Figure 2

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

Analysis of the Solution

Note that the order we choose is very important. If, for example, we use [latex]\frac{{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].

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

[latex]\frac{292 - 10}{6 - 0} =\frac{282}{6} =47[/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.

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

[latex]\begin{align}f\left(2\right)&={2}^{2}-\frac{1}{2} &&& f\left(4\right)&={4}^{2}-\frac{1}{4} \\ &=4-\frac{1}{2} &&&& =16-{1}{4} \\ &=\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}&=\frac{f\left(4\right)-f\left(2\right)}{4 - 2} \\[1mm] &=\frac{\frac{63}{4}-\frac{7}{2}}{4 - 2} \\[1mm] &=\frac{\frac{49}{4}}{2} \\[1mm] &=\frac{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 }&=\frac{F\left(6\right)-F\left(2\right)}{6 - 2} \\[1mm] &=\frac{\frac{2}{{6}^{2}}-\frac{2}{{2}^{2}}}{6 - 2} && \text{Simplify}. \\[1mm] &=\frac{\frac{2}{36}-\frac{2}{4}}{4} \\[1mm] &=\frac{-\frac{16}{36}}{4} &&\text{Combine numerator terms}. \\[1mm] &=-\frac{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}&=\frac{g\left(a\right)-g\left(0\right)}{a - 0}&&\text{Evaluate} \\[1mm] &​=\frac{\left({a}^{2}+3a+1\right)-\left({0}^{2}+3\left(0\right)+1\right)}{a - 0}&&\text{Simplify}​ \\[1mm] &=\frac{{a}^{2}+3a+1 - 1}{a}&&\text{Simplify and factor}​ \\[1mm] &=\frac{a\left(a+3\right)}{a}&&\text{Divide by the common factor }a​ \\[1mm] &=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].

We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from [latex]t=1[/latex] to [latex]t=3[/latex] and from [latex]t=4[/latex] on.

In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval [latex]\left(4,\infty \right)[/latex].

Analysis of the Solution

Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at [latex]t=1[/latex] , [latex]t=3[/latex] , and [latex]t=4[/latex] . These points are the local extrema (two minima and a maximum).

Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between [latex]x=2[/latex] and [latex]x=3[/latex], and a mirror-image high point, or local maximum, somewhere between [latex]x=-3[/latex] and [latex]x=-2[/latex].

Figure 7

Analysis of the Solution

Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 7 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.

Figure 8

Based on these estimates, the function is increasing on the interval [latex]\left(-\infty ,-{2.449}\right)[/latex]
and [latex]\left(2.449\text{,}\infty \right)[/latex]. Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at [latex]\pm \sqrt{6}[/latex], but determining this requires calculus.)

Observe the graph of [latex]f[/latex]. The graph attains a local maximum at [latex]x=1[/latex] because it is the highest point in an open interval around [latex]x=1[/latex]. The local maximum is the [latex]y[/latex] -coordinate at [latex]x=1[/latex], which is [latex]2[/latex].

The graph attains a local minimum at [latex]\text{ }x=-1\text{ }[/latex] because it is the lowest point in an open interval around [latex]x=-1[/latex]. The local minimum is the y-coordinate at [latex]x=-1[/latex], which is [latex]-2[/latex].