1.2 Rates of change and behavior of graphs (Page 2/15)

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How To

Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values $x_{1}$ and $x_{2} .$

Calculate the difference $y_{2} - y_{1} = Δ y .$
Calculate the difference $x_{2} - x_{1} = Δ x .$
Find the ratio $\frac{Δ y}{Δ x} .$

Computing an average rate of change

Using the data in [link] , find the average rate of change of the price of gasoline between 2007 and 2009.

In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is

\begin{array}{l} \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{$ 2.41 - $ 2.84}{2009 - 2007} \\ = \frac{- $ 0.43}{2 years} \\ = - $ 0.22 per year \end{array}

Try It

Using the data in [link] , find the average rate of change between 2005 and 2010.

$\frac{$ 2.84 - $ 2.31}{5 years} = \frac{$ 0.53}{5 years} = $ 0.106$ per year.

Computing average rate of change from a graph

Given the function $g (t)$ shown in [link] , find the average rate of change on the interval $[- 1, 2] .$

At $t = - 1,$ [link] shows $g (−1) = 4.$ At $t = 2,$ the graph shows $g (2) = 1.$

Graph of a parabola with a line from points (-1, 4) and (2, 1) to show the changes for g(t) and t.

The horizontal change $Δ t = 3$ is shown by the red arrow, and the vertical change $Δ g (t) = - 3$ is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

\frac{1 - 4}{2 - (- 1)} = \frac{- 3}{3} = −1

Computing average rate of change from a table

After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in [link] . Find her average speed over the first 6 hours.

t (hours)	0	1	2	3	4	5	6	7
D ( t ) (miles)	10	55	90	153	214	240	282	300

Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of

\begin{array}{l} \begin{array}{l} \frac{292 - 10}{6 - 0} = \frac{282}{6} \end{array} \\ = 47 \end{array}

The average speed is 47 miles per hour.

Computing average rate of change for a function expressed as a formula

Compute the average rate of change of $f (x) = x^{2} - \frac{1}{x}$ on the interval $[2, 4].$

We can start by computing the function values at each endpoint of the interval.

\begin{array}{l} f (2) = 2^{2} - \frac{1}{2} \begin{matrix}  \end{matrix} & f (4) = 4^{2} - \frac{1}{4} \\ = 4 - \frac{1}{2} & = 16 - \frac{1}{4} \\ = \frac{7}{2} & = \frac{63}{4} \end{array}

Now we compute the average rate of change.

\begin{array}{l} Average rate of change = \frac{f (4) - f (2)}{4 - 2} \\ = \frac{\frac{63}{4} - \frac{7}{2}}{4 - 2} \\ ​ = \frac{\frac{49}{4}}{2} \\ = \frac{49}{8} \end{array}

Try It

Find the average rate of change of $f (x) = x - 2 \sqrt{x}$ on the interval $[1, 9] .$

$\frac{1}{2}$

Finding the average rate of change of a force

The electrostatic force $F,$ measured in newtons, between two charged particles can be related to the distance between the particles $d,$ in centimeters, by the formula $F (d) = \frac{2}{d^{2}} .$ Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm.

We are computing the average rate of change of $F (d) = \frac{2}{d^{2}}$ on the interval $[2, 6] .$

\begin{array}{l} Average rate of change = \frac{F (6) - F (2)}{6 - 2} \\ = \frac{\frac{2}{6^{2}} - \frac{2}{2^{2}}}{6 - 2} & Simplify . \\ = \frac{\frac{2}{36} - \frac{2}{4}}{4} \\ = \frac{- \frac{16}{36}}{4} & Combine numerator terms . \\ = - \frac{1}{9} & Simplify \end{array}

The average rate of change is $- \frac{1}{9}$ newton per centimeter.

Finding an average rate of change as an expression

Find the average rate of change of $g (t) = t^{2} + 3 t + 1$ on the interval $[0, a] .$ The answer will be an expression involving $a .$

We use the average rate of change formula.

\begin{array}{l} \begin{array}{l} Average rate of change = \frac{g (a) - g (0)}{a - 0} & Evaluate . \\ = \frac{(a^{2} + 3 a + 1) - (0^{2} + 3 (0) + 1)}{a - 0} & Simplify . \\ = \frac{a^{2} + 3 a + 1 - 1}{a} & Simplify and factor . \\ = \frac{a (a + 3)}{a} & \begin{array}{l} Divide by the common factor a . \end{array} \end{array} \\ = a + 3 \end{array}

This result tells us the average rate of change in terms of $a$ between $t = 0$ and any other point $t = a .$ For example, on the interval $[0, 5],$ the average rate of change would be $5 + 3 = 8.$

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Source: OpenStax, Essential precalculus, part 1. OpenStax CNX. Aug 26, 2015 Download for free at http://legacy.cnx.org/content/col11871/1.1

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