Question

Please help me all the tutor can’t answer my question if you can’t answer don’t accept

Answer 1

Let's determine the compund functions, firts.

1) already solved.

2) h(m(x)):

$h(m(x))=h(x^2-4)=\frac{1}{\sqrt[]{(x^2-4)^{}}}$

3) m(h(x)):

$m(h(x))=m(\frac{1}{\sqrt[]{x}})=(\frac{1}{\sqrt[]{x}})^2-4$

Let's work with equation 2. We need to simplify it.

$\begin{gathered} \frac{1}{\sqrt[]{(x^2-4)^{}}}\cdot\frac{\sqrt{x^2-4}}{\sqrt{x^2-4}} \\ =\frac{1\cdot\sqrt{x^2-4}}{\sqrt{x^2-4}\sqrt{x^2-4}} \\ =\frac{\sqrt{x^2-4}}{x^2-4} \end{gathered}$

Now, let's stablish the conditions to find its domain:

Cond. 1)

$x^2-4\ne0$

Cond. 2)

$x^2-4>0$

condition 2 implicitly includes condition 1, then we will work from it

$\begin{gathered} x^2-4>0 \\ x^2-4+4>0+4 \\ x^2>4 \\ x<-\sqrt{4}\quad \mathrm{or}\quad \: x>\sqrt{4} \\ x<-2\quad \mathrm{or}\quad \: x>2 \end{gathered}$

in interval notation:

$\mleft(-\infty\: ,\: -2\mright)\cup\mleft(2,\: \infty\: \mright)$

Now, let's work with the 3rd compound function

$(\frac{1}{\sqrt[]{x}})^2-4$

In this case, x must be greater than and different from zero. Therefore, the domain will be:

$x>0$

in interval notation:

$\mleft(0,\: \infty\: \mright)$