Mean Value Theorem: Proof & Claim

[fishturtle1]

Homework Statement

a) Suppose $f$ is twice differentiable on an open interval $I$ and $f''(x) = 0$ for all $x \in I$. Show $f$ has the form $f(x) = ax + b$ for suitable constants $a$ and $b$.

b) Suppose $f$ is three times differentiable on an open interval $I$ and $f''' = 0$ on $I$. What form does $f$ have? Prove your claim.

Relevant Equations

Mean Value theorem: Let $f$ be a continuous function on $[a, b]$ that is differentiable on $(a, b)$. Then there exists [at least one] $x$ in $(a, b)$ such that:

$$f'(x) = \frac{f(a) - f(b)}{a - b}$$


Theorem: Let $f$ be a differentiable function on $(a,b)$ such that $f'(x) = 0$ for all $x \in (a,b)$. Then $f$ is a constant function on $(a,b)$.

a) Proof: By theorem above, there exists a $a \in \mathbb{R}$ such that for all $x \in I$ we have $f'(x) = a$. Let $x, y \in I$. Then, by Mean Value Theorem,

$$a = \frac{f(x) - f(y)}{x - y}$$

This can be rewritten as $f(x) = ax - ay + f(y)$. Now, let $g(y) = -ay + f(y)$. Then $g'(y) = \lim_{t \rightarrow y} \frac{g(t) - g(y)}{t - y} = \lim_{t \rightarrow y}\frac{-at + f(t) +ay - f(y)}{t - y} = \lim_{t\rightarrow y}\frac{-at + ay}{t - y} + \lim_{t\rightarrow y}\frac{f(t) - f(y)}{t - y} = -a + a = 0$ So, $g$ is constant on $I$. So there exists $b \in \mathbb{R}$ such that for all $y \in I$, $g(y) = b$. We can conclude $f(x) = ax + b$. []

b) I claim that there is $a, b, c$ such that for all $x \in I$: $f(x) = ax^2 + bx + c$

Proof: By part a) there exists $a, b \in \mathbb{R}$ such that for all $x \in I$ we have $f'(x) = ax + b$. Let $x, y \in I$. By MVT there exists $t \in I$ such that $$at + b = \frac{f(x) - f(y)}{x - y}$$

This can be rewritten as $f(x) = (at + b)x - y(at + b) + f(y)$. Let $g(y) = -y(at + b) + f(y)$. Then, $g'(y) = -(at + b) + (ay + b) = a(y - t)$.Did I use MVT incorrectly?

 

[PeroK]

A couple of points. It might have been simpler to take some fixed $x_0 \in I$, rather than a second variable $y$.

I can't see the conclusion to part b).

 

[fishturtle1]

A couple of points. It might have been simpler to take some fixed $x_0 \in I$, rather than a second variable $y$.

I can't see the conclusion to part b).

Thanks for the reply.

Where would I be able to fix $x_0$? It seems for MVT, we choose an $x, y \in I$ and then are guaranteed a $c \in I$ such that $f'(c) = \frac{f(x) - f(y)}{x - y}$?

Also, for part b) I meant I'm not sure how to continue. It seems $g'(y) \neq 0$ so $g(y)$ is not constant. (and I was expecting $g(y)$ to be constant). Also, I'm not sure what to do with $(at + b)x = atx + bx$. It seems it would be easier if $t = x$ but I'm not sure how to get that.

I thought to use definition of derivative: $\lim_{y \rightarrow x} \frac{f(y) - f(x)}{y - x} = f'(x) = ax + b$ but haven't made progress with this either.

 

[PeroK]

I would have taken $x_0 \in I$ and then from the MVT shown that For all $x \in I$:

$a = \frac{f(x) - f(x_0)}{x - x_0}$

 

[fishturtle1]

I would have taken $x_0 \in I$ and then from the MVT shown that For all $x \in I$:

$a = \frac{f(x) - f(x_0)}{x - x_0}$

OH ok so it would be something like this?: Fix $x_0 \in I$ and let $x$ be any element in $I$. By MVT,
$$a = \frac{f(x) - f(x_0)}{x - x_0}$$
$$a(x - x_0) + f(x_0) = f(x)$$
$$ax - ax_0 + f(x_0) = f(x)$$
$$ax + b = f(x)$$
where $b = -ax_0 + f(x_0)$ is a constant since $a, x_0, f(x_0)$ are constants.[]

So this solves my problem with $g$ not being constant in part b.

