Activity 3.3.2.
Recall from Section 1.3 that the average rate of change of a function \(f\) on an interval \([a,b]\) is
\begin{equation*}
AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}\text{.}
\end{equation*}
In Section 1.6, we also saw that if we instead think of the average rate of change of \(f\) on the interval \([a,a+h]\text{,}\) the expression changes to
\begin{equation*}
AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}\text{.}
\end{equation*}
In this activity we explore the average rate of change of \(f(t) = e^t\) near the points where \(t = 1\) and \(t = 2\text{.}\)
In a new Desmos worksheet, let \(f(t) = e^t\) and define the function \(A\) by the rule
\begin{equation*}
A(h) = \frac{f(1+h)-f(1)}{h}\text{.}
\end{equation*}
(a)
(b)
Compute the value of \(A(h)\) for at least \(6\) different small values of \(h\text{,}\) both positive and negative. For instance, one value to try might be \(h = 0.0001\text{.}\) Record a table of your results.
(c)
What do you notice about the values you found in (b)? How do they compare to an important number?
(d)
Explain why the following sentence makes sense: “The function \(e^t\) is increasing at an average rate that is about the same as its value on small intervals near \(t = 1\text{.}\)”
(e)
Adjust your definition of \(A\) in Desmos by changing \(1\) to \(2\) so that
\begin{equation*}
A(h) = \frac{f(2+h)-f(2)}{h}\text{.}
\end{equation*}
How does the value of \(A(h)\) compare to \(f(2)\) for small values of \(h\text{?}\)
