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We know that if we have two real differentiable functions $\;f,g\;$ on an interval $\;I\subset \Bbb R\;$ s.t. $\;h'(x)=g'(x)\;\;\forall\,x\in I\;$ , we then get that $\;f(x)=g(x)+C\;$ on $\;I\;$ , $\;C=$ a constant, and from here we get, under the same assumptions, that with a twice differentiable function $\;f\;$ on an interval $\;I\;$

$$(1)\;\;f''(x)=a=\text{ a constant}\;\implies\;\exists\,\text{constants}\;\;b,c\;\;s.t.\;\;f(x)=\frac a2x^2+bx+c\;\;(2)$$

The above is pretty simple, but my problem now is to find a counterexample as simple as possible to the last claim above if $\;I\;$ is not an interval , meaning: a function $\;f\;$ differentiable twice and fulfilling (1) does not necessarily have the form (2) if we don't assume $\;I\;$ is an interval .

By simple I mean that this is Calculus I and any argument about connectedness or the like must be avoided if possible.

Thanks (Yes, I know: "Don't thank anyone!", but my mom's education kicks in here...)

DonAntonio
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    If $I$ is the disjoint union of two intervals, then the constants $b$ and $c$ can differ on the two separate intervals (it. $f$ would be a "piece-wise quadratic") – Prahlad Vaidyanathan Dec 31 '13 at 13:10
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    Thanks for the idea, @PrahladVaidyanathan .+1 – DonAntonio Dec 31 '13 at 13:17
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    @PrahladVaidyanathan. But not an interval does not imply that you are dealing with the union of two disjoint intervals. So you could still have cases where $I$ is not an interval and form (2) is necessary. – drhab Dec 31 '13 at 13:19
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    Yes, but that is a much more involved question. The OP just asked for a counterexample! – Prahlad Vaidyanathan Dec 31 '13 at 13:24
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    How about the step function, which illustrates the first order case as well? – Macavity Dec 31 '13 at 13:26
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    Thanks @drhab. I think Prahlad's counterexample can work fine to the question "show that if $;I;$ is not an interval then the above isn't true". – DonAntonio Dec 31 '13 at 13:27

1 Answers1

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$f: \mathbb{R}\setminus\{0\} \to \mathbb{R} $ $$ f(x) = \begin{cases}1 & x>0 \\ -1 & x<0 \end{cases} $$

TBrendle
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