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