Question

6. Fie funcţia ƒ: R⇒R, care verifică relația f(x-1) + f(1) = 3x-8, oricare xe R. Calculaţi: a) f(1); b) f(2); c) ƒ(3); d) ƒ(4); e) ƒ(5); ƒ(-2).​

Answer 1

$f: \mathbb{R} \to \mathbb{R}, \ \ f(x - 1 ) + f(1) = 3x - 8$

• a) x - 1 = 1 ⇒ x = 2

$f(2 - 1 ) + f(1) = 3 \cdot 2 - 8$

$f(1) + f(1) = 6 - 8$

$2 \cdot f(1) = -2 \implies \boxed{f(1) = -1}$

• b) x - 1 = 2 ⇒ x = 3

$f(3 - 1 ) + f(1) = 3 \cdot 3 - 8$

$f(2) -1 = 9 - 8 \implies \boxed{ f(2) = 2}$

• c) x - 1 = 3 ⇒ x = 4

$f(4 - 1 ) + f(1) = 3 \cdot 4 - 8$

$f(3) - 1 = 12 - 8 \implies \boxed{f(3) = 5}$

• d) x - 1 = 4 ⇒ x = 5

$f(5 - 1 ) + f(1) = 3 \cdot 5 - 8$

$f(4) - 1 = 15 - 8 \implies \boxed{f(4) = 8}$

• e) x - 1 = 5 ⇒ x = 6

$f(6 - 1 ) + f(1) = 3 \cdot 6 - 8$

$f(5) - 1 = 18 - 8 \implies \boxed{f(5) = 11}$

• f) x - 1 = -2 ⇒ x = -1

$f(-1 - 1 ) + f(1) = 3 \cdot (-1) - 8$

$f(-2) - 1 = - 3 - 8 \implies \boxed{f(-2) = -10}$