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Given that for two functions $f$ and $g$ it holds that $f'' = g''$ for all $x \in \mathbb{R}$, how can it be shown that the difference of $f$ and $g$ is afin, i.e. that $f - g = ax+b$, for some a and some b.

I do not expect solutions, hints are enough for starting, I will try it then myself.

Thanks

Carlos Eugenio Thompson Pinzón
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TestGuest
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1 Answers1

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Well you know that $f^{\prime\prime}(x)-g^{\prime\prime}(x) = 0$ for all $x$. What functions have derivative equal to zero everywhere? What does this tell you about $f^\prime(x)-g^\prime(x)$?

Casteels
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  • ah of course their slope must be constant – TestGuest Oct 29 '13 at 10:22
  • Ok, then I can define h:= f-g, and the second derivative of h is zero, hence h must be a linear function, right? – TestGuest Oct 29 '13 at 12:52
  • Well that's what you want to prove, so be careful about circular reasoning. Do you know about integrals and the fundamental theorem of calculus? – Casteels Oct 29 '13 at 12:54
  • yes, I do. Could I then define h'' := f''-g'', and then int(h'') = h' = a, and thus int(h') = ax+b, for constants a,b ? – TestGuest Oct 29 '13 at 13:00
  • That's the idea, but just be careful using the FTC since it deals with definite integrals. But I think you can get away with just using antiderivatives. – Casteels Oct 29 '13 at 13:18