Bonjour,
1)
f(0)=3
f'(0)=1
f'(1)=0
2)
f(x)=(ax+b)*e^x+c
f'(x)=a*e^x+(ax+b)*e^x=e^x(ax+a+b)
f(0)=3 ==> (a*0+b)*e^0+c=3 ==> b+c=3 (1)
f'(0)=1 ==> e^0*(0+a+b)=1 ==> a+b=1 (2)
f'(1)=0 ==> e^1*(a+a+b)=0 ==> 2a+b=0 (3)
(3)-(2) ==> a=-1
(2) ==> b=1-(-1)=2
(1) ==> c=3-2=1
