Question

Sendo F e G funçoes de dominio Real com F(x)= x² + 2x e G(x)=1-3x, determine:
G[ F(x) ]=?
F[ F (x) ]=?
G[ G (x) ]=?

Answer 1

Calculando g(f(x)):

$g(x)=1-3x\\g(f(x))=1-3f(x)\\g(f(x))=1-3(x^{2}+2x)\\g(f(x))=1-3x^{2}-6x\\g(f(x))=-3x^{2}-6x+1$

Calculando f(f(x)):

$f(x)=x^{2}+2x\\f(f(x))=[f(x)]^{2}+2f(x)\\f(f(x))=[x^{2}+2x]^{2}+2(x^{2}+2x)\\f(f(x))=[(x^{2})^{2}+2*x^{2}*2x+(2x)^{2}]+2x^{2}+4x\\f(f(x))=x^{4}+4x^{3}+4x^{2}+2x^{2}+4x\\f(f(x))=x^{4}+4x^{3}+6x^{2}+4x$

Calculando g(g(x)):

$g(x)=1-3x\\g(g(x))=1-3g(x)\\g(g(x))=1-3(1-3x)\\g(g(x))=1-3+9x\\g(g(x))=9x-2$