Question

Na função ƒ(x) = 2x – x, o valor de fof(0) + fof(1) + fof(2) + fof(3) é:

Answer 1

f(x) = 2x - x
fof(0)= 2.(2.0 - 0) - 0
fof(0) = 0
fof(1)= 2.(2.1 - 1) - 1  >>>> 2.(2 - 1) -1 >>>> 2.1 - 1 >>>> 2 - 1
fof(1)= 1
fof(2)= 2.(2.2 - 2) - 2 >>>> 2.(4 - 2) - 2 >>>> 2.2 -2 >>>> 4 - 2
fof(2)= 2
fof(3)= 2.(2.3 - 3) - 3 >>>> 2.(6 - 3) -3 >>>> 2.3 - 3 >>>> 6 - 3
fof(3)= 3

fof(0) +fof(1) + fof(2) + fof(3) = 0 + 1 + 2 + 3 = 6

Answer 2

Resposta:

30

Explicação passo-a-passo:

f(f(0)+f(f(1))+f(f(2))+f(f(3))

e f(x)=2x-x

resolvemos o de dentro pra resolvermos o de fora

assim

f(f(0)) = 2(2.0-0)-0 => (4.0-0)-0 => 0 = 0

f(f(1)) = 2(2.1-1)-1 => (4.2-2)-1 => (8-2)-1 = 5

f(f(2)) = 2(2.2-2)-2 => (4.4-4)-2 => (16-24)-2 = 10

f(f(3)) = 2(2.3-3)-3 => (4.6-6)-3 => (24-6)-3 = 15

soma tudo

0+5+10+15 = 30