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

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In summary: Thinking)In summary, we discussed the conditions for a function to have a local extremum at a point $\xi$ in an interval $I$, and how to find $a$ and $b$ such that $f(a)=f(b)$ using the intermediate value theorem. We also explored the case where $f'(\xi)=0$ and showed that in this case, $\xi$ must be a local extremum.
  • #1
evinda
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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)
 
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  • #2
evinda said:
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)

evinda said:
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)
 
  • #3
Klaas van Aarsen said:
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)
Klaas van Aarsen said:
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)
 
  • #4
evinda said:
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)
 
  • #5
Klaas van Aarsen said:
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)
 
  • #6
evinda said:
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)

evinda said:
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)
 
  • #7
Klaas van Aarsen said:
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)
 
  • #8
evinda said:
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)
 
  • #9
Klaas van Aarsen said:
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)
 
  • #10
evinda said:
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)

evinda said:
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)
 
  • #11
Klaas van Aarsen said:
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)
 
  • #12
evinda said:
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)
 

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

1. What does it mean for $f'(\xi)$ to be equal to zero?

When $f'(\xi)$ is equal to zero, it means that the derivative of the function $f(x)$ at the point $\xi$ is equal to zero. This indicates that the slope of the tangent line to the graph of $f(x)$ at $\xi$ is equal to zero, and therefore the function is not changing at that point.

2. Why is it important to show the existence of $a,b$ where $f'(\xi) \neq 0$?

Showing the existence of $a,b$ where $f'(\xi) \neq 0$ is important because it proves that the function $f(x)$ is not constant in the interval $[a,b]$. This means that the function is changing and has a non-zero slope at every point in the interval, which is necessary for many mathematical and scientific applications.

3. How can we prove that $f'(\xi) \neq 0$ for a given function?

To prove that $f'(\xi) \neq 0$, we can use the definition of the derivative and the properties of limits to show that the limit of the difference quotient is not equal to zero. This can be done by finding the derivative of the function and evaluating it at the point $\xi$, or by using other methods such as the Mean Value Theorem.

4. Can $f'(\xi)$ be equal to zero at some points in the interval $[a,b]$?

Yes, $f'(\xi)$ can be equal to zero at some points in the interval $[a,b]$. However, to show the existence of $a,b$ where $f'(\xi) \neq 0$, we only need to prove that there exists at least one point in the interval where $f'(\xi) \neq 0$. This means that $f(x)$ can have points of zero slope, but it must also have points of non-zero slope in the interval.

5. What are the implications of $f'(\xi) \neq 0$ for the behavior of the function $f(x)$?

If $f'(\xi) \neq 0$ for all points in the interval $[a,b]$, then the function $f(x)$ is continuously changing and has a non-zero slope at every point. This means that the function is not constant and has a distinct behavior in the interval. It also allows us to make conclusions about the concavity and extrema of the function, as well as its behavior near the points $\xi$ and $a,b$.

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