Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 292: 41

See below.

Given $f(x)=\frac{2x+3}{x+4}$ and $g(x)=\frac{4x-3}{2-x}$, we have: 1. $f(g(x))=\frac{2(\frac{4x-3}{2-x})+3}{(\frac{4x-3}{2-x})+4}=\frac{8x-6+6-3x}{4x-3+8-4x}=x$, with $x\ne2$ 2. $g(f(x))=\frac{4(\frac{2x+3}{x+4})-3}{2-(\frac{2x+3}{x+4})}=\frac{8x+12-3x-12}{2x+8-2x-3}=x$, with $x\ne-4$