#### Answer

$\frac{f(x)-f(1)}{x-1} =\frac{-1}{(x+1)}$

#### Work Step by Step

$f(x) = \frac{x+3}{x+1}$
(*) $f(1) = \frac{1+3}{1+1}=\frac{4}{2}=2$
$\frac{f(x)-f(1)}{x-1} = \frac{\frac{x+3}{x+1}-2}{x-1}=\frac{\frac{x+3-2(x+1)}{x+1}}{x-1}=\frac{\frac{x+3-2x-2}{x+1}}{x-1}=\frac{x+3-2x-2}{(x+1)(x-1)}=\frac{-x+1}{(x+1)(x-1)}=\frac{-(x-1)}{(x+1)(x-1)}=\frac{-1}{(x+1)}$