Question

Buna seara !Am nevoie de ajutor la acest exercitiu .

Answer 1

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: