Réponses

  • Omnes
  • Modérateur confirmé
2014-03-17T00:18:56+01:00
Salut,

On a : f(x) = -1/2(x-1)² + 2

1a.
f(x) = -1/2(x-1)² + 2
f(x) = -1/2(x² -2x + 1) + 2
f(x) = -1/2x² + x - 1/2 + 2
f(x) = -1/2x² + x -1/2 + 4/2
f(x) = -1/2x² + x + 3/2

1b.

f(x) = -1/2(x+1)(x-3)
f(x) = (-1/2x -1/2)(x-3)
f(x) = -1/2x² + 3/2x - 1/2x + 3/2
f(x) = -1/2x² +x + 3/2

2.
a)
-1/2(x+1)(x-3) < 0

(x+1)(x-3) < 0
x + 1 < 0
x < -1
x - 3 < 0
x < 3

S = }-inf; 3[

b.

f(-2) = -1/2(-2-1)² + 2
f(-2) = -1/2*9 + 2
f(-2) = -9/2 + 2
f(-2) = -9/2 + 4/2
f(-2) = -5/2

f(3) = -1/2(3-1)² + 2
f(3) = -1/2*2² + 2
f(3) = -1/2*4 + 2
f(3) = -2 + 2
f(3) = 0

c.
f est croissante sur ]-inf; 1] puis décroissante sur [1;+inf[

d.

f(x) = 3/2
-1/2x² +x + 3/2 = 3/2
-1/2x² + x + 3/2 - 3/2 = 0
-1/2x² + x = 0
x(-1/2x + 1) = 0
x = 0
-1/2x + 1 = 0
1/2x = -1
x = -2

e.

Je cherche et reviens vers toi.

f.

f(x) = 0
-1/2(x+1)(x-3) = 0
(x+1) = 0
x = -1
x-3 = 0
x = 3

Bonne soirée (je reviens pour la e.)