Question

Consideram functia f:R->R, f(x)=-3x+2. Determinati numarul real a, daca f(a-1)+f(a)+f(a+1)=0.

Answer 1

f(x) = -3x + 2

Atunci:

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

Dar f(a-1) + f(a) + f(a+1) = 0, adica:

-3a + 5 -3a + 2 -3a - 1 = 0

-9a + 6 = 0

9a = 6

a = 6/9

a = 2/3

R: a = 2/3