Bonsoir,
2)
A=(-1,-1)
f(x)=x^3
f'(x)=3x^2
f'(-1)=3
y+1=(x+1)*3==>y=3x+2
3)
[tex]x^3-(3x+2)=x^3+x^2-x^2-x-2x-2\\
=x^2(x+1)-x(x+1)-2(x+1)\\
=(x+1)(x^2-x-2)\\
=(x+1)(x^2+x-2x-2)\\
=(x+1)(x(x+1)-2(x+1))\\
=(x+1)^2(x-2)[/tex]
4)
[tex] \left \{ {{y=x^3} \atop {y=3x+2}} \right. \\\\
x^3=3x+2\\
x^3-(3x+2)=0\\\\(x+1)^2(x-2)=0
[/tex]
Les points d'intersection sont (-1,-1) : un point double (une tangente)
et (2,2^3=8)
