we have the functions
[tex]\begin{gathered} f(x)=\frac{1}{x-2} \\ \\ g(x)=\sqrt{x+2} \end{gathered}[/tex]Find out f(g(x))
[tex]f\mleft(g\mleft(x\mright)\mright)=\frac{1}{\sqrt{x+2}-2}[/tex]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
[tex]\begin{gathered} x+2\ge0 \\ x\operatorname{\ge}-2 \end{gathered}[/tex]the solution to the first inequality is the interval [-2, infinite)
step 2
Solve the equation
[tex]\begin{gathered} \sqrt{x+2}-2\ne0 \\ \sqrt{x+2}\operatorname{\ne}2 \\ therefore \\ x\operatorname{\ne}2 \end{gathered}[/tex]The domain is the interval
[–2, 2) ∪ (2, ∞)
