Question

Fie a apartine R si functiile f,g:R->R ,f(x)=3x+2 si g(x)=2x+a. Determinati valorile lui a pentru care f°g=g°f

Answer 1

f•g=3(2x+a)+2=6x+3a+2
g•f=2(3x+2)+a=6x+4+a

6x+3a+2= 6x+4+a
3a+2=4+a
3a-a=4-2
2a=2
a=1

Answer 2

f°g=f(g(x)=f(2x+a) =3(2x+a)+2=6x+3a+2
g°f=g(f(x))=g(3x+2)=2(3x+2)+a=6x+a+4

6x+3a+2=6x+a+4
3a+2=a+4
2a=2
 a=1

verificare
f(x) =3x+2
g(x)=2x+1

f°g=3(2x+1)+2=6x+5
g°f=2(3x+2)+1=6x+5
adevarat , bine rezolvat