Question

QUAL É O DOMÍNIO DA FUNÇÃO?​

Answer 1

$\boxed{\begin{array}{l}\sf f(x)=\dfrac{5x}{x^3-4x}-\dfrac{2}{\sqrt{x+1}}\\\sf supondo~x\ne0,vamos~simplificar~a~1^a~frac_{\!\!,}\tilde ao:\\\sf f(x)=\dfrac{5\backslash\!\!\!x}{\backslash\!\!\!x\cdot(x^2-4)}-\dfrac{2}{\sqrt{x+1}}\\\\\sf f(x)=\dfrac{5}{x^2-4}-\dfrac{2}{\sqrt{x+1}}\\\underline{\rm interpretemos~cada~parcela~de~f(x)}\\\underline{\rm como~novas~func_{\!\!,}\tilde oes}\\\sf fazendo~g(x)=\dfrac{5}{x^2-4}~~h(x)=\dfrac{2}{\sqrt{x+1}}\\\sf ent\tilde ao\\\sf f(x)=g(x)-h(x).\end{array}}$

$\boxed{\begin{array}{l}\sf g(x)=\dfrac{5}{x^2-4}\\\sf essa~func_{\!\!,}\tilde ao~s\acute o~existe~quando~o~denominador\\\sf \acute e~n\tilde ao~nulo.\\\sf portanto\\\sf x^2-4\ne0\\\sf x^2\ne4\\\sf x\ne\pm\sqrt{4}\\\sf x\ne\pm2\end{array}}$

$\boxed{\begin{array}{l}\sf h(x)=\dfrac{2}{\sqrt{x+1}}\\\sf devemos~garantir~que~o~denominador\\\sf n\tilde ao~se~anule\\\sf e~o~radicando~n\tilde ao~seja~negativo.\\\sf portanto\\\sf x+1>0\\\sf x>-1\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm fazendo~a~diferenc_{\!\!,}a~entre~g(x)~e~h(x)~temos:}\\\sf Dom~f(x)=\{x\in\mathbb{R}/-2<x<1\}\end{array}}$