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Question Number 34485 by candre last updated on 07/May/18
is there a function such that:  ∀x≥0;f(x+T_1 )=f(x)  ∀x≤0;f(x−T_2 )=f(x)  f is diferentiabre in x=0
$$\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}: \\ $$$$\forall{x}\geqslant\mathrm{0};{f}\left({x}+{T}_{\mathrm{1}} \right)={f}\left({x}\right) \\ $$$$\forall{x}\leqslant\mathrm{0};{f}\left({x}−{T}_{\mathrm{2}} \right)={f}\left({x}\right) \\ $$$${f}\:\mathrm{is}\:\mathrm{diferentiabre}\:\mathrm{in}\:{x}=\mathrm{0} \\ $$
Answered by MJS last updated on 07/May/18
f(x): y=c; c∈R
$${f}\left({x}\right):\:{y}={c};\:{c}\in\mathbb{R} \\ $$

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