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

Please help me all the tutor can’t answer my question if you can’t answer don’t accept-example-1
User Tayler
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1 Answer

4 votes

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(√(x^2-4))/(√(x^2-4)) \\ =(1\cdot√(x^2-4))/(√(x^2-4)√(x^2-4)) \\ =(√(x^2-4))/(x^2-4) \end{gathered}

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

Cond. 1)


x^2-4\\e0

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<-√(4)\quad \mathrm{or}\quad \: x>√(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)

User Aamir Shah
by
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