Question

Fie: f : R—>R , f(x) = 3x + 2
Sa se calculeze:
a). f(-1)+f(-2)
b). $(f(\frac{1}{3}) -1) * f(\frac{1}{2})$
c). f(1) + f (2) +....f(40)
d). f(x)-5=x+3
e). f(x-1)+1=0
f). f(2x-3)≤0

Answer 1

Explicație pas cu pas:

succes..... .......

Answer 2

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