Respuesta :

Alfpfeu

Answer:

Step-by-step explanation:

Hello,

[tex](goh)(x)=g(h(x))=2\left( \sqrt{x^2+1}\right)^2=2(x^2+1)\\\\x=(goh)((goh)^{-1}(x))=2((goh)^{-1}(x)^2+1)\\ \\\left((goh)^{-1}(x)\right)^2=\dfrac{x}{2}-1=\dfrac{x-2}{2}\\ \\(goh)^{-1}(x)=\sqrt{\dfrac{x-2}{2}}[/tex]

And this is a function defined for x-2 [tex]\geq[/tex] 0, meaning x [tex]\geq[/tex] 2

Thanks

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