Answer
$f(\frac{1}{x+3})= \frac{x+4}{x+2}$.
Work Step by Step
The given function is
$\Rightarrow f(x)=\frac{1+x}{1-x}$
Replace $x$ with $\frac{1}{x+3}$.
$\Rightarrow f(\frac{1}{x+3})=\frac{1+\frac{1}{x+3}}{1-\frac{1}{x+3}}$
Multiply the numerator and the denominator by $(x+3)$.
$\Rightarrow f(\frac{1}{x+3})= \frac{(x+3)\left (1+\frac{1}{x+3}\right )}{(x+3)\left (1-\frac{1}{x+3}\right )}$
Use the distributive property.
$\Rightarrow f(\frac{1}{x+3})= \frac{1(x+3)+\frac{1(x+3)}{x+3}}{1(x+3)-\frac{1(x+3)}{x+3}}$
Simplify.
$\Rightarrow f(\frac{1}{x+3})= \frac{x+3+1}{x+3-1}$
$\Rightarrow f(\frac{1}{x+3})= \frac{x+4}{x+2}$.
