and $f''(\xi) \neq 0$, we have that $\xi$ is a local extremum.

Yep. (Nod)

Then let $a, b \in \mathbb{R}$ with $a<b$ such that $a< \xi<b$ and $[a,b] \subseteq I$. Then $f$ is continuous on $[a,b]$.
From the Intermediate Value Theorem we have that for every $y_0$ between $f(a)$ and $f(b)$, there exists a number $x_0 \in [a,b]$ such that $f(x_0)=y_0$.

First of all can we pick $a,b$ such that $a< \xi<b$ ? Because here $\xi$ is specific.

Secondly, the condition $f(x_0)=y_0$ does not help to conclude that $f(a)=f(b)$, does it?

Sure.
Assuming that $\xi$ is not on the boundary of $I$ (which does not seem to be given, but we do need that), we can always find $a,b\in I$ such that $a<\xi<b$. (Thinking)

We can't be sure that for such $a,b$ we will have $f(a)=f(b)$ though. (Worried)

So let's pick $a_1,b_1\in I$ instead such that $a_1<\xi<b_1$.
Let's assume that $f$ has a local maximum at $\xi$ and $f(a_1)<f(b_1)$ for now.
Then we must be able to reach any intermediate value between $f(a_1)$ and $f(\xi)$ mustn't we? (Thinking)

 

[evinda]

Sure.
Assuming that $\xi$ is not on the boundary of $I$ (which does not seem to be given, but we do need that), we can always find $a,b\in I$ such that $a<\xi<b$. (Thinking)

We can't be sure that for such $a,b$ we will have $f(a)=f(b)$ though. (Worried)

So let's pick $a_1,b_1\in I$ instead such that $a_1<\xi<b_1$.
Let's assume that $f$ has a local maximum at $\xi$ and $f(a_1)<f(b_1)$ for now.
Then we must be able to reach any intermediate value between $f(a_1)$ and $f(\xi)$ mustn't we? (Thinking)

Yes, since $f$ is continuous on $[a_1, \xi]$, we have from the intermediate value theorem that for evey $y_0$ between $f(a_1)$ and $f(\xi)$ there exists a number $x_0 \in [a_1, \xi]$ such that $f(x_0)=y_0$.

But how does this help? (Thinking)

 

[I like Serena]

Yes, since $f$ is continuous on $[a_1, \xi]$, we have from the intermediate value theorem that for evey $y_0$ between $f(a_1)$ and $f(\xi)$ there exists a number $x_0 \in [a_1, \xi]$ such that $f(x_0)=y_0$.

But how does this help?

Doesn't that mean that there is an $a_2\in[a_1, \xi]$ such that $f(a_2)=f(b_1)$? (Wondering)

 

[evinda]

Doesn't that mean that there is an $a_2\in[a_1, \xi]$ such that $f(a_2)=f(b_1)$? (Wondering)

Ah yes, right! Thus we have that $\frac{f(b_1)-f(a_2)}{b_1-a_2}=0=f'(\xi)$, i.e. we have found the desired $a,b$, right? (Thinking)

Then, we suppose that $f'(\xi) \neq 0$. We consider the function $g(x)=f(x)-f'(\xi)x$.

We have that $g'(x)=f'(x)-f'(\xi)$.

So, $g'(\xi)=0$.

Furthermore, $g''(x)=f''(x)$ and so $g''(\xi) \neq 0$.

So from the previous case we have that there are $a, b \in I$ such that $\frac{g(b)-g(a)}{b-a}=g'(\xi)$.

From this we get that $\frac{f(b)-f'(\xi)b-f(a)+f'(\xi)b}{b-a}=g'(\xi)=0$.

But from this we get that $\frac{f(b)-f(a)}{b-a}=0$.

Have I done something wrong? (Thinking)

 

[I like Serena]

Ah yes, right! Thus we have that $\frac{f(b_1)-f(a_2)}{b_1-a_2}=0=f'(\xi)$, i.e. we have found the desired $a,b$, right?

Yep. (Nod)

Then, we suppose that $f'(\xi) \neq 0$. We consider the function $g(x)=f(x)-f'(\xi)x$.

We have that $g'(x)=f'(x)-f'(\xi)$.

So, $g'(\xi)=0$.

Furthermore, $g''(x)=f''(x)$ and so $g''(\xi) \neq 0$.

So from the previous case we have that there are $a, b \in I$ such that $\frac{g(b)-g(a)}{b-a}=g'(\xi)$.

From this we get that $\frac{f(b)-f'(\xi)b-f(a)+f'(\xi)b}{b-a}=g'(\xi)=0$.

But from this we get that $\frac{f(b)-f(a)}{b-a}=0$.

Have I done something wrong?

Shouldn't it be $\frac{f(b)-f'(\xi)b-f(a)+f'(\xi)\,{\color{red}a}}{b-a}=g'(\xi)=0$? (Thinking)

 

[evinda]

Yep. (Nod)
Shouldn't it be $\frac{f(b)-f'(\xi)b-f(a)+f'(\xi)\,{\color{red}a}}{b-a}=g'(\xi)=0$? (Thinking)

Ah I see, then we get the desired result! (Cool)

Do we have to look also at the case when $f$ has a local minimum at $\xi$ ? (Thinking)

 

[I like Serena]

Ah I see, then we get the desired result!

Do we have to look also at the case when $f$ has a local minimum at $\xi$ ?

Strictly speaking, yes.
And also the case that $f(a_1)\ge f(b_1)$. (Thinking)

I would usually hand wave it away saying that we can show the same result in those cases in similar fashion. (Bandit)
We can can't we? (Wondering)