Răspuns:
f(x)=lx-3l(x²-5x+6)=
{(x-3)(x²-5x+6)x≥3
{-(x-3)(x²-5x+6 x<3
Continuitatea in x=3
Ld=x->3 x>3 lim (x-3)(x²-5x+6)=(3-3)(9-15+6)=0
Ls x->3 x<3 -(x-3)(x²-5x+6)=-(3-3)(9-15+6)=0
f(3)=(3-3)(9-15+6)=0
Ls=Ld=f(3)=0
Functia e continua in x=3
Verificam derivatele in x=3
x≥0 f d`(x)={(x-3) `(x²-5x+6)+(x-3)(x²-5x+6)`=(x²-5x+6)+(x-3)(2x-5)=(x-2)(x-3)+(x-3)(2x-5)=(x-3)(x-2+2x-5)=(x-3)(3x-7)
x<0 f s`(x)=(3-x)`(x²-5x+6)+(3-x)(x²-5x+6) `=-(x²-5x+6)+(3-x)(2x-5)=-(x-2)(x-3)+(3-x)(2x-5)=(x-2)(3-x)+(3-x)(2x-5)=(3-x)(x-2+2x-5)=(3-x)(3x-7)
Se observa ca f `s(3)=f `d(3)=0=>
Functia nu admite puncte unghiulare
Ld x->3 x>3 lim
Explicație pas cu pas:
