Articles
Register
Sign In
Search
Puchino
@Puchino
January 2021
1
44
Report
Partie B seulement
Bonjour, j'aurai besoin de votre aide sur toute la partie B. Merci beaucoup d'avance :)
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
aur70
Verified answer
Bonjour
f(x)=(2-x)(x+3)+3(2-x)
1/ f(x)=2x+6-x²-3x+6-3x=-x²-4x+12
2/ on factorise
f(x)=(2-x)[(x+3)+3] = (2-x)(x+3+3)=(2-x)(x+6)
3/ 16-(x+2)²=4²-(x+2)²=[4-(x+2)][4+(x+2)]=(4-x-2)(4+x+2)=(2-x)(x+6)=f(x)
4/ a/ f(-2)=16-(-2+2)²=16-0=16
b/ f(1+√2)= -(1+√2)²-4(1+√2)+12 = -(1+2√2+2)-4-4√2+12 = -3-2√2-4-4√2+12 = 5-6√2
c/ f(x)=0 ⇔(2-x)(x+6)=0 ⇔2-x=0 ou x+6=0 ⇔ x=2 ou x=-6
5/ a/
x -∞ -6 2 + ∞
(2-x) + + 0 -
(x+6) - 0 + +
produit - + -
b/ f(x)<0 ⇔(2-x)(x+6)<0 ⇔ x<-6 ou x>2 (lecture dans le tableau de signe)
2 votes
Thanks 2
More Questions From This User
See All
Puchino
February 2021 | 0 Respostas
Responda
Puchino
February 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
[TRIGONOMETRIE] J'ai besoin de votre aide, merci de bien vouloir justifier :)
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
Puchino
January 2021 | 0 Respostas
Responda
×
Report "Partie B seulement Bonjour, j'aurai besoin de votre aide sur toute la partie B. Merci beaucoup d'av.... Pergunta de ideia de Puchino"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
Bonjourf(x)=(2-x)(x+3)+3(2-x)
1/ f(x)=2x+6-x²-3x+6-3x=-x²-4x+12
2/ on factorise
f(x)=(2-x)[(x+3)+3] = (2-x)(x+3+3)=(2-x)(x+6)
3/ 16-(x+2)²=4²-(x+2)²=[4-(x+2)][4+(x+2)]=(4-x-2)(4+x+2)=(2-x)(x+6)=f(x)
4/ a/ f(-2)=16-(-2+2)²=16-0=16
b/ f(1+√2)= -(1+√2)²-4(1+√2)+12 = -(1+2√2+2)-4-4√2+12 = -3-2√2-4-4√2+12 = 5-6√2
c/ f(x)=0 ⇔(2-x)(x+6)=0 ⇔2-x=0 ou x+6=0 ⇔ x=2 ou x=-6
5/ a/
x -∞ -6 2 + ∞
(2-x) + + 0 -
(x+6) - 0 + +
produit - + -
b/ f(x)<0 ⇔(2-x)(x+6)<0 ⇔ x<-6 ou x>2 (lecture dans le tableau de signe)