Ex 11.
o functie fractie admite asimptote verticale in punctele care anuleaza numitorul
observi ca pt a=1 si b=1 vei avea numitorul x²-2x+1=(x-1)²/.Pt x=1 numitorul e 0 , deci x=1 a.v.
ex16.
cos 2x=cos²x-sin²x => sin²x=(1-cos2x)/2>integrezi
I=∫[(1-cos2x)/2]*dx=1/2∫xdx-1/2∫cos2xdx=x/2-1/4sin 2x x∈[0.π/4]
aplici formula Leibnitz Newton si obtii
I=π/4
Ex 17.
F(x)=∫[(x²/(x^6+1)]*dx
x³=y=> 3x²dx=dy=.x²dx=dy/3
integrala devine
F(y)=1/3*∫dy./(y²+1)=1/3 arctg y=1/3 arctg x³+C
Ex 12
Daca x=0 punct de extrem atunci f `(0) =0
f `(x)=(x²-2x-a-b)/(x-1)²
f `(0)=(-a-b)/1=0 => a=-b
Deoarece A(0,1) ∈Gf atunci f(0)=1
f(0)=-b/1=1=> b=-1 =>a =1
