1. f(x)=ax+b
a) 3*f(x) + 2= 3x+5
3*(ax+b) +2= 3x+5
3ax+3b +2= 3x+5
3ax=3x =>a=1
3b+2=5 => 3b=3=> b=1
=> f(x) = x+1
b)f(1) + f(2) + f(3) = 6a+2b si f(4)=8
a+b+2a+b+3a+b= 6a+2b si 4a+b=8
6a+3b=6a+ 2b => b=0 si 4a+0=8 => a=2
=> f(x)=2x
2. f(x)=ax+b
a) A(2,7) B(-1,-2) ∈ Gf
deci 2a+b=7
-a+b=-2
si calculam sistemul; inmultim a doua relatie cu 2 si obtinem
2a+b=7
-2a+2b=-4
adunam si obtinem 3b=3 adica b=1
2a+1=7=> 2a=6 => a=3
=> f(x)=3x+1
b)A(-1,-1) B(1,3)∈Gf
deci -a+b=-1
a+b=3
adunam si obtinem 2b=2 de unde b=1
a+1=3=> a=2
=> f(x)=2x+1
c)A(2,-1) B(-2,7) ∈ Gf
deci 2a+b=-1
-2a+b=7
adunam si obtinem 2b=6 deci b=3
2a+3=-1 => 2a=-4 => a=-2
=> f(x)=-2x+3
d) A(2,8) B(-2,4) ∈Gf
deci 2a+b=8
-2a+b=4
adunam si obtinem 2b=12 => b=6
2a+6=8 =>2a=2 =>a=1
=> f(x)=x+6
3.f(x)=(2a-3)x+[tex]\frac{a-1}{3}[/tex]
stiind ca A(1,-8) ∈ Gf => x=1 si f(1)=-8
asadar (2a-3)+[tex]\frac{a-1}{3}[/tex]= -8 aducem la acelasi numitor si il eliminam; numitorul
fiind 3
3(2a-3)+ a-1=-8*3
6a-9+a-1=-24
7a=-24+9+1
7a=-14
a=-2
