Question

13. Seja f: R->R uma função tal que: f(x + 1) = 2f(x) – 5f(x + 1) e f(0) = 6. Calcule f(2). 

Answer 1

f(x+1)=2.f(x) - 5.f(x+1)
6.f(x+1)=2.f(x)
f(x+1) = f(x)/3

Pondo x = 0 , temos :

f(0+1)=f(0)/3
f(1)=6/3
f(1)=2

Pondo x=1 , temos :

f(1+1)=f(1)/3
f(2)=2/3

Answer 2

Acompanhe o raciocínio:

$f(x+1)=2f(x)-5f(x+1)\\ \\ 2f(x)=f(x+1)+5f(x+1)\\ \\ 2f(x)=6f(x+1)\\ \\ \boxed{f(x)=3f(x+1)}\\ \\ f(0)=3f(0+1)=6\\ \\ \boxed{f(1)=2}\\ \\ f(1)=3f(1+1)\\ \\ 3f(2)=2\\ \\ \boxed{\boxed{f(2)=\frac{2}{3}}}$