In the next few sections we're going to learn a bit about how Geogebra calculates derivatives. But before we do that, let's do a quick application. I want you to see that the derivative is a valuable concept that is worth knowing about. [br][br][url=https://www.geogebra.org/m/x39ys4d7#material/sn4u4fqv]Earlier[/url] we tried to determine the peak altitude of the missile by plotting points. Essentially we "guessed" the maximum height. The best I could get was that the maximum altitude was somewhere between 29 and 31 seconds. A good guess was that the maximum height occurred at 30 seconds, but it wasn't clear whether this was exactly correct. [br][br]With derivatives we no longer need to "guess" to find maximums; indeed, derivatives provide a much better way of precisely determining the maximum of a function. The key insight is that at a maximum of a function, a function's growth rate goes from [i]increasing[/i] to [i]decreasing[/i]. Thus the slope of tangent line switches from [i]positive[/i] to [i]negative[/i]. [br][b][br]Thus the derivative switches from positive to negative, and the only way to go from a positive to a negative number continuously is through 0.[/b] [br][br]In other words, a huge clue about the location of a maximum is to find where the derivative is equal to 0. We'll discuss this in more detail later, but for now, let's just try it out. Type the following code into the input bar:[br][br][code]Intersect(g', y = 0)[/code]

The above code should create the point [code]G=(30.23792,0)[/code]. This is simply the point where the derivative, [code]g'[/code], intersects the line [code]y=0[/code] (which is just the x-axis). The x coordinate, 30.23792, is the [i]exact[/i] time when the rate of change of the original model [code]g(x)[/code] is equal to 0, and thus based on our understanding of [code]g(x)[/code], it is also the [i]exact[/i] time that [code]g(x)[/code] achieves a maximum. [br][br]But we're not quite done with our analysis! Don't forget, this is just the [i]time[/i] that the maximum height of the missile occurs. To find the maximum [i]height[/i] of the missile, we would still need to plug this [i]time[/i] back into the original model, [code]g(x)[/code] to obtain a [i]height[/i]. [br][br]Do it now! Type [code]g(30.23792)[/code], or if you're feeling clever, you can type [code]g(x(G))[/code] which is a little faster. What's going on in the second code snippet is similar to a composite function: [code]x(G)[/code] is code that will obtain the x-coordinate of point [code]G[/code], and then [code]g(x(G))[/code] directly feeds that x-coordinate into the function [code]g[/code] without ever stopping to take a look. Either way you should get the number 5119.20733. [br][br]We can interpret these calculations pretty easily: we've used the model to estimate both the maximum height of the missile (5119.20733 meters) and the time it will occur (30.23792 seconds). Note: You can find both of these numbers [i]without[/i] Geogebra by solving the algebraic equation formed by setting [code]g'(x)[/code] equal to 0. It's not hard at all, and only requires a two step algebraic solution, but the large numbers can be burdensome. Regardless, as always in this course, if we can let Geogebra do it, we always do so![br][br]Before we move on I just want to make it clear that what you've done in this activity is [b]one of the most important applications of the derivative[/b]. Using derivatives to pick off maximums and minimums of mathematical models [i]exactly[/i] is a really powerful application! Congrats on getting a feel for it so quickly! We'll discuss this topic more later. In the next lesson we'll talk about the units of the derivative, and then after that it's onto a sequence of actives that invite you to learn about how Geogebra calculates derivatives so quickly.