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LaMissD
@LaMissD
January 2021
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Bonjour!! :)
Est ce que quelqu'un pourrait me donner un petit coup de pouce pour cet exercice svp??
...mercredi d'avance
Bonne journée
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scoladan
Verified answer
Bonjour,
1) f(x) = ln(ax + b)
f(2) = 0
<=> ln(2a+b) = 0
==> 2a + b = 1 (Eq. 1)
f'(x) = a/(ax+b)
f''(3) = 3/4
<=> a/(3a+b) = 3/4
<=> 4a = 3(3a+b) (Eq. 2)
(1) ==> a = (1 - b)/2
==> (2) devient : 2(1-b) = 3(3(1-b)/2 + b)
<=> 2(1-b) = 3(3/2 - b/2)
<=> 4(1-b) = 9 - 3b
<=> -4b + 3b = 9 - 4
==> b = -5 ==> a = (1-b)/2 = 3
==> f(x) = ln(3x - 2)
Ensemble de définition :
3x - 2 > 0
==> x > 2/3 ==> Df = ]2/3, +infini[
Sens de variations
f'(x) = 3/(3x-2) > 0 sur Df.
Donc f est croissante sur Df.
2) A(2;0) appartient à C
==> 0 = ln(2a + b)
==> 2a + b = 1 (Eq. 1)
La tgte en A a pour coefficient directeur -2
==> f'(2) = -2
==> a/(2a+b) = -2
==> a = -2(2a+b)
==> 5a = -2b
==> a = -2b/5 (Eq. 2)
(1) devient : -4b/5 + b = 1
soit b/5 = 4 ==> b = 5
et donc a = -2b/5 = -2
==> f(x) = ln(-2x+5)
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Verified answer
Bonjour,
1) f(x) = ln(ax + b)
f(2) = 0
<=> ln(2a+b) = 0
==> 2a + b = 1 (Eq. 1)
f'(x) = a/(ax+b)
f''(3) = 3/4
<=> a/(3a+b) = 3/4
<=> 4a = 3(3a+b) (Eq. 2)
(1) ==> a = (1 - b)/2
==> (2) devient : 2(1-b) = 3(3(1-b)/2 + b)
<=> 2(1-b) = 3(3/2 - b/2)
<=> 4(1-b) = 9 - 3b
<=> -4b + 3b = 9 - 4
==> b = -5 ==> a = (1-b)/2 = 3
==> f(x) = ln(3x - 2)
Ensemble de définition :
3x - 2 > 0
==> x > 2/3 ==> Df = ]2/3, +infini[
Sens de variations
f'(x) = 3/(3x-2) > 0 sur Df.
Donc f est croissante sur Df.
2) A(2;0) appartient à C
==> 0 = ln(2a + b)
==> 2a + b = 1 (Eq. 1)
La tgte en A a pour coefficient directeur -2
==> f'(2) = -2
==> a/(2a+b) = -2
==> a = -2(2a+b)
==> 5a = -2b
==> a = -2b/5 (Eq. 2)
(1) devient : -4b/5 + b = 1
soit b/5 = 4 ==> b = 5
et donc a = -2b/5 = -2
==> f(x) = ln(-2x+5)