Răspuns:
Explicație pas cu pas:
f : R—>R , f(x) = 3x + 2
a) f(-1) + f(-2) = 3·(-1)+2+3·(-2)+2 = 3·(-3)+4 = -9+4 = -5
b) [f(1/3)-1] * f(1/2)
* = inmultire =>
[f(1/3)-1] = 3·1/3+2-1 = 2
f(1/2) = 3/2+2 = 7/2
[f(1/3)-1] x f(1/2) = 2 x 7/2 = 7
c) f(1)+f(2)+...f(40) = 3+2+6+2+9+2+....+120+2 =
= 40·2 + 3·(1+2+3+....+40) = 80+3·40·41:2 =
= 80+60·41 = 80+2460 = 2540
d) f(x) -5 = x+3 <=>
3x+2-5 = x+3 => 2x = 6 => x = 3
e) f(x-1) +1 = 0 <=>
3(x-1)+2+1 = 0 <=>
3(x-1) +3 = 0 <=> 3x = 0 => x = 0
f) f(2x-3) ≤ 0 <=>
3(2x-3)+2 ≤ 0 <=>
6x-7 ≤ 0 => 6x ≤ 7 => x ≤ 7/6
