Bonsoir,
Exercice 4:
f(x) = (2x+6)-(x+3)²
1) f(x) = (2x+6)-(x+3)² = 2x+6-(x²+6x+9) = 2x+6-x²-6x-9 = x²-4x-3
f(x) = (2x+6)-(x+3)² = 2(x+3) - (x+3)² = (x+3)(2-x-3) = (x+3)(-1-x)
2) f(0) = 0²-4×0-3 = -3
f(√2) = (√2+3)(-1-√2) = -√2-2-3-3√2 = -5-4√2
f(-½) = (-½)²-4×(-1/2)-3 = ¼+2-3 = ¼-1 = -¾
3) f(x) = 0
(x+3)(-1-x) = 0 ==> équation produit nul
soit x+3 = 0 donc x = -3
soit -1-x = 0 donc x = -1
f(x) = -3
x²-4x-3 = -3
x²-4x = 0
x(x-4) = 0
soit x = 0
soit x-4 = 0, donc x = 4
Exercice 5:
g(x) = (x-3) / (x+5)
1) x+5 ≠ 0 , donc x ≠ -5
x € ]-∞;-5[ et ]-5;+∞[
2) f(0) = (0-3)/(0+5) = -3/5
f(√2) = (√2-3)/(√2+5) = [(√2-3)(√2+5)]/[(√2+5)² = (2+5√2-3√2-15)/(2+10√2+25) = (-13+2√2)/27+10√2)
f(-½) = (-½-3)/(-½+5) = (-½-6/2)/(-½+10/2) = (-7/2)/9/2 = -7/2×2/9 = -7/9
3) f(x) = 0
(x-3)/(x+5) = 0
Puisque x+5≠0, donc x-3 = 0, x = 3
f(x) = -3
(x-3)/(x+5) = -3
x-3 = -3x-15
x+3x = -15+3
4x = -12
x = -12/4 = -3
f(x) = 1
(x-3)/(x+5) = 1
(x-3)-(x+5)/(x+5) = 0
x-3-x-5 = 0
-8 = 0 impossible
je pense que la fonction f n'a pas de point ayant pour ordonnée 1.
