Question

determine o dominio das seguintes funções
A) $f(x)=x^{3}+x$
B)$f(x)=\frac{2x-1}{3x+4}$
c)$f(x)=\sqrt{-3x+15}$
d)$f(x)=\frac{x+1}{\sqrt{4x+4} }$
E)$f(x)=x^{2} -3x+2$
f)$f(x)=\sqrt{x-6}$
G)$f(x)=\sqrt[3]{x^{2} -1}$
H)$f(x)=\frac{-3x+1}{7x+10}$
i)$f(x)\sqrt[3]{x}$
j)$f(x)=\sqrt{x}$
k)$f(x)=x^{2} -12$

Answer 1

$\large\boxed{\begin{array}{l}\tt a)~\sf f(x)=x^3+x\\\sf Como~n\tilde ao~h\acute a~restric_{\!\!,}\tilde ao,o~dom\acute inio\\\sf \acute e~toda~a~reta~real, ou~seja,\\\sf Df(x)=\{x\in\mathbb{R}\}\\\tt b)~\sf f(x)=\dfrac{2x-1}{3x+4}\\\sf para~que~esta~func_{\!\!,}\tilde ao~fac_{\!\!,}a~sentido\\\sf \acute e~preciso~que~o~denominador~seja~n\tilde ao~nulo.\\\sf portanto~3x+4\ne0\\\sf 3x\ne-4\\\sf x\ne=-\dfrac{4}{3}\\\sf DA\acute I\\\sf Df(x)=\bigg\{ x\in\mathbb{R}/ x\ne-\dfrac{4}{3}\bigg\}\end{array}}$$\large\boxed{\begin{array}{l}\tt b)~\sf f(x)=\sqrt{-3x+15}\\\sf para~que~a~func_{\!\!,}\tilde ao~fac_{\!\!,}a~sentido\\\sf \acute e~preciso~que~o~radicando\\\sf seja~positivo~ou~nulo.\\\sf desse~modo\\\sf -3x+15\geqslant0\\\sf 3x\geqslant15\\\sf x\geqslant\dfrac{15}{3}\\\sf x\geqslant5\\\sf DA\acute I\\\sf Df(x)=\{x\in\mathbb{R}/x\geqslant5\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf f(x)=\dfrac{x+1}{\sqrt{4x+4}}\\\sf Aqui~devemos~garantir~que\\\sf o~denominador~n\tilde ao~se~anule\\\sf e~o~radicando~seja~positivo\\\sf portanto\\\sf4x+4>0\\\sf 4x>-4\\\sf x>-\dfrac{4}{4}\\\sf x>-1\\\sf Df(x)=\{x\in\mathbb{R}/x>-1\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt e)~\sf f(x)=x^2-3x+2\\\sf Aqui~n\tilde ao~h\acute a~restric_{\!\!,}\tilde ao,logo~o~dom\acute inio\\\sf \acute e~toda~reta~real\\\sf Df(x)=\{x\in\mathbb{R}\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt f)~\sf f(x)=\sqrt{x-6}\\\sf Aqui~devemos~garantir~que\\\sf o~radicando~seja~positivo~ou~nulo.\\\sf x-6\geqslant0\\\sf x\geqslant6\\\sf Df(x)=\{x\in\mathbb{R}/x\geqslant6\}\\\tt g)~\sf f(x)=\sqrt[\sf3]{\sf x^2-1}\\\sf Aqui~n\tilde ao~h\acute a~restric_{\!\!.}\tilde ao~portanto\\\sf Df(x)=\{x\in\mathbb{R}\}|\end{array}}$

$\large\boxed{\begin{array}{l}\tt h)~\sf f(x)=\dfrac{-3x+1}{7x+10}\\\sf Aqui~devemos~garantir~que~o~denominador\\\sf n\tilde ao~se~anule\\\sf logo\\\sf 7x+10\ne0\\\sf 7x\ne-10\\\sf x\ne-\dfrac{10}{7}\\\\\sf Df(x)=\bigg\{x\in\mathbb{R}/x\ne-\dfrac{10}{7}\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt i)~\sf f(x)=\sqrt[\sf3]{\sf x}\\\sf Aqui~n\tilde ao~h\acute a~restric_{\!\!,}\tilde ao~portanto\\\sf Df(x)=\{x\in\mathbb{R}\}\\\tt j)~\sf f(x)=\sqrt{x}\\\sf Aqui~devemos~garantir~que~o~radicando\\\sf seja~positivo~ou~nulo.\\\sf portanto\\\sf Df(x)=\{x\in\mathbb{R}/x\geqslant0\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt k)~\sf f(x)=x^2-12\\\sf Aqui~n\tilde ao~h\acute a~restric_{\!\!.}\tilde ao~portanto\\\sf Df(x)=\{x\in\mathbb{R}\}\end{array}}$