# The Derivative Function - Problem 3

By using the definition of the derivative, it is possible to find a formula for the derivative function. Recall that the definition of the derivative at some point x is the limit as h approaches 0 of (f(x+h)-f(x))/h. Substitute the x in f(x) with x+h and evaluate f(x) at this point. By plugging this value into the definition of the derivative and simplifying, you will find a new expression. When h approaches 0, the h's in this expression will become 0, so that the result is solely dependent on x. To find zeros of the derivative, look at the graph of the derivative function. The zeros will be the points at which the derivative crosses the x-axis. Using a graphing calculator's trace button, you can find the exact locations of x when the function is 0.

We're looking at the derivative function. Here is another function. I have f(x) equals -¼x³ plus x² plus ¼x minus 1. I want to find a way to roughly approximate the derivative of this function f'(x).

You might want to do something like this, in an instance where you don't know how to differentiate a function. Differentiate means to find the derivative. If you don't know how to actually differentiate a function, the second best thing, is to approximate the derivative, and that's what we're going to do today.

So let's start with the definition of the derivative. Now it's the limit as h approaches 0 of f(x+h) minus f(x) over h. Approximating the derivative means approximating this limit. So what I'm going to do is approximate it with a value of this difference quotient.

Now I'm taking the limit as h approaches 0. So it makes sense that I can approximate this limit, with a value of the difference quotient using a sufficiently small h, with h very small. So for example, let's say h is 1000th. So x plus point 001 minus f(x) over point 001. This will give us a reasonable approximation of the derivative.

Now in part b, I have to do exactly this. It asks me to graph f and f' on the TI-84, and find the zeros of the f' to the nearest hundredth. So I'm going to go on to the TI-84, and approximate this derivative on my calculator.

I'm looking at my TI-84. Now I've already entered f(x) here as Y1. So what I want enter here as y2 as my approximation for the derivative. Now this is not called f(x) on my calculator. It's called Y1. So how do I enter y1 in the space for y2?

Well, if you go the variables menu, hit the button VARS. Then go to Y-VARS, and hit enter. You have this list of y variables. So I can hit enter again for y1. I get y1. So I want Y1 of x plus point 001. Then I have to subtract Y1 of x.

So I go into VARS again to the right. Let me go back VARS, Y-VARS, and then hit enter. I want Y1. Then I want X, Y1(X). I know that has to be divided by this point 001. So let me go all the way over. I need to use parenthesis here. So second, insert the parenthesis. I go all the way to the end parenthesis divided by point 001.

Let's see what that looks like. Hit graph. Now it will take a second, and then it will draw this approximation for the derivative. Then here it is. It looks like a downward opening parabola. Now remember, my problem in part b asks me to find the zeros of this derivative. That means where the derivative crosses the x axis. So let's find this 0 first. It looks like it's a little to the left of 0. Somewhere between -1, and 0.

So your calculator can actually do that for you. You could just hit second, trace, and get into the CALCULATE menu. It's the second entry, number 2. You need to enter a Left Bound, and Right Bound. Now first you need to make sure that you're on the right function. I don't want to find zeros of Y1. That's my original function. I want to find zeros of Y2. So I have to use the up, or down arrow key to switch functions.

Now I'm looking at Y2. For our Left Bound, all I have to do is cursor a little bit to the left. So now I'm into the left of my zero, and hit enter. Then I have to cursor to the right Right Bound. It's basically asking you for an interval in which to search. Now you want to give it a guess. So cursor over really close to the actual zero, there is my value negative point 12. So I'll just remember that.

Now I want to search for the zero over here. The right hand zero. So second, CALC, cursor down to 0, hit enter. Now I'm on the wrong function again. I'm on Y1. I want to switch to Y2, so I use the up, and down arrows, and now I'm in Y2. Remember I'm looking for this 0 here. It looks like it's between 2, and 3. I could use 2 as a left down. I can actually type that in. 2, enter. Then as a Right Bound, I could use 3. Just type it in 3, enter. It looks like the 0 is around 2.8, so let me type that in for a guess. There it is 2.79. So that's my second 0.

Now that I have both 0s, let me go back to the board and finish up the problem. I graphed f, and f' already. I want to find the zeros. Well, f'(x) equals 0 when x was approximately negative point 12 or when x was approximately 2.79. These were the two places where a parabola crossed the axis here and here. So that's how you graph a derivative on your calculator.

If you don't actually know the formula for the derivative, you can always use a difference quotient, and take a value of h that's fairly small. This will give you a pretty good approximation of the derivative. Remember that's basically what you're doing. Is you're taking the limit as h goes to 0, so if you use a small enough value of h, you'll get a good approximation for your derivative.

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