simplifying complex fractions when proving inverses with function composition? I'm working with two functions, $f(x)=\frac{x-3}{x+4}$ and $g(x)=\frac{4x+3}{1-x}$.
I need to simplify $f(g(x))$ and its opposite, but i'm not sure of the procedures regarding the more complicated forms like 
$$\frac{\frac{4x+3}{1-x}-3}{\frac{4x+3}{1-x}+4}$$
 A: Multiply that fraction by $\frac{1 - x}{1 - x}$ to simplify the fraction, simplify the numerator and denominator to binomials, then see if there are any common factors that can be canceled.
A: with the substitution $x \rightarrow g(x)$ the numerator of $f(x)$ becomes:
$$
x-3 \rightarrow \frac{4x+3}{1-x}-3=\frac{7x}{1-x}
$$
and the denominator becomes
$$
x+4 \rightarrow \frac{4x+3}{1-x}+4=\frac{7}{1-x}
$$
so
$$
f(g(x))= \frac{\frac{7x}{1-x}}{\frac{7}{1-x}}= x
$$
A: Keep in mind that when you add or subtract fractions, you must form a common denominator.  Therefore,
\begin{align*}
f(g(x)) & = f\left(\frac{4x + 3}{1 - x}\right)\\
        & = \frac{\dfrac{4x + 3}{1 - x} - 3}{\dfrac{4x + 3}{1 - x} + 4}\\
        & = \frac{\dfrac{4x + 3}{1 - x} - 3 \cdot \dfrac{1 - x}{1 - x}}{\dfrac{4x + 3}{1 - x} + 4 \cdot \dfrac{1 - x}{1 - x}}\\
        & = \frac{\dfrac{4x + 3}{1 - x} - \dfrac{3 - 3x}{1 - x}}{\dfrac{4x + 3}{1 - x} + \dfrac{4 - 4x}{1 - x}}\\
        & = \frac{\dfrac{7x}{1 - x}}{\dfrac{7}{1 - x}}\\
        & = \frac{7x}{1 - x} \cdot \frac{1 - x}{7}\\
        & = x
\end{align*}
provided that $x \neq 1$.  Notice that $1$ is not in the domain of 
$$g(x) = \frac{4x + 3}{1 - x}$$
so we have shown that $f(g(x)) = x$ for each $x$ in the domain of $g$.
To show that $f$ and $g$ are inverse functions, it remains for you to show that $g(f(x)) = x$ for each $x$ in the domain of $f$.
