f(x) = ax +b +c ln(x)
f(1) = a +b +cln(1) = a+b ( car ln(1) =0)
a+b = 1 => b= 1-a
f(2) = 2a +b+c ln(2) = 2ln(2)
2a +b = 2ln2 - c ln2 = ln2( 2 -c)
(2a +b) /(2-c) = ln 2
la dérivée f' = a+ ( c /x)
f'(2) = a + c/2
a+c/2 = 0 => c/2 = - a => c = -2a
on remplace méthode par substitution
(2a +b) /(2-c) = ln 2 =>
2a + (1-a) / 2- (-2a) = ln2
2a + 1- a = ln2 (2 +2a) =
a +1 = 2ln2 + 2a ln2
a - 2a ln 2 = 2 ln 2 -1
a ( 1 -2ln2) = 2 ln2 -1
a =( 2ln2 -1 ) / ( 1- 2ln2)
a =( 2ln2 -1 ) /- (2 ln2 -1 )
a = -1
b= 1-a = 1+1 =2
b=2
c = -2a =>
c = 2
donc la fonction
f(x) = - x + 2 +2 ln(x)
