Question

sendo f(×)= x^2-2×+1 g(x)= 2x+1 calcule:
f(f(x)) f(g(x))

Answer 1

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


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