Essential Knowledge 2.1B1

So let's just take this point for a little journey. (I'm going to omit the word "about" in this narrative, because all numbers are estimates so I would literally have to say "about" before every single number.)

It starts at x = -2, which has a y value of -71, which is off our graph. I try going left of x = -2, but the slope just gets even more negative. So I go right. It looks like around x = -1.83, the slope is -50.05, so I'll make a note of this.

x = -1.83, f'(x) = -50.05

I'll just keep going and finding values

x = -1.5, f'(x) = -18.02

x = -1.19, f'(x) = 0

x = -1.01, f'(x) = 5.78

x = -0.77, f'(x) = 9.43

x = -0.49, f'(x) = 8.91

x = -0.25, f'(x) = 5.5

x = 0, f'(x) = 0

x = 0.26, f'(x) = -6.45

x = 0.5, f'(x) = -11.99

x = 0.75, f'(x) = -16.26

x = 0.99, f'(x) = -18

x = 1.25, f'(x) = -16.01

x = 1.5, f'(x) = -9.07

x = 1.69, f'(x) = 0

x = 2, f'(x) = 24.13

x = 2.2, f'(x) = 45.78

and after that, f'(x) has values larger than 50 and thus we couldn't graph them anyways.

Now, if we wanted to know the derivative at any of those values, those would be good estimates (but remember, we got a LOT of help from GeoGebra finding those). What if I wanted to find the function f'(x)? Well, there are ways, but that's not part of AP Calculus AB. What if I wanted to make a graph of f'(x)? We could estimate that by plotting the points that we found above.

Wow, these really look like they're making a nice smooth curve. I'll try to draw it.

So that's our estimation. Now, I'll have GeoGebra make the curve of f'(x) using calculus, and see how good my estimation was.

Pretty impressive. Again, remember this was only made possible because GeoGebra calculated our slopes for us very accurately (in fact GeoGebra probably used calculus!)

Tables

Now let's estimate using tables. We could use f(x) = 3x⁴ - 2x³ - 12x² to make a table.

Now let's try to find some derivatives. Let's say we want to find the derivative at x = -1. The official method would be the limit of the difference quotient.

This wouldn't work because we only have values at specific times, so there is no way to take a limit. We would need an equation for that. So, we can only use the equation for average rate of change, which means the difference quotient without the limit.

But we can make it more accurate by choosing a value close to -1, such as -0.9. If we use a = -1 and x = -0.9, we get the following.

We could do this for every value on the table. For example, to find dy/dx at x = -0.9, use a =- 0.9 and x = -0.8 in the difference quotient. To find find dy/dx at x = -0.8, use a =- 0.8 and x = -0.7 in the difference quotient. We would get a table like this.

These, when graphed, would make the following curve

Now let's compare these to the actual curve.

This one isn't as exact as the one estimated graphically. But of course, for the millionth time, that one was only so accurate because GeoGebra was finding the slopes for us.

So we have two methods for estimating. Now of course, how did I get the actual curve, represented as purple in the graph above? I used the limit of the difference quotient. Let's use the one with x and a in it

Now we can replace a with x, since x is the variable in our equation we've been using (sorry that's pretty confusing but that's just how this equation is set up.)

f '(x) = 12x³ -6x² -24x

We could use this equation to give the actual values for the derivative. Let's add these on to our existing table to see how they compare

It's important to know how to solve derivatives using all three methods.