The Derivative Function - Problem 2 4,480 views
To sketch the graph of the derivative function, find the slopes of tangent lines of the original function at various points. The derivative function at those points will be equal to the slope of the respective tangent lines. For example, if at x=4, f(x)=7 (i.e. f(4)=7), and the slope of the tangent line at that point is 3, then the derivative function f'(x) will be 3 when x=4 (i.e. f'(4)=3). Some important tangent lines to look at are when the slope of the tangent line is 0. This is where the derivative will cross the x-axis, or where the slope of the function changes from positive to negative, or negative to positive.
I have another problem here. It says given the graph of f(x), sketch it's derivative function f '(x). Well, here is my graph. F(x) is in red. I want to sketch f'(x). The way I'm going to do this is I'm going to pick points. I'm going to draw a tangent lines, and measure their slopes, because the slope of the tangent line gives me the derivative.
So let's start with x equals -1. This point right here. So I need to carefully draw a tangent line, and then measure its slope. You want to draw a long enough tangent line that you can pick points that are reasonably far apart. So let me pick this point here, and let's say this point here. Actually let me choose this point here. I chose these two points, because they're at integer values of x. This one's at -2, this one's at 0. So the question is what are the y values?
Well, this point here if these increments are a ¼ each. This looks like it might be 4.1. So one of my points is -2, 4.1. The second point here is at x equals 0, and it looks like it might be 5.9. Now let's remember what we're trying to do here. We're measuring the slope of the tangent line at -2. This is going to give us f'(-1) so that's approximately this slope 5.9 minus 4.1 over 0 minus -2. So 5.9 minus 4.1 is 1.8 over 2, that's point 9.
So what I do now so that I want to graph f' is at -1, I plot the value point 9. So looking at my graph here -1, point 9 right there. Then at x equals 1, I want to do the same thing. So let me draw a tangent line, and then I'll measure its slope. I'll make it nice and long so you can pick points that are far apart. Well that's nice. What's nice is that it goes right though 4, 0 which is a nice point to use.
So I'm going to use 4, 0, and I'll use this point here. 0, 5.3. So the two points are 0, 5.3, and 4, 0. Now this is going to give me an approximate value for the derivative of 1, because I drew the tangent line at x equals 1. So this is going to be 0 minus 5.3, 4 minus 0. That's -5.3 over 4. I'm going to need my calculator here.
So -5.3 divided by 4, -1.325. So let's plot that. It's about -1 and a 1/3. So at 1 we have that's -1.5 so -1 and a 1/3 is about here. You'll notice that there are a lot of other blue points. I've actually taken the liberty of plotting these ahead of time. So we just did these two, but I plotted the rest of them. All you have to do is connect these points with the curve, and you get the derivative function.
Now what's interesting about the derivative function is when you compare the graph of the derivative to the graph of the original function. You can see a lot of interesting things. Like for example where the derivative function crosses through the x axis, y equal 0, the original function is going to have a horizontal tangent. The tangent with slope 0. The same thing happens over here.
You see where there is a horizontal tangent, that's precisely where the derivative crosses the x axis. Where the function is decreasing most steeply, that's where you get a minimum. So this is the graph of the derivative f'(x) for this function f(x).