In 2020, the price of gasoline was [latex]$2.168[/latex]. In 2022, the cost was [latex]$3.951[/latex]. The average rate of change is:

[latex]\begin{align*} \frac{\Delta y}{\Delta x} &= \frac{f(x_2) - f(x_1)}{x_2 - x_1} \\ &= \frac{$3.951 - $2.168}{2022 - 2020} \\ &= \frac{$1.783}{2 \text{ years}} \\ &= $0.8915 \text{ per year} \end{align*}[/latex]

Analysis of the Solution

Note that an increase is expressed by a positive change or “positive increase.” A rate of change is positive when the output increases as the input increases. In this case, we see a significant positive rate of change, indicating a sharp increase in gasoline prices between 2020 and 2022.

Here, the average speed is the average rate of change. She traveled [latex]282[/latex] miles in [latex]6[/latex] 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 [latex]47[/latex] miles per hour.

Analysis of the Solution

Because the speed is not constant, the average speed depends on the interval chosen. For the interval [latex][2,3][/latex], the average speed is [latex]63[/latex] 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].