Mean Value Theorem

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
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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).

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).

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

member 587159
Hint. If ##f'(x) = g'(x)##, what can you say about ##(f-g)(x)##?
We can say ##(f-g)(x)## is constant.

PeroK
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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.

fishturtle1
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.

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fishturtle1
PeroK
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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##