Answer
$(f\circ g)(x)=2sec(\frac{1}{2}x)$ and $(g\circ f)(x)=sec(x)$.
See graph.
Work Step by Step
1. Given $f(x)=2sec(x)$ and $g(x)=\frac{1}{2}x$, we have $(f\circ g)(x)=2sec(\frac{1}{2}x)$ and $(g\circ f)(x)=\frac{1}{2}(2sec(x))=sec(x)$.
2. See graph for $(f\circ g)(x$ (red) and $(g\circ f)(x)$ (blue).
