B.5 Lab 1 Activity: Approximate & Bound Error of df/dx

We assume that you have completed the "Lab 1: Introduction" on the preceding screen. If you have not, please do so now in order to understand the steps below.

As we proceed through the activities below, please keep in mind that computing the slope of the secant line is the same as calculating the average rate of change for the interval of interest.

Title text: Computing Average Rate of Change. Graph of y = f(x) versus x, with the points (a, f(a)) and (b, f(b)) labelled, and a secant line passes through them. Text says: average rate of change = f(b) - f(a), divided by b - a, which equals the slope of the line that passes through the interval's initial and final points. Text at the bottom says: Graphically, finding the average rate of change on the interval [a, b] means computing the slope of the secant line that passes through the interval's initial and final points.

This lab is based on work done by the CLEAR Calculus Project at Oklahoma State University. From the Project's website: "Project CLEAR Calculus is a research-based effort to make calculus conceptually accessible to more students while simultaneously increasing the coherence, rigor, and applicability of the content learned in the courses."
We at Matheno appreciate the work CLEAR Calculus has done to help students learn Calculus better, and are happy to build off of their efforts.

As a reminder, at the end of Lab 1 Introduction on the preceding screen we summarized the work of the lab through using Oerhtman's Five Questions for Approximations: [Ref]

Let's continue from where we left off on the preceding screen and collect more data. We've already included the data from that previous screen here so you don't have to re-do those first entries.

Lab 1, Activity:
Estimate $𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 1$ for $𝑓 (𝑥) = 2 𝑥$

Part I: Estimate from the right

The graph below is the same as that on the preceding screen, except you now control the value of $𝑥 2$ and hence the size of the interval we're examining. Here in Part I, we require $𝑥 2 > 𝑥 1 (= 1)$ so we know we're currently generating only overestimates of $𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 1 .$

Start by moving $𝑥 2$ to be closer to $𝑥 1 .$ Then click the box underneath the graph "to use the selected interval and calculate the line segment's slope value." Make sure the resulting calculation makes sense to you.

Then when you're ready, add the resulting data to your table with the "Add data to table" button.

You can then change the value of $𝑥 2$ and add more data, or continue to the next part. You can return to this calculator and add more data whenever you'd like.

Data Collection for Slope of Line Segments with variable right end-point
Data Point	$(𝑥 2, 𝑦 2)$	$Δ 𝑥$	Average Rate of Change over Interval = Line Segment's Slope: $Δ 𝑦 Δ 𝑥 = 𝑓 (𝑥 2) - 𝑓 (1) 𝑥 2 - 1$
1	(2.20000000, 4.59479342)	1.20000000	2.16232785
2	(1.50000000, 2.82842712)	0.50000000	1.65685425

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

Proceed as you did before — now with the second "free end" point of the line segment constrained to values less than 1, so we're shrinking our interval from the left.

Data Collection for Slope of Line Segments with variable left end-point
Data Point	$(𝑥 2, 𝑦 2)$	$Δ 𝑥$	Average Rate of Change over Interval = Line Segment's Slope: $Δ 𝑦 Δ 𝑥 = 𝑓 (𝑥 2) - 𝑓 (1) 𝑥 2 - 1$
1	(-0.80000000, 0.57434918)	-1.80000000	0.79202823

Part III: Bounding the error

We explained what this graph shows on the preceding screen.

Conclusion

[Concluding text will appear here after you've determined the value
of $𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 1$ within the desired error bound.]

[End of lab]

The Upshot

A function's average rate of change over an interval is equal to the slope of the secant line that passes through the endpoints of the interval: $a v e r a g e r a t e o f c h a n g e [𝑥 1, 𝑥 2] = s l o p e o f l i n e s e g m e n t = Δ 𝑦 Δ 𝑥 = 𝑓 (𝑥 2) - 𝑓 (𝑥 1) 𝑥 2 - 𝑥 1$
You can make the average rate of change arbitrarily close to the instantaneous rate of change at $𝑥 1$ by making $𝑥 2$ sufficiently close to $𝑥 1$ (and hence $Δ 𝑥$ sufficiently small).

These points raise a natural question that led to the development of Calculus itself:

How small can we make $Δ 𝑥$ ??

We know it can't be exactly zero, because we can't divide by zero in the slope calculation. So how small can we make it?

We will take up this key question – arguably THE key question – when we explore THE key tool in Calculus in the next Chapter, on "Limits".