## College Algebra (11th Edition)

$f$ and $g$ are inverses
The two functions are inverses if $(f\circ g)(x)=x$ and if $(g\circ f)(x)=x$. First, $f(g(x))=\frac{\frac{4x+3}{1-x}-3}{\frac{4x+3}{1-x}+4}$. After multiplying both the nominator and the denominator by $(1-x)$, we get: $\frac{(\frac{4x+3}{1-x}-3)\cdot (1-x)}{(\frac{4x+3}{1-x}+4) \cdot(1-x)}=\frac{4x+3-3\cdot1+(-3)\cdot(-x)}{4x+3+4\cdot1+4\cdot(-x)}=\frac{7x}{7}=x$. Second, $g(f(x))=\frac{4\frac{x-3}{x+4}+3}{1-\frac{x-3}{x+4}}$. After multiplying both the numerator and the denominator by $x+4$, we get $\frac{(4\frac{x-3}{x+4}+3)\cdot(x+4)}{(1-\frac{x-3}{x+4})\cdot(x+4)}=\frac{4\cdot x-3\cdot4+3\cdot x+3\cdot4}{x\cdot 1+4\cdot 1-x+3}=\frac{7x}{7}=x$ Therefore, $f$ and $g$ are inverses.