 

[PeroK]

OH ok so it would be something like this?: Fix $x_0 \in I$ and let $x$ be any element in $I$. By MVT,
$$a = \frac{f(x) - f(x_0)}{x - x_0}$$
$$a(x - x_0) + f(x_0) = f(x)$$
$$ax - ax_0 + f(x_0) = f(x)$$
$$ax + b = f(x)$$
where $b = -ax_0 + f(x_0)$ is a constant since $a, x_0, f(x_0)$ are constants.[]

So this solves my problem with $g$ not being constant in part b.

To finish off part b) you might have to use more than the MVT. I don't immediately see a way to finish it off using the same technique as you used for part a).

 

[PeroK]

Hint. If $f'(x) = g'(x)$, what can you say about $(f-g)(x)$?

 

[fishturtle1]

Hint. If $f'(x) = g'(x)$, what can you say about $(f-g)(x)$?

We can say $(f-g)(x)$ is constant.

 

[PeroK]

We can say $(f-g)(x)$ is constant.

Okay. So, if you could find a single function for which $f'''(x) = 0$, then you can find them all?

I'm going offline now.

 

[member 587159]

Here is how I would approach the problem:

Prove the following claim:

Let $f,g: I \to \mathbb{R}$ be differentiable functions where $I$ is an interval. If $f'=g'$ on $I$, then there is a constant $c$ such that $f=g+c$.

Proof: We have $(f-g)'=0$ and by your claim $f-g$ is constant. $\quad \square$

Alright, let's see how this helps.

First, you have $0'=0=f''=(f')'$ so by the claim there is a constant $a$ such that $f'=a$. Note that for all $x\in I$ we have $(ax)'= a =f'(x)$ so applying the claim a second time, you get that there is a constant b such that $ax+b=f(x)$ for all $x\in I$.

You can do the other question by applying this claim three times. Let me know what you get.

EDIT: basically, the exercise is a good preparation for what you will see soon: indefinite integration.

 

[PeroK]

Okay. So, if you could find a single function for which $f'''(x) = 0$, then you can find them all?

What I said was imprecise. I should have said that if you can find a function for which $f'(x) = ax + b$, then you can find them all.

 

[fishturtle1]

Here is how I would approach the problem:

Prove the following claim:

Let $f,g: I \to \mathbb{R}$ be differentiable functions where $I$ is an interval. If $f'=g'$ on $I$, then there is a constant $c$ such that $f=g+c$.

Proof: We have $(f-g)'=0$ and by your claim $f-g$ is constant. $\quad \square$

Alright, let's see how this helps.

First, you have $0'=0=f''=(f')'$ so by the claim there is a constant $a$ such that $f'=a$. Note that for all $x\in I$ we have $(ax)'= a =f'(x)$ so applying the claim a second time, you get that there is a constant b such that $ax+b=f(x)$ for all $x\in I$.

You can do the other question by applying this claim three times. Let me know what you get.

EDIT: basically, the exercise is a good preparation for what you will see soon: indefinite integration.

Proof: We have $f''' = 0'''$. So, there exists $f'' = 2a$ for some $2a \in \mathbb{R}$. Now, $f'' - (2ax)' = 0$. So there exists $b \in \mathbb{R}$ such that $f' - 2ax = b$ i.e. $f' - 2ax - b = 0$. Now, $(ax^2 + bx)' = 2ax + b$. So, $f' - (ax^2 + bx)' = 0$. Thus, there exists $c \in \mathbb{R}$ such that $f - (ax^2 + bx) = c$. This can be rewritten as $f(x) = ax^2 + bx + c$. $\square$