To find g o f(x), we need to replace each x in g(x) for f(x), this is:
![\begin{gathered} g\circ f(x)=g(f(x)) \\ g\circ f(x)=\sqrt[]{-(f(x))-2} \\ g\circ f(x)=\sqrt[]{-(2x-3)-2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ycnj8i5xq4uy4tzpsue39pyn8eriyb9puj.png)
Its domain must contain the numbers that make the function be real. In this case the function can be indefinite when having a negative root, it means that the domain of this function is:

Do the same procedure to find f o g(x):
![\begin{gathered} f\circ g(x)=f(g(x)) \\ f\circ g(x)=2\cdot(g(x))-3 \\ f\circ g(x)=2\cdot(\sqrt[]{-x-2})-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/6hjg5ayde9ajstry3rxg9gyk5at6qpjqt5.png)
The domain of this function is:
