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kity99
@kity99
January 2021
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Bonsoir,
Je suis en terminale S , j'ai besoin de l'aide pour l'exercice 2
Merci pour ceux qui pourront m'aider :)
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scoladan
Verified answer
Bonjour,
g(x) = (5 - x²)/(x + 3)
1) g'(x) = [(-2x)(x + 3) - (5 - x²)]/(x + 3)²
= (-2x² - 6x - 5 + x²)/(x + 3)²
= (-x² - 6x - 5)/(x + 3)²
2) Signe de -x² - 6x - 5 :
Δ = (-6)² -4x(-1)x(-5) = 36 - 20 = 16 = 4²
⇒ 2 racines : x = (6 - 4)/(-2) = -1 et x = (6 + 4)/(-2) = -5
Donc (-x² - 6x - 5) = -(x + 1)(x + 5)
x -∞ -5 -3 -1 +∞
(x+1) - - || - 0 +
(x+5) - 0 + || + +
g'(x) - 0 + || + 0 -
3) On en déduit :
g croissante sur ]-5;-3[U]-3;-1[
g décroissante sur ]-∞;-5]U[-1;+∞[
4) g(-2) = 1 et g'(-2) = 3
Donc (T) : y = g'(-2)(x + 2) + g(2) soit y = 3x + 7
5) voir ci-joint (pas à l'échelle demandée)
3 votes
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Lista de comentários
Verified answer
Bonjour,g(x) = (5 - x²)/(x + 3)
1) g'(x) = [(-2x)(x + 3) - (5 - x²)]/(x + 3)²
= (-2x² - 6x - 5 + x²)/(x + 3)²
= (-x² - 6x - 5)/(x + 3)²
2) Signe de -x² - 6x - 5 :
Δ = (-6)² -4x(-1)x(-5) = 36 - 20 = 16 = 4²
⇒ 2 racines : x = (6 - 4)/(-2) = -1 et x = (6 + 4)/(-2) = -5
Donc (-x² - 6x - 5) = -(x + 1)(x + 5)
x -∞ -5 -3 -1 +∞
(x+1) - - || - 0 +
(x+5) - 0 + || + +
g'(x) - 0 + || + 0 -
3) On en déduit :
g croissante sur ]-5;-3[U]-3;-1[
g décroissante sur ]-∞;-5]U[-1;+∞[
4) g(-2) = 1 et g'(-2) = 3
Donc (T) : y = g'(-2)(x + 2) + g(2) soit y = 3x + 7
5) voir ci-joint (pas à l'échelle demandée)