Réponses

2013-01-29T22:07:52+01:00
2013-01-29T22:24:17+01:00

f'(x) = 2ax+b/ax²+bx+c

 

On sait que f'(0) = 0

f'(0) = 2a*0+b/a*0²+b*0+c = b/c

 

D'où f'(0) = b/c = 0