bonjour
f (x) = ( 0.5 x + 1 ) ( x - 3 )
1. f (2) = ( 0.5 *2 + 1 ) ( 2 - 3 ) = 2 * - 1 = - 2
f ( - 3) = ( 0.5* -3 + 1 ) ( - 3 - 3) = 2.5 * - 6 = - 15
f ( 0) = 1 * - 3 = - 3
2 ) ( 0.5 x + 1 ) ( x - 3 ) = 0
il suffit que l'in des facteurs soit nul
0.5 x + 1 = 0 ⇔ x = - 2
ou bien x - 3 = 0 ⇔ x = 3
S ( - 2 ; 3 )
f (x) = 0.5 x² - 1.5 x + x - 3 = 0.5 x² - 0.5 x - 3
0.5 x² - 0.5 x - 3 = - 3
0.5 x² - 0.5 x - 3 + 3 = 0
0.5 x² - 0.5 x = 0
0.5 x ( x - 1 ) = 0
x = 0 ou 1
