We’re now going to work through our first “lab,” in which we’ll systematically determine the instantaneous rate of change for a particular function, $f(x) = 2^x,$ at $x=1,$ by considering the average rate of change of the function for various intervals around that point. On this screen we’ll introduce the systematic steps you’ll use, and then on the following screen you’ll cycle through those steps to reach whatever level of accuracy you would like. The steps we’ll use lie at the heart of Calculus, and we will return to them again and again.

You’ll notice throughout the lab a light lab-notebook-type grid behind everything. This background is meant to remind you that we’re exploring a single context throughout these activities.

---

Lab 1.1, Introduction: Estimate $\left. \dfrac{df}{dx} \right|_{x=1}$ for $f(x) = 2^x$

 

On this screen we’re going to work through some initial, sample data to illustrate the lab’s procedure. Then on the next screen you’ll have full-functionality to collect data of your choosing.

Part I: Estimate from the right

The interactive Desmos graph below shows the graph of our function of interest, $f(x)=2^x.$ Our goal is to estimate the function’s instantaneous rate of change at $x=1,$ $\left. \dfrac{df}{dx} \right|_{x=1}.$ As before, we have anchored the left point of the secant line at the point of interest: $x_1 = 1,$ and $y_1 = f(x_1) = 2^1 = 2.$

One big change from the estimates we did in the preceding section is that we are now anchoring the second “free end” of the secant line to the curve, rather than having you try to place that point as close to the curve as you could manage. That is, for every $x_2$ value we’ll automatically calculate the corresponding y-value: $y_2 = f(x_2).$ The point $(x_2, y_2)$ thus lies on the function’s curve.

On this screen we will set some sample values of $x_2$ to illustrate the steps you’re going to take, and thus get your data collection underway. Then on the next screen you’ll be able to set $x_2$ to whatever you’d like to continue your investigation, and we’ll save you some work and continue to automatically calculate $y_2 = f(x_2)$ for you.

Table I: Data Collection for Slope of Line Segments, for $x_2 \gt 1$

Data Point	$(x_2, y_2)$	$\Delta x = x_2\, -\, x_1$	Average Rate of Change over Interval = Line Segment’s Slope: $\dfrac{\Delta y}{\Delta x} = \dfrac{f(x_2)\, -\, f(1)}{x_2\, -\,1}$

Perfect. Now that you’ve seen how to calculate the average rate of change for an interval and add the resulting data to a table, please continue to Part II immediately below.

---

Part II: Underestimate or Overestimate?

Notice that both of our values for $x_2$ were to the right of our point of interest, $x_1 = 1:$ Our first data set had $x_2 = 2.2,$ while our second data set had $x_2 = 1.5.$

Looking at the graph for the function, do you think the computed average rates of change for the sample intervals in Part I are an underestimate of the function’s instantaneous rate of change at $=1,$ or an overestimate? What’s your reasoning?

Tap on your answer below:

---

Part III: Develop a lower bound for the estimate (estimate from the left)

We’ve reproduced the graph and slope-calculator from above — but now the second “free end” point of the line segment is to the left of $x_1=1.$ In particular, for this sample data collection we have set the value $x_2 = -0.8$. From the reasoning of the animation above, we know that the resulting secant line’s slope is an underestimate of the instantaneous rate of change at $x=1.$

Since we know we are now collecting data that produce estimated values that are less than the function’s instantaneous rate of change, we’ll put the new information in a different table.

Proceed as you did in Part 1 above and, after you’ve made sense of the slope calculation, add this initial data to the new table below.

---

Part IV: The Error Bound

The graph below intially shows sample data; we’ll update it with your data in a minute. But first let’s understand what the graph shows:

The green line, with the larger slope, shows a possible result from Part I above, with the line-segment now extended to a full-line. Its slope thus represents an overestimate of $\left. \dfrac{df}{dx} \right|_{x=1}.$

The blue line, with the shallower slope, shows a possible result from Part III above. Its slope thus represents an underestimate of $\left. \dfrac{df}{dx} \right|_{x=1}.$

If you’d like, you can check the box to “Show instantaneous rate (orange tangent line),” which has slope equal to the function’s true value of $\left. \dfrac{df}{dx} \right|_{x=1}.$ At this point in the course we don’t know how to calculate this true-value yet, and so we’re providing this line merely for comparison purposes.

