Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 23

Evaluate the difference quotient $\displaystyle \frac{f (3+h)-f(3)}{h}$ for the function $f(x) = 4 + 3x - x^2$


$
\begin{equation}
\begin{aligned}
\displaystyle \frac{f (3+h)-f(3)}{h} &= \frac{4+3(3+h)-(3+h)^2 - [4+3(3)-(3)^2]}{h}
&& ( \text{Substitute $f(3+h)$ and $f(3)$ to the function $f(x)$, then divide it by $h$} ) \\
\\
&= \frac{4+9+3h-[9+6h+h^2]-4-9+9}{h}
&&( \text{ Simplify the equation})\\
\\
&= \frac{3h-6h-h^2}{h}
&&( \text{ Combine like terms})\\
\\

&= \frac{-3h-h^2}{h} = \frac{\cancel{h}(-3 - h)}{\cancel{h}}
&&( \text{ Factor the numerator and cancel out like terms})\\
\\
& = -3 -h

\end{aligned}
\end{equation}
$