Question

ME AJUDEM PFVR
20 Pontos

2. Achar o domínio das funções: f • g e g • f

a) f(x) = x^2 + 1
g(x) = 3^√x

b) f(x) = 4/x
g(x) = x^3 + 2x

Answer 1

item a)

$\text{f(x)} = \text x^2+1 \\\\ \text {g(x)}=\sqrt[3]{\text x} \\\\ \text{D} : \text f \ \text o \ \text g \to \text{D} : \text{f(g(x)) } \\\\ \therefore \\\\ \text {f(g(x))} = (\sqrt[3]{\text x})^2+1 \\\\ \text{f(g(x)) }=\sqrt[3]{\text x^2} +1 }$

Para qualquer valor de x a função pertence aos reais, portanto :

$\huge\boxed{\text{D} = -\infty <\text x < \infty \ }}\checkmark \\\\\\ \text{ou} \\\\\\ \huge\boxed{\text {D} = (-\infty,\infty ) }$

Agora façamos a  $\huge{\text g \ \text o \ \text f}$ :  

$\text D : \text g \ \text o \ \text f = \text D : \text{g(f(x))} \\\\ \text{g(f(x))} = \sqrt[3]{\text x^2+1}$

Para qualquer valor de x a função pertence aos reais, portanto :

$\huge\boxed{\text{D} = -\infty <\text x < \infty \ }}\checkmark \\\\\\ \text{ou} \\\\\\ \huge\boxed{\text {D} = (-\infty,\infty ) }$

item b)

$\displaystyle \text{f(x)} = \frac{4}{\text x} \ \ ; \ \ \text {g(x)} = \text x^3+2\text x\\\\\\ \text f \ \text o \ \text g = \text{f(g(x) )} \\\\\\ \text{f(g(x))} = \frac{4}{\text x^3+2\text x}$

O denominador não pode ser 0, ou seja :

$\text x^3+2\text x \neq 0 \to \text x(\text x^2+2)\neq 0 \to \text x \neq 0$

Portanto :

$\huge\boxed{\text D = \mathbb{R} - \{ 0\}}\checkmark$

OU

$\huge\boxed{\text D : (-\infty,0)\ \cup\ (0,\infty)\ }\checkmark$

(Domínio igual aos reais menos o 0)

Agora façamos : $\huge{\text g \ \text o \ \text f}$

$\displaystyle \text{g o f = g(f(x))} \\\\ \text{g(f(x)) } = [\text{f(x)}]^3+2[\text{f(x)}] \\\\\\ \text{g(f(x))}= (\frac{4}{\text x})^3+2\frac{4}{\text x} \\\\\\ \text{g(f(x)) } = \frac{64}{\text x^3}+\frac{8}{\text x} \\\\\\ \text{g(f(x))}=\frac{8\text x^3+64\text x}{\text x^4}$

Novamente o denominador não pode ser 0, ou seja :

$\huge\boxed{\text D = \mathbb{R} - \{ 0\}}\checkmark$

OU

$\huge\boxed{\text D : (-\infty,0)\ \cup\ (0,\infty)\ }\checkmark$