Question

2. Sejam as funções f(x) = x² – 2x + 1 e g(x) = 2x + 1, calcule:
a) f (g(1))
b) g (f(2))
c) f (f(1))

Answer 1

Explicação passo-a-passo:

A)

$f(g(x)) = {(2x + 1)}^{2} - 2(2x + 1) + 1 \\f(g(x)) =4 {x}^{2} + 4x + 1 - 4x - 2 + 1 \\ f(g(x)) =4 {x}^{2} \\ f(g(1)) =4. {1}^{2} = 4$

B)

$g(f(x)) =2( {x}^{2} - 2x + 1) + 1 \\ g(f(x)) =2 {x}^{2} - 4x + 2 + 1 \\ g(f(x)) =2 {x}^{2} - 4x + 3 \\ g(f(2)) =2. {2 }^{2} - 4.2 + 3 \\ g(f(2)) =8 - 8 + 3 = 3$

C)

$f(f(x)) = {( {x}^{2} - 2x + 1)}^{2} - 2( {x}^{2} - 2x + 1) + 1 \\ f(f(x)) = {x}^{4} - 2 {x}^{3} + {x}^{2} - 2 {x}^{3} + 4 {x}^{2} - 2x + {x}^{2} - 2x + 1 - 2 {x}^{2} + 4x - 2 + 1 \\ f(f(x)) = {x}^{4} - 4 {x}^{3} + 4 {x}^{2} \\ f(f(1)) = {1}^{4} - 4. {1}^{3} + 4. {1}^{2} \\ f(f(1)) =1 - 4 + 4 = 1$