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cc maraichers 17 QCM3 E


JJA
Go to solution Solved by Le_Chapelier_Fou,

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  • Solution

Hey !

 

L'appromixation affine ça correspond à la partie affine du DL1 c'est à dire g(t) + t*g'(t) :

  • g(0) = 100/2 + 50
  • g'(t) = 100*50/(1+e^-50t)²
  • g'(0) = 5000/4 = 1250

Soit g(t) + t*g'(t) = 50 + 1250t, pour t = 0 😉

Edited by Le_Chapelier_Fou
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