Question

CONSIDERAM FUNCTIA f:R ⇒R ,f(x)= -3x+2 .DETERMINATI  NUMARUL REAL A ,DACA
f(a-1)+f(a)+f(a+1)=0
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Answer 1

f(a-1)=-3(a-1)+2= -3a +3 +2= -3a+5
f(a)=-3a+2
f(a+1)=-3(a+1)+2= -3a -3 +2= -3a-1

f(a-1)+f(a)+f(a+1)=0 inlocuim
 -3a+5-3a+2 -3a-1=0
-9a+6=0
-9a=-6
9a=6
a=6/9
a=2/3

Answer 2

f(a-1) = -3(a-1)+2 = -3a+3+2 = -3a+5 ;
f(a)    = -3(a) +2   =        "       = -3a+2 ;
f(a+1)= -3(a+1)+2= -3a-3+2   = -3a-1    adunate , obtinem:
                -9a +6 = 0
               a = -6 / -9 = 2/3
                       a = 2/3