A.6 Estimating $𝑑 𝑓 𝑑 𝑥$ 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, $𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 𝑎,$ 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 $𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 3 = 6 .$ In the following activity, we're going to learn how to approximate this rate to a high degree of accuracy.

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.

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 $𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 3 = 6 .$

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 $Δ 𝑥 = 0$ 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 $Δ 𝑥$ 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.

Use estimate of the rate $𝑑 𝑓 𝑑 𝑥$ 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

𝑑 𝑓 𝑑 𝑥

for the function

𝑓 (𝑥) = \sqrt 𝑥

𝑥 = 9,

and then (b) use that value to approximate the value of

\sqrt 9.1 .

Practice Problem 1: Approximate

\sqrt 9.1

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

(a) Use the approach we developed above to estimate the rate $𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 9 .$

(b)

Then use that value and the tools we developed in a preceding topic to approximate

\sqrt 9.1 .

View/Hide Solution

Graph of possible line segment used to estimate df/dx at x = 9, and the resulting line-segment slope calculation.

(a) As shown in the figure, after several trials we decided that when the line segment closely mimics the curve's behavior it has slope equal to 0.1668. Hence we estimate

𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 9 \approx 0.1668 ✓

You probably didn't obtain exactly the same value, but your value should be close to ours.

(b) We now have the two pieces of information we need to approximate the value of

\sqrt 9.1 :

(I) we know

\sqrt 9 = 3,

and (II) we have the estimate

𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 9 \approx 0.1668 .

Our point of interest at

𝑥 = 9.1

is a horizontal distance

𝑑 𝑥 = 0.1

away from the point we know about at

𝑎 = 9 .

Hence

𝑓 v a l u e a t 9 + 𝑑 𝑥 ⏞ 𝑓 (9 + 𝑑 𝑥) \approx 𝑓 v a l u e a t 9 ⏞ 𝑓 (9) + (r a t e a t 𝑥 = 9) * (𝑑 𝑥) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑓 𝑓 (9 + 0.1) \approx 3 + (0.1668) (0.1) \approx 3 + 0.01668 \approx 3.01668 ✓

The figure below shows how the small change df adds to the height

𝑓 (9) = 3

to yield the final result.

Hence our approximation is

\sqrt 9.1 \approx 3.01668 .

(For comparison purposes, the actual value to 5 decimal places is

\sqrt 9.1 = 3.01662 .

Pretty good, huh?!?)

Your answer will of course be different from ours if you obtained a different estimate for

𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 9,

but your calculation should provide a similar result – and remember this is all only an approximation, albeit a good one. Graph of the small change df adds to the height f(9)=3 to yield the final result.

Graph of the small change df adds to the height f(9)=3 to yield the final result.

[close solution]

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.

Practice Problem 2: Estimate

c o s (𝜋 4 + 0.03)

The interactive graphing calculator below shows the graph of the curve $𝑔 (𝑥) = c o s 𝑥$ versus x. We're going to look at the function's behavior near $𝑥 = 𝜋 4 .$ You know that
$c o s (𝜋 4) = 1 \sqrt 2 \approx 0.7071$
and the point $(𝜋 4, 0.7071)$ is highlighted on the curve.

(a) Use the approach we developed above to estimate the rate $𝑑 𝑔 𝑑 𝑥 ∣ a t 𝑥 = 𝜋 / 4 .$
Note that unlike in earlier problems, now you must calculate the slope of the line segment yourself. A calculator can help.

(b) Then use that value and the tools we developed earlier to approximate $c o s (𝜋 4 + 0.03) .$

View/Hide Solution

(a) The line segment we used to mimic the curve's behavior is shown in the left-hand figure below. The right-hand figure shows the value of the segment's second point that we can use to compute the segment's slope. Your values will of course be different, but your calculation should be of the same form.

(i) Graph of the line segment we used, layered on top of the original curve. (ii) The curve now removed, and the coordinates of the endpoints of the line segment used to compute the slope of the line segment, which in turn is our estimated rate of change for the function at pi/4.

We have as two points of our line segment $(𝜋 4, 0.7071)$ and $(0.7941, 0.701) .$ Hence the slope of our line segment is

Δ 𝑦 Δ 𝑥 = 0.701 - 0.7071 0.7941 - 𝜋 / 4 = - 0.701

That value is our approximation for $𝑑 𝑔 𝑑 𝑥 ∣ a t 𝑥 = 𝜋 / 4 :$ $𝑑 𝑔 𝑑 𝑥 ∣ a t 𝑥 = 𝜋 / 4 \approx - 0.701 ✓$ Your value is probably different, but should be close to ours.

As a check, does it make sense that the value if negative? The answer is yes: the function's values are decreasing as x increases from $𝑥 = 𝜋 / 4,$ and so the rate of change must be negative.

(b) We now have the two pieces of information we need to approximate the value of
$c o s (𝜋 4 + 0.03) :$

(I) we know $c o s (𝜋 4) \approx 0.7071,$ and

(II) we have the estimate $𝑑 𝑔 𝑑 𝑥 ∣ a t 𝑥 = 𝜋 / 4 \approx - 0.701 .$

Our point of interest is a horizontal distance $𝑑 𝑥 = 0.034$ away from the point we know about at $𝑎 = 𝜋 4 .$ Hence $𝑔 v a l u e a t 𝜋 / 4 + 0.03 ⏞ ¯¯ ⏞ ¯¯ ⏞ 𝑔 (𝜋 4 + 𝑑 𝑥) \approx 𝑔 v a l u e a t 𝜋 / 4 ⏞ 𝑔 (𝜋 4) + (r a t e a t 𝑥 = 𝜋 / 4) * (𝑑 𝑥) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑔 𝑔 (𝜋 4 + 0.03) \approx 0.7071 + (- 0.701) (0.03) \approx 0.7071 - 0.021 \approx 0.686 ✓$

Again, your final answer is probably different than ours since you likely chose a different line segment in part (a) than we did. But your answer should be close to ours – or, actually, close to the true value of, to five decimal places, 0.68558.

[close solution]

Practice Problem 3: Estimate

t a n (- 𝜋 3 - 0.08)

The interactive graphing calculator below shows the graph of the curve $𝑘 (𝑥) = t a n (𝑥)$ versus x. We're going to look at the function's behavior near $𝑥 = - 𝜋 3 .$ You know that
$t a n (- 𝜋 3) = - \sqrt 3 \approx - 1.7321$
The point $(- 𝜋 3, - 1.7321)$ is highlighted on the curve.

Estimate the value of $t a n (- 𝜋 3 - 0.08)$ .

View/Hide Solution

In order to estimate $t a n (- 𝜋 3 - 0.02)$ , we need two pieces of information: (I) We already know $t a n (- 𝜋 3) = - \sqrt 3 \approx - 1.7321 .$ (II) We also need to have an estimate for $𝑑 𝑘 𝑑 𝑥 ∣ a t 𝑥 = - 𝜋 / 3 .$

Using the same approach as we did in the problems above, we use the interactive graph to zoom in on the curve and create a line segment that mimics the curve's behavior near $𝑥 = - 𝜋 / 3 .$ The line segment we used and the coordinates of its endpoints are shown in the figure below. As always, your values will probably be different, but your final value for the line segment's slope should be close to ours.

We have as two points of our line segment $(- 𝜋 3, - 1.7321)$ and $(- 1.0435, - 1.7175) .$ Hence the slope of our line segment is $Δ 𝑦 Δ 𝑥 = - 1.7175 - (- 1.7321) - 1.0435 - (- 𝜋 / 3) = 3.949$

Hence our approximation for $𝑑 𝑘 𝑑 𝑥 ∣ a t 𝑥 = - 𝜋 / 3$ is $𝑑 𝑘 𝑑 𝑥 ∣ a t 𝑥 = - 𝜋 / 3 \approx 3.949$

Now, our point of interest is a horizontal distance $𝑑 𝑥 = - 0.02$ away from the point we know about at $𝑎 = - 𝜋 3 .$ Hence $𝑘 v a l u e a t - 𝜋 / 3 - 0.02 ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ 𝑘 (- 𝜋 3 + 𝑑 𝑥) \approx 𝑘 v a l u e a t - 𝜋 / 3 ⏞ 𝑘 (- 𝜋 3) + (r a t e a t 𝑥 = - 𝜋 / 3) * (𝑑 𝑥) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑘 𝑘 (- 𝜋 3 + (- 0.02)) \approx - 1.7321 + (3.949) (- 0.02) \approx - 1.7321 - 0.0790 \approx - 1.811 ✓$

[close solution]

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 $𝑑 𝑓 𝑑 𝑥 ∣ a t 𝑥 = 𝑎,$ 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

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.

Please head over to the Forum to post any questions or comments about anything on this screen!

A.6 Estimating 𝑑𝑓𝑑𝑥 at 𝒙 =𝒂 Graphically

Use estimate of the rate 𝑑𝑓𝑑𝑥 in our linear approximations

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A.6 Estimating $𝑑 𝑓 𝑑 𝑥$ at $𝒙 = 𝒂$ Graphically

Use estimate of the rate $𝑑 𝑓 𝑑 𝑥$ in our linear approximations