Question

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Answer 1

Răspuns:

Explicație pas cu pas:

1) f:R --> R  ; f(x) = x - 3

f(-4) · f(-3)·......f(3)·f(4) = 0  ; deoarece exista f(3) = 3-3 = 0

2) f:R --> R  ; f(x) = 2x + 1

f(-2) + f(-1) + f(0) + f(1) = 2·(-2)+1 + 2·(-1)+1+2·0+1+2·1+1 =

= -3-1+1+3 = 0

3)  f:R --> R  ; f(x) = x + 3  ; g(x) = 2x - 1

2f(x) +3g(x) = -5 <=>

2(x+3) +3(2x-1) = -5 <=>

2x+6+6x-3 = -5 <=> 8x = -2 => x = -2/8 => x = -1/4

4) f:R --> R  ; f(x) = 3 - 4x

f(x) - 1 ≥ 4x <=> 3 - 4x ≥ 4x  <=> 3 ≥ 8x => 8x ≤ 3 => x ≤ 3/8 ; x∈(-∞ ; 3/8]

5) f:R --> R  ; f(x) = 2x + 1  ; x = f(x) => x = 2x + 1  => x = -1 => A(-1 ; -1)

6) f:R --> R  ; f(x) = 2x - 1 ; g(x) = -4x + 1

f(x) ∩ g(x) <=> 2x-1 = -4x +1  => 6x = 2  => x = 2/6 => x = 1/3

f(1/3)  = 2/3 - ³⁾1 = (2-3)/3 = -1/3  => A (1/3 ; -1/3)

7) f(x) = ax+b  A(0 ; -2)  ; B(2 ; 0)  ∈ Gf

A(0 ; -2) ∈Gf <=> f(0) = -2  => a·0+b = -2 => b = -2

B(2 ; 0)  ∈ Gf <=> f(2) = 0  => a·2 +b = 0  => 2a-2 = 0 => 2a = 2 =>

a = 1  => f(x) = x - 2