Calculus and Its Applications, Chapter 1, 1.3, Section 1.3, Problem 8

For the function $\displaystyle f(x) = \frac{9}{x}$
(a) Determine the simplified form of the difference quotient
(b) Complete the table.

a.) For $\displaystyle f(x) = \frac{9}{x}$
$\displaystyle f(x + h) = \frac{9}{x + h}$
Then,

$
\begin{equation}
\begin{aligned}
f(x + h) - f(x) &= \frac{9}{x + h} - \frac{9}{x}\\
\\
&= \frac{9x - 9 (x + h)}{x(x +h)}\\
\\
&= \frac{9x - 9x - 9h}{x^2 + xh}\\
\\
&= \frac{-9h}{x^2 + xh}
\end{aligned}
\end{equation}
$


Thus,
$\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{\frac{-9h}{x^2 + xh}}{h} = \frac{-9h}{h(x^2 + xh)} = \frac{-9}{x^2 + xh}$

b.)


$
\begin{array}{|c|c|c|}
\hline
x & h & \displaystyle \frac{f(x+h)-f(x)}{h} \\
\hline
5 & 2 & -0.2571 \\
\hline
5 & 1 & -0.30 \\
\hline
5 & 0.1 & -0.3529 \\
\hline
5 & 0.01 & -0.3593 \\
\hline
\end{array}
$