Calculus and Its Applications, Chapter 1, 1.3, Section 1.3, Problem 54

Determine the simplified difference quotient of the function $\displaystyle f(x) = \sqrt{3 - 2x}$
For $\displaystyle f(x) = \sqrt{3 - 2x}$

$
\begin{equation}
\begin{aligned}
f(x + h) &= \sqrt{3 - 2 (x +h)}\\
\\
&= \sqrt{3 - 2x - 2h}
\end{aligned}
\end{equation}
$

Then,

$
\begin{equation}
\begin{aligned}
f(x +h) - f(x) &= \sqrt{3- 2x - 2h} - \sqrt{3 - 2x}
\end{aligned}
\end{equation}
$


Thus,

$
\begin{equation}
\begin{aligned}
\frac{f(x+h)-f(x)}{h} &= \frac{\sqrt{3- 2x - 2h} - \sqrt{3 - 2x}}{h}\\
\\
&= \frac{\sqrt{3-2x -2h} - \sqrt{3-2x}}{h} \left( \frac{\sqrt{3-2x-2h}+\sqrt{3-2x}}{h} \right)
&& \text{Multiply by 1 by using the conjugate of the numerator}\\
\\
&= \frac{3 -2 x - 2h - (3-2x)}{h ( \sqrt{3 - 2x - 2h} + \sqrt{3 - 2x}) }\\
\\
&= \frac{3-2x-2h+3+2x}{h( \sqrt{3 - 2x - 2h} + \sqrt{3 - 2x}}\\
\\
&= \frac{-2h}{h ( \sqrt{3 - 2x - 2h} + \sqrt{3 - 2x})}\\
\\
&= \frac{-2}{\sqrt{3-2x -2 h } + \sqrt{3 - 2x}}
\end{aligned}
\end{equation}
$