College Algebra, Chapter 3, 3.4, Section 3.4, Problem 20

A function $\displaystyle f(t) = \sqrt{t}$. Determine the average rate of change of the function between $t = a$ and $t = a + h$.


$
\begin{equation}
\begin{aligned}

\text{average rate of change } =& \frac{f(b) - f(a)}{b - a}
&& \text{Model}
\\
\\
\text{average rate of change } =& \frac{f(a + h) - f(a)}{a + h - a}
&& \text{Substitute } a = a \text{ and } b = a + h
\\
\\
\text{average rate of change } =& \frac{\sqrt{a + h} - \sqrt{a}}{h}
&& \text{Simplify}
\\
\\
\text{average rate of change } =& \frac{\sqrt{a + h} - \sqrt{a}}{h} \cdot \frac{\sqrt{a + h} + \sqrt{a}}{\sqrt{a + h} + \sqrt{a}}
&& \text{Multiply numerator and denominator by conjugate radical}
\\
\\
\text{average rate of change } =& \frac{a + h - a}{h (\sqrt{a + h} + \sqrt{a})}
&& \text{Combine like terms}
\\
\\
\text{average rate of change } =& \frac{\cancel{h}}{\cancel{h} (\sqrt{a + h} + \sqrt{a})}
&& \text{Cancel out like terms}
\\
\\
\text{average rate of change } =& \frac{1}{\sqrt{a + h} + \sqrt{a}}
&& \text{Answer}

\end{aligned}
\end{equation}
$