1) Let's begin with that by finding the composite functions.
2) So, let's begin with the ones to the left:
[tex]\begin{gathered} (f(g(x))=\frac{1}{2\cdot\frac{1}{2x}}=\frac{1}{\frac{1}{x}}=x \\ \\ (g(f(x))=\frac{1}{2\cdot\frac{1}{2x}}=\frac{1}{\frac{1}{x}}=x \\ ----- \\ \\ f^(x)=\frac{1}{2x} \\ \\ y=\frac{1}{2x} \\ \\ x=\frac{1}{2y} \\ \\ 2yx=1 \\ \\ 2y=\frac{1}{x} \\ \\ y=\frac{1}{x}\div\text{ 2} \\ \\ y=\frac{1}{2x} \end{gathered}[/tex]
Hence, we can tell that:
3) Let's now proceed with that one on the right:
[tex]\begin{gathered} f\left(g\left(x\right)\right)=x+3+3\Rightarrow f(g(x))=x+6 \\ \\ g(f(x))=(x+3)+3\Rightarrow g(f(x))=x+6 \end{gathered}[/tex]
And now the inverse:
[tex]\begin{gathered} f(x)=x+3 \\ \\ y=x+3 \\ \\ x=y+3 \\ \\ x-y=3 \\ \\ -y=3-x \\ \\ y=x-3 \\ \\ f^{-1}(x)=x-3 \end{gathered}[/tex]
And then, the answer is:
