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A.6 Estimating df/dx at x=a Graphically

On this screen we’re going to take another step toward seeing how we can estimate the rate df/dx at the point x=a graphically.

Recall that when we introduced linear approximations, and then did practice problems on linear approximations, we had to provide the rate at which each function changes at the point of interest. Then on the preceding screen we started to see how we can use Desmos to estimate this rate for any function by using a Leibniz triangle at the point of interest.

Let’s now put some pieces together and see how we can use a graph to first estimate the rate of change,
$\left.\dfrac{df}{dx}\right|_\text{at $x = a$},$ and then use that rate to estimate the function’s value a short distance away from a point we know about.

To begin we’ll again going use the function $f(x)=x^2,$ and pretend that we don’t yet know that $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$} = 6.$ In the following activity, we’re going to learn how to find this rate – or at least an approximation to it.

ACTIVITY 1: Line segment to mimic $f(x)=x^2$ near the point $(3,9)$

The interactive graphing calculator below shows the graph of the curve $f(x) = x^2$ versus x, with the point $(3,9)$ highlighted.

You’ll also see a line segment that has one end anchored at that point of interest $(3,9).$ You can drag the other end of the line segment to be anywhere within the calculator’s window.

Step 1: Zoom in on the graph a bit, and then drag the free end of the line segment so that the segment mimics the function’s behavior. You don’t have to try to be exact; you’re going to iterate this process as many times as you’d like.

Graph of $f(x) =x^2$ versus x, and alterable line segment

 
Step 2. When you’re ready, check the box below to show the calculation
of the line segment’s slope = $\dfrac{\text{rise}}{\text{run}}=\dfrac{\Delta y}{\Delta x}$


The value of the line segment’s slope is an approximation to the function’s rate of change, which we know in this case is $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$} = 6.$

How close to that true value is your result? Try zooming in further and repeat the process to see if your approximation improves.

You probably observe that the more you zoom in, the closer your approximation is to the true value.

You might also notice that if the two points of the line segment overlap, or if your line segment is vertical, then $\Delta x = 0$ and we can’t calculate the line segment’s slope. Instead, our slope calculation returns “NaN (Not a Number: can’t divide by 0!).” That is, you get an undefined result. Hence while we can make $\Delta x$ as small as we like, it cannot be zero.

Let’s repeat this process for a different function, one for which you don’t already know the value of the rate of change at the point of interest.

ACTIVITY 2: Line segment to mimic $g(x)$ near the point $(5,80)$

The interactive graphing calculator below shows the graph of the curve $y=g(x)$ versus x, with the point $(5,80)$ highlighted.

You’ll also see a line segment that has one end anchored at that point of interest $(5,80).$ You can drag the other end of the line segment to be wherever you want.

Step 1: Zoom in on the graph a bit, and then drag the free end of the line segment so that the segment mimics the function’s behavior. You don’t have to try to be exact; you’re going to iterate this process as many times as you’d like.

Graph of $g(x)$ versus x, and alterable line segment

Step 2. When you’re ready, check the box below to show the calculation of the line segment’s slope $=\dfrac{\Delta y}{\Delta x}$


Repeat this process as much as you’d like until you have a clear estimate in mind for the value of $\left.\dfrac{dg}{dx}\right|_\text{at $x = 5$}.$ You’ll probably find that the result you get is sometimes a bit above a particular number, and sometimes a bit below, but that each value is close to this particular number. What is that number?

Use estimate of the rate df/dx in our linear approximations

Let’s tie a few pieces together, and use the tools we’ve developed so far to (a) estimate the rate df/dx for the function $f(x) = \sqrt{x}$ at $x=9,$ and then (b) use that value to approximate the value of $\sqrt{9.1}.$

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In this next problem, we’re going to remove one piece of scaffolding: you’ll also have to compute the slope of the line segment, which is the value of our estimated rate, as part of the calculation.

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As you have seen a number of times now, to obtain a better approximation for a function’s rate of change at the point $x=a,$ we can use a graph of the function in Desmos and zoom in as much as we can, and then find the slope of a line segment that looks to mimic the function’s behavior in a small region near $x=a.$ The closer we move the “free” end of the line segment toward $x=a,$ the better our estimate seems to become.

This method is clearly limited, though, beginning with the fact that we don’t always have a function programmed into Desmos. Furthermore, while we’re able to find an estimate of $\left.\dfrac{df}{dx}\right|_\text{at $x = a$},$ we don’t yet have a way to determine the true value. Indeed, we have been relying quite heavily on graphical representations of functions — for good reason, since Desmos gives us a great way to start to see some of the key ideas in Calculus in action.

But we know that there are other ways to convey information about functions in the real world, including verbal descriptions of a physical situation, and through data sets.

In the next Section, we’re going to use those other representations, along with the continuing use of graphs, to develop a systematic way to find the rate at which a given function changes. We will begin, once again, by considering some simple motion.


The Upshot

  1. We can use Desmos to place a line segment “by eye” to mimic a function’s behavior near a point of interest, and hence estimate the function’s rate of change near that point. We can then use that estimated rate in our linear approximation calculations.


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