we have the functions
![\begin{gathered} f(x)=(1)/(x-2) \\ \\ g(x)=√(x+2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pxay7wmhqjx76hcyveygfp2gp7k0ieeopr.png)
Find out f(g(x))
![f\mleft(g\mleft(x\mright)\mright)=(1)/(√(x+2)-2)](https://img.qammunity.org/2023/formulas/mathematics/college/t4xf1kibwj4cl3k31oithuown1m1xlotsf.png)
Remember that
The radicand must be greater than or equal to zero and the denominator cannot be equal to zero
so
step 1
Solve the inequality
![\begin{gathered} x+2\ge0 \\ x\operatorname{\ge}-2 \end{gathered}]()
the solution to the first inequality is the interval [-2, infinite)
step 2
Solve the equation
![\begin{gathered} √(x+2)-2\\e0 \\ √(x+2)\operatorname{\\e}2 \\ therefore \\ x\operatorname{\\e}2 \end{gathered}]()
The domain is the interval
[–2, 2) ∪ (2, ∞)
The answer is the option C