Answer:
A. H(x) is an inverse of F(x)
Step-by-step explanation:
The given functions are:
[tex]F(x)=\sqrt{x-2}[/tex]
[tex]G(x)=(x-2)^2[/tex]
[tex]H(x)=x^2+2[/tex]
We compose F(x) and G(x) to get:
[tex](F\circ G)(x)=F(G(x))[/tex]
[tex](F\circ G)(x)=F((x-2)^2)[/tex]
[tex](F\circ G)(x)=\sqrt{(x-2)^2-2}[/tex]
[tex](F\circ G)(x)=\sqrt{x^2-4x+4-2}[/tex]
[tex](F\circ G)(x)=\sqrt{x^2-4x+2}[/tex]
[tex](F\circ G)(x)\ne x[/tex]
Hence G(x) is not an inverse of F(x).
We now compose H(x) and G(x).
[tex](F\circ H)(x)=F(H(x))[/tex]
[tex](F\circ H)(x)=F(x^2+2)[/tex]
[tex](F\circ H)(x)=\sqrt{x^2+2-2}[/tex]
We simplify to get:
[tex](F\circ H)(x)=\sqrt{x^2}[/tex]
[tex](F\circ H)(x)=x[/tex]
Since [tex](F\circ H)(x)=x[/tex], H(x) is an inverse of F(x)
