Calculus and Its Applications, Chapter 1, 1.3, Section 1.3, Problem 14

For the function $\displaystyle f(x) = x^2 - 4x$
(a) Determine the simplified form of the difference quotient
(b) Complete the table.

a.) For $\displaystyle f(x) = x^2 - 4x$

$
\begin{equation}
\begin{aligned}
f(x+h) &= (x + h)^2 - 4(x + h)\\
\\
&= x^2 + 2xh + h^2 - 4x - 4h
\end{aligned}
\end{equation}
$


Then,

$
\begin{equation}
\begin{aligned}
f(x + h) - f(x) &= x^2 + 2xh + h^2 - 4x - 4h - (x^2 - 4x)\\
\\
&= x^2 + 2xh + h^2 - 4x - 4h - x^2 + 4x\\
\\
&= 2xh + h^2 - 4h
\end{aligned}
\end{equation}
$


Thus,

$
\begin{equation}
\begin{aligned}
\frac{f(x +h)- f(x)}{h} &= \frac{2xh + h^2 - 4h}{h}\\
\\
&= \frac{h(2x + h - 4)}{h}\\
\\
&= 2x + h - 4
\end{aligned}
\end{equation}
$


b.)


$
\begin{array}{|c|c|c|}
\hline
x & h & \displaystyle \frac{f(x+h)-f(x)}{h} \\
\hline
5 & 2 & 8\\
\hline
5 & 1 & 7 \\
\hline
5 & 0.1 & 6.1 \\
\hline
5 & 0.01 & 6.01 \\
\hline
\end{array}
$

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