Răspuns:
b) fie f(x)=ax+b
f(x-3)=a(x-3)+b=ax-3a+b
f(x+3)=a(x+3)+B=ax+3a+b
Inlocuiesti
(ax+b-3a)(ax+b+3a)-12x+36=4x²+9
(ax+b)²-(3a)²-12x+36=4x²+9
(ax)²+2abx+b²-9a²-12x+36=4x²+9
(ax)²+2x(ab-6)+b²+36-9a²=4x²+9=>
a²=4=>a=±2
ab-6=0
a1= -2
-2b-6=0
-2b= 6
b= -3=>
f(x)=-2x-3
Caz2
a=2
2b-6=0
2b=6
b=3
f(x)=2x+3
d) (fof)x-fof(x-1)=1
f(x)=ax+b
fof)(x)=a(ax+b)+b=a²x+ab+b
f(x-1)=a(x-1)+b=ax-a+b
(fof)(x-1)=a(x-1)-a+b=ax-a-a+b=
ax-2a+b
fof(x)-(fof)(x-1)=
a²x+ab+b-ax-2a+b=1
x(a²-a)+ab-2a+b=1
=>
a²-a=0 a1=0 a2=1
ab-2a+b=1
Pt a=0
0*b-2*0+b=1=>
b=1=>
f1(x)=1
a=1
1*b-2*1+b=1
2b-2=1
2b=3
b=3/2
f2(X)=x+3/2
Explicație pas cu pas:
