Question

Dadas f(x)=4x-4 e g(x)=-2x2+x-1 calcule g(f(x))=0

Answer 1

Vamos começar determinando a função composta g(f(x)):

$g(f(x))~=~-2\,.\,\left(f(x)\right)^{\,2}~+~f(x)~-~1\\\\\\g(f(x))~=~-2\,.\,(4x-4)^2~+~(4x-4)~-~1\\\\\\g(f(x))~=~-2\,.\,(4^2.x^2-4\,.\,4x~-~4\,.\,4x~+~4^2)~+~(4x-5)\\\\\\g(f(x))~=~-2\,.\,(16x^2-16x-16x+16)~+~(4x-5)\\\\\\g(f(x))~=~-(2\,.\,16x^2-2\,.\,32x+2\,.\,16)~+~(4x-5)\\\\\\g(f(x))~=~-32x^2+64x-32+4x-5\\\\\\\boxed{g(f(x))~=~-32x^2+68x-37}$

Agora, utilizando Bhaskara, podemos determinar as raízes da função:

$\Delta~=~68^2-4.(-32).(-37)\\\\\\\Delta~=~4624~-~4736\\\\\\\boxed{\Delta~=~-112}$

Não há soluções Reais para a equação g(f(x))=0.

Caso já tenha conhecimento sobre números complexos, continue lendo, caso contrario, sua resposta é "Não há soluções Reais".

Nos complexos, teremos:

$x~=~\dfrac{-68\pm\sqrt{-112}}{2\,.\,(-32)}\\\\\\\\x~=~\dfrac{-68\pm\sqrt{-1~.~112}}{-64}\\\\\\\\x~=~\dfrac{-68\pm\sqrt{-1}~.~\sqrt{112}}{-64}\\\\\\\\x~=~\dfrac{-68\pm\,i\,.\,\sqrt{4^2~.~7}}{-64}\\\\\\\\x~=~\dfrac{-68\pm i\,.\,4\sqrt{7}}{-64}$

$\boxed{x~=~\dfrac{17}{16}~\pm~i\,\dfrac{\sqrt{7}}{16}}$