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

[latex]\begin{align}\frac{\Delta y}{\Delta x}&=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\[2mm]&=\dfrac{$2.41-$2.84}{2009 - 2007}\\[2mm]&=\dfrac{-$0.43}{2\text{ years}}\\[2mm]&={-$0.22}\text{ per year}\end{align}[/latex]

Analysis of the Solution

Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases.

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

[latex]\begin{align}\dfrac{292 - 10}{6 - 0}&=\dfrac{282}{6}\\[2mm]&=47 \end{align}[/latex]

The average speed is 47 miles per hour.

Analysis of the Solution

Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2,3], the average speed is 63 miles per hour.

At [latex]t=-1[/latex], the graph shows [latex]g\left(-1\right)=4[/latex]. At [latex]t=2[/latex], the graph shows [latex]g\left(2\right)=1[/latex].

The horizontal change [latex]\Delta t=3[/latex] is shown by the red arrow, and the vertical change [latex]\Delta g\left(t\right)=-3[/latex] is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

[latex]\dfrac{1 - 4}{2-\left(-1\right)}=\dfrac{-3}{3}=-1[/latex]

Analysis of the Solution

Note that the order we choose is very important. If, for example, we use [latex]\dfrac{{y}_{2}-{y}_{1}}{{x}_{1}-{x}_{2}}[/latex], we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as [latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex].

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

[latex]\begin{align}f(2)&=2^{2}-\frac{1}{2} &\,\,f(4)&=4^{2}-\frac{1}{4}\\[2mm]&=4-\frac{1}{2}&\,\,&=16-\frac{1}{4}\\[2mm]&=\frac{7}{2}&\,\,&=\frac{63}{4}\end{align}[/latex]

Now we compute the average rate of change.

[latex]\begin{align}\text{Average rate of change}&=\dfrac{f\left(4\right)-f\left(2\right)}{4 - 2} \\[2mm]&=\dfrac{\frac{63}{4}-\frac{7}{2}}{4 - 2} \\[2mm]&=\dfrac{\frac{49}{4}}{2}\\[2mm]&=\dfrac{49}{8}\end{align}[/latex]

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

[latex]\begin{align}\text{Average rate of change}&=\dfrac{F\left(6\right)-F\left(2\right)}{6 - 2}\\[2mm]&=\dfrac{\frac{2}{{6}^{2}}-\frac{2}{{2}^{2}}}{6 - 2}&\text{Simplify} \\[2mm]&=\dfrac{\frac{2}{36}-\frac{2}{4}}{4}\\[2mm]&=\dfrac{-\frac{16}{36}}{4}&\text{Combine numerator terms}\\[2mm]&=-\dfrac{1}{9}&\text{Simplify}&\end{align}[/latex]

The average rate of change is [latex]-\frac{1}{9}[/latex] newton per centimeter.

We use the average rate of change formula.

[latex]\begin{align}\text{Average rate of change}&=\dfrac{g\left(a\right)-g\left(0\right)}{a - 0}&\text{Evaluate}\\[2mm]&=\dfrac{\left({a}^{2}+3a+1\right)-\left({0}^{2}+3\left(0\right)+1\right)}{a - 0}&\text{Simplify}\\[2mm]&=\dfrac{{a}^{2}+3a+1 - 1}{a}&\text{Simplify and factor}\\[2mm]&=\dfrac{a\left(a+3\right)}{a}&\text{Divide by the common factor }a\\[2mm]&=a+3\end{align}[/latex]

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