Question

Se f(x) = $\sqrt{a-x^{2} }$ ,g(x)= $\sqrt{b-x}$ e f(g(2)) = 2, então f(g)(0)):

a) $\sqrt{3}$
b) 2
c) $\sqrt{2}$
d) 3
e) 1

Answer 1

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

$\sf f(g(x))~=~\sqrt{a~-~\big(g(x)\big)^2}\\\\\\\sf f(g(x))~=~\sqrt{a~-~\left(\sqrt{b~-~x}~\right)^2}\\\\\\\sf f(g(x))~=~\sqrt{a~-~\left(\sqrt[\not2]{b~-~x}~\right)^{\not2}}\\\\\\\sf f(g(x))~=~\sqrt{a~-~(b~-~x)}\\\\\\\boxed{~\sf \sf f(g(x))~=~\sqrt{a~-~b~+~x}~}$

O enunciado nos diz que f(g(2))=2, portanto:

$\sf f(g(2))~=~2~~\longrightarrow~~\sqrt{a~-~b~+~2}~=~2\\\\\\\sqrt{a~-~b~+~2}~=~2\\\\\\Elevando~os~dois~lados~da~equacao~ao~quadrado\\\\\\\sqrt{a~-~b~+~2}^{~2}~=~2^2\\\\\\a~-~b~+~2~=~4\\\\\\a~-~b~=~4~-~2\\\\\\\boxed{\sf a~-~b~=~2}$

Podemos agora determinar o que nos foi solicitado, f(g(0)):

$\sf f(g(0))~=~\sqrt{a~-~b~+~0}\\\\\\\sf f(g(0))~=~\sqrt{a~-~b}\\\\\\\boxed{\sf f(g(0))~=~\sqrt{2}}~~\Rightarrow~Letra ~C$

$\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$