1) a. cos(BAM) = AH/AM = AM/AB
AM² = AH x AB = 6x
AM = V(6x)
b. AMB est un triangle rectangle
(HK) perpendiculaire à (MB)
donc (AM)//(HK)
Thales entraine HK/AM = BH/AB
f(x) = HK = (6-x)V(6x) /6 = V6 /6 (6-x)Vx
2) a. f'(x) = V6/6 ((6-x)/2Vx - Vx ) = V6/6 (2-x)/Vx
b. 0 2 6
f'(x) + --
f(x) / \
