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Sorry for maybe duplicate but I searched a lot in this website but can't find any answer for my question:
Without using integrals and stuff consider $f:\mathbb{R}\rightarrow \mathbb{R}$ and there is some $a\in \mathbb{R}$, $\forall x\:,\:f'\left(x\right)\:=\:a\:\:$.
Prove that exist some $b$ such that $$\forall x,\:f\left(x\right)\:=\:ax+b$$

It seems very trivial but i don't know how to write the proof.
What exactly i need to show? any ideas would be very helpfull.
Thanks in advance!

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One simple way (assuming the right things, certainly), is to define

$$h(x):=f(x)-ax\implies h'(x)=f'(x)-a=a-a=0\implies h(x)=b\,(\text{a constant})=\;\forall\,x$$

The above assume that we know $\;f'(x)=0\;$ over an interval $\;\implies f(x)\;$ is a constant there.

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  • $\begingroup$ Which comes from the Mean Value Theorem. $\endgroup$ – Aaron Maroja Feb 23 '15 at 11:06

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