You can also choose to see the function’s curve with “Show $f(x) = 2^x$ (light grey line).” By doing so, you can see how the orange tangent line just grazes the curve at $x=1.$ You can also see, by contrast, how the blue lower-bound secant line intersects the function’s grey curve at two points. The green upper-bound secant line also intersects the curve at two points.

Once you understand what the graph shows, please proceed to. . .

Update the graph with my data
When you’re ready, click the “Show my error bounds” button beneath the graph. This action will update the slope of the green line with the value from the bottom row of your Data Table I above, showing your current upper-bound estimate for the instanteous rate of change. Simultaneously it will update the slope of the blue line with the value from your Data Table II above, showing your current lower-bound estimate. You’ll see some additional a summary of your current estimates below the graph as well.

Can’t do this yet: each data table above needs at least one row.
Please enter your data above and then try again.

Conclusion to this introduction

[A conclusion will appear here when you have completed the parts above.]

---

Overview of the approximation process

Before we leave this screen, let’s summarize the approximation process we are using — especially since, as we said, we’ll use variations of this process at crucial places several times in this course. We’re going to use the framework developed by Michael Oerhtman of the CLEAR Calculus Project [Ref]. It consists of these five questions:

In this lab, the answers to those questions are:
1. What are you approximating?
We are approximating the instantaneous rate of change $\left. \dfrac{df}{dx} \right|_{x=1}$ for $f(x) = 2^x,$ which equals the slope of the tangent line to the curve $y=f(x)$ at $x=1$ (the orange tangent line in the graph immediately above).

2. What are the approximations?
To approximate the instantaneous rate of change, we use the average rates of change for the function over an interval that starts or ends with $x_1 =1.$
\[\text{average rate of change}_{[x_1,\, x_2]} = \text{slope of line segment} = \frac{\color{purple}{\Delta
y}}{\color{green}{\Delta x}} = \frac{\color{purple}{f(x_2)\,-\, f(x_1)}}{\color{green}{x_2\, -\, x_1}}\] If we take the other endpoint of the interval to the right of $x=1$ (so $x_2 \gt 1$), then we obtain an overestimate (the slope of the blue secant line in the graph above). And if we take the other endpoint to the left of $x=1$ (so $x_2 \lt 1$), then we obtain an underestimate (the slope of the green secant line in the graph above).

3. What are the errors?
The error in a given approximation is the difference between the instantaneous rate of change $\left. \dfrac{df}{dx} \right|_{x=1}$ and our current estimates of the average rate of change. Above, we determined that

\[0.79202823 \le \left. \dfrac{df}{dx} \right|_{x=1} \le 1.65685425 \]
Hence the size (absolute value) of our current error for the underestimate and the error for our overestimate are
\begin{align*}
\text{current error for the underestimate } &= \left| \overbrace{\left( \left. \frac{df}{dx} \right|_{x=1}\right)}^\text{true value}\, -\, \overbrace{0.79202823}^\text{our underestimate}\right| \\[8px] \text{current error for the overestimate } &= \quad\left| \overbrace{\left( \left. \frac{df}{dx} \right|_{x=1}\right)}^\text{true value}\, -\, \overbrace{1.65685425}^\text{our overestimate}\right|
\end{align*}
Note that we do not know the exact value of either of those errors; if we did, we would know the value of $\left. \frac{df}{dx} \right|_{x=1}$ itself!

More generally, for any average rate of change we compute as an estimate, the size (absolute value) of the error is
\[\text{error } = \left|\left( \left. \frac{df}{dx} \right|_{x=1}\right)\, -\, \dfrac{\color{purple}{f(x_2)\,-\,f(x_1)}}{\color{green}{x_2\, -\, x_1}}\right| \] 4. What is the bound on the size of the errors?
While we don’t know the size of either of the errors, we do know that they can be no larger than the overall error bound we found above, which is the difference between the overestimate value and the underestimate value. Currently that error bound value is

Hence we know

$\left| \left( \left. \frac{df}{dx} \right|_{x=1}\right)\, -\, 0.79202823\right| \le 0.86482602 \quad$ and $\quad\left| \left( \left. \frac{df}{dx} \right|_{x=1}\right)\, -\, 1.65685425\right| \le 0.86482602$

5. How can the error be made smaller than any predetermined bound?
If we want a smaller error bound, we simply shrink the size of the interval $\left[x_1, x_2 \right]$ over which we compute the average rate of change. We will take this very approach on the next screen.

Please proceed when you’re ready.