Respuesta :
Given the functions:
[tex]\begin{gathered} g(x)=\frac{1}{\sqrt[]{x}} \\ m(x)=x^2-4 \end{gathered}[/tex]I would like to find their domain as well and then complete the answers:
[tex]\begin{gathered} D_g=(0,\infty) \\ D_m=(-\infty,\infty) \end{gathered}[/tex]For the first question: g(x) / m(x)
[tex]\begin{gathered} \frac{g(x)}{m(x)}=\frac{\frac{1}{\sqrt[]{x}}}{x^2-4}=\frac{1}{\sqrt[]{x}\cdot(x^2-4)}=\frac{1}{x-4\sqrt[]{x}} \\ x-4\sqrt[]{x}\ne0 \\ x\ne4\sqrt[]{x} \\ x^2\ne4x \\ x\ne4 \end{gathered}[/tex]As we can see, the domain of this function cannot take negative values nor 4, 0. So, its domain is
[tex]D_{\frac{g}{m}}=(0,4)\cup(4,\infty)[/tex]For the second domain g(m(x)), let's find out what is the function:
[tex]\begin{gathered} g(m(x))=\frac{1}{\sqrt[]{x^2-4}} \\ \sqrt[]{x^2-4}>0 \\ x^2>4 \\ x>2 \\ x<-2 \end{gathered}[/tex]This means that x cannot be among the interval -2,2:
[tex]D_{g(m)}=(-\infty,-2)\cup(2,\infty)[/tex]For the last domain m(g(x)) we perfome the same procedure:
[tex]m(g(x))=(\frac{1}{\sqrt[]{x}})^2-4=\frac{1}{x}-4[/tex]For this domain it is obvious that x cannot take the zero value but anyone else.
[tex]D_{m(g)}=(-\infty,0)\cup(0,\infty)_{}[/tex]