Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 26

Evaluate the difference quotient $\displaystyle \frac{f(x) - f(1)}{x - 1}$ for the function $\displaystyle f(x) = \frac{x + 3}{x + 1}$


$
\begin{equation}
\begin{aligned}


\displaystyle \frac{f(x) - f(1)}{x - 1} &= \frac{\displaystyle \frac{x + 3}{x + 1} - \left(\frac{1 + 3 }{1 + 1} \right)}{x - 1}
&& (\text{ Substitute $f(x)$ and $f(1)$ to the function $f(x)$, then divide it by $(x-1)$ })\\

\\

&= \frac{\displaystyle \frac{x + 3}{x + 1} - 2}{x - 1}
&&( \text{ Get the LCD of the numerator})\\
\\

&= \frac{x + 3 - 2x - 2}{(x + 1)(x - 1)}
&&(\text{ Combine like terms})\\

\\
&= \frac{-x + 1}{(x + 1)( x - 1)}
&& (\text{ Factor -1 in the numerator})\\
\\

&= \frac{-1 \cancel{(x -1)}}{(x + 1)\cancel{(x - 1)}}
&& (\text{ Cancel out like terms})\\
\\
&= \frac{-1}{x + 1}

\end{aligned}
\end{equation}
$