College Algebra, Chapter 3, 3.4, Section 3.4, Problem 16

A function $\displaystyle f(x) = 4 - x^2$. Determine the average rate of change of the function between $x = 1$ and $x = 1 + h$.


$
\begin{equation}
\begin{aligned}

\text{average rate of change } =& \frac{f(b) - f(a)}{b - a}
&& \text{Model}
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\text{average rate of change } =& \frac{f(1 + h) - f(1)}{1 + h - 1}
&& \text{Substitute } a = 1 \text{ and } b = 1 + h
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\text{average rate of change } =& \frac{4 - (1 + h)^2 - [4 - (1)^2] }{h}
&& \text{Simplify}
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\text{average rate of change } =& \frac{4 - (1 + 2h + h^2) - (3)}{h}
&& \text{Apply Distributive Property}
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\text{average rate of change } =& \frac{4 - 1 - 2h - h^2 - 3}{h}
&& \text{Combine like terms}
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\text{average rate of change } =& \frac{-2h - h^2}{h}
&& \text{Factor $h$ from each term}
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\text{average rate of change } =& \frac{\cancel{h} (-2 - h)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
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\text{average rate of change } =& -2 - h
&& \text{Answer}

\end{aligned}
\end{equation}
$