Question

sabendo que f(x)=8 elevado a x -2 elevado a x+1 e g(x)=-4 elevado a X,para quais valores de x temos f(x)+g(x)>0

Answer 1

Olá Gabriel,


Organizando as funções:


$\mathsf{f(x)=8^{x} -2^{x+1}}\\\mathsf{g(x)=-4^{x}}\\\\\mathsf{f(x)+g(x)\ \textgreater \ 0}\\\\\\\mathsf{8^x-2^{x+1}+(-4^x)\ \textgreater \ 0}\\\mathsf{(2^3)^x-2^x.2-(2^2)^x\ \textgreater \ 0}\\\mathsf{2^{3x}-2^x.2-2^{2x}\ \textgreater \ 0}\\\\\\\mathsf{Chamaremos~de~(y)~o~(2^x)~para~facilitar~o~calculo}\\\\\mathsf{y^3-2y-y^2\ \textgreater \ 0}\\\\\mathsf{Organizando:}\\\\\mathsf{y^3-y^2-2y\ \textgreater \ 0}\\\mathsf{y.(y^2-y-2)\ \textgreater \ 0}\\\mathsf{y^2-y-2\ \textgreater \ 0}\\\\\\\mathsf{\Delta=b^2-4.a.c}\\\mathsf{\Delta=(-1)^2-4.1.(-2)}\\\mathsf{\Delta=1+8}\\\mathsf{\Delta=9}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}$

$\mathsf{y=\dfrac{-(-1)\pm\sqrt{9}}{2.1}}\\\\\\\mathsf{y=\dfrac{1+3}{2}}\\\\\mathsf{y=\dfrac{4}{2}}\\\\\boxed{\mathsf{y=2}}\\\\\\\mathsf{y'=\dfrac{1-3}{2}}\\\\\mathsf{y'=\dfrac{-2}{2}}\\\\\boxed{\mathsf{y'=-1}}~\mathsf{\gets~N\~ao~serve~pois~y~n\~ao~pode~ser~negativo}$

$\mathsf{y\ \textgreater \ 2}\\\\\mathsf{y=2^x}\\\\\mathsf{2^x\ \textgreater \ 2}\\\\\mathsf{x\ \textgreater \ 1}\\\\\\\boxed{\mathsf{S=\{x\in \mathbb{R}~|~x\ \textgreater \ 1\}}}$

Dúvidas? comente