 Alright, so remember that the first derivative tells us something about the slope of the line tangent to the graph of a function, which means that we can use the first derivative to tell us something about what the graph looks like. Now in these days of graphing calculators it kind of seems pointless to actually talk about graphing, but it's actually a fairly valuable exercise because many times what we have corresponds to information about the first or even the second derivative, and what we're actually interested in is what the function looks like. And to that end, we'll consider the, we'll side-step the question of what is the derivative and focus on what it actually tells us. So here's a general approach to graphing functions using derivatives. So we might begin with, well, finding the derivative. Somehow either we start with the derivative or we apply the derivative rules to obtain it. Find where the derivative is positive or negative, and we want to sketch a graph that corresponds to a graph that's rising whenever we have a positive derivative and falling whenever we have a negative derivative. And to produce our first sketch, we'll just use a connect-the-dots-straight-line graph. Later on we'll refine our graph using higher derivatives. So let's take a look at this. We want to sketch a possible graph for y equals f of x, where f prime of x is x squared minus 50 plus 15x minus 35. So here's our function derivative, and note that we already have the derivative. We don't have to go out and find it. So we do want to know where the derivative is positive and negative, and in general, if you want to try and solve an inequality greater than zero, less than zero, you start by trying to solve the equality derivative equal to zero, and so that solves the equation x squared plus 15x minus 35 equal to zero, and that's a fairly standard quadratic equation. So we'll use the quadratic formula to find our solutions, and it's helpful if we actually get a numerical value for that, so x is around negative 17 and around positive 2. So we'll go ahead and graph these points on our number line, and there's around 17 negative, and there's around 2 positive, and we are finding the places where the derivative is equal to zero, but what is of interest is where the derivative is positive or negative. And again, the key idea here is that the derivative cannot change from positive to negative or vice versa without either being zero or being undefined. So let's take a look at that. So we'll use a test point method, and we'll note that our two points have divided our number line into one, two, three distinct regions, and within each region, the sine is going to be the same. So we can try a test point at that first interval someplace over here, minus 20 works, minus a hundred works, minus a billion works, if you can reason your way through what the sine of that's going to be, and if x equals negative 20, substituting that into our derivative equation, we find slope is equal to 65, so the derivative and the slope is positive in that first interval, so we'll go ahead and make a note of that. Likewise, we can pick a point in the middle of this next interval, zero happens to be right in there, so we can try zero, and at that point, if I substitute zero into my derivative function, I get negative 35, so the derivative and the slope is going to be negative in that middle interval. And then finally our last interval, a good test point, x equals someplace to the right here, how about x equals a hundred, I'll substitute that into my derivative formula, and find that the derivative is 11,465, importantly positive, and so our derivative is going to be positive in the last interval. And so in our first interval, our slope is positive, in the middle interval, our slope is negative, and in the last interval, our slope is positive once again, so for reference, it's useful to just sketch some representative lines that have positive slope, negative slope, positive slope, and this is mainly to give us some sense of what our graph is going to look like, and actually our graph doesn't look too different from that, so in this first interval, I'm going to sketch a graph that is rising, then falling, then rising again, and my straight line stick figure sketch that is going to look something like that. And I'll suppose I take a look at another function, so again here, I have the derivative x plus 3 quantity squared over x minus 4, and first thing to notice here is that because of this x minus 4 and the denominator x, absolutely cannot equal to 4, so the other things I have to do is solve for when the derivative is equal to 0, so I have a quotient, so the only time I can get a quotient equal to 0 is when the numerator is 0, so the only solution is x equal to negative 3. And we'll graph our solution x equals negative 3, also our forbidden point, x equal to 4, we'll put those on the number line, and we'll throw a dashed line through here to indicate this is a line we may not touch. And we're looking for these signs of the derivative in the 1, 2, 3 intervals that our points have identified, so again we'll pick a test point. So again our first interval, way out here, x equals negative 10, perfectly good test point, substituting that into our derivative expression, we get f of negative 10 is 49 over negative 6, and that's a negative number, so our graph is falling in that first interval. In this middle interval here, x equal to 0 is a perfectly good test point, if I substitute that into the derivative, the derivative at 0 is negative once again, our graph continues to fall, and then in our last interval here, x equals positive 10, perfectly good test point, and we find that our derivative at 10, 169 over 6, that's positive, so the graph is rising in that third interval. The important thing here is I can't cross this line, this line at x equals 4 is a forbidden line, so we'll leave a little bit of a gap between those two. Now as a straight line sketch, this isn't too bad, but we can refine it a little bit. This line, this forbidden line here, something interesting happens there, so let's see if we can figure out what's going on. So for that we want to take a closer look at the derivative and look at the one-sided limits as x approaches 4. So if I'm coming up from below, as x approaches 4 from below, the derivative is going to approach negative infinity, so our numerator is always positive, our denominator goes to a small negative number, so numerator close to 7 squared, close to 49, divided by small number makes large number, and that's negative, so the negative is going to be in the quotient as well. So as x gets close to 4 from below, our derivative gets close to negative infinity, and so that means our graph is falling, that's the negative, rapidly that's the infinity, so our graph falls very rapidly as we approach 4 from below. The same sort of analysis as x gets close to 4 from above, our numerator, still close to 49, our denominator close to small but positive number, so 49 over small positive gets us large and positive, so the graph is rising positive very quickly, infinity, as we move past x equals 4, and so what's that going to look like? Well rather than having these rather modest, falling or rising lines, I'm going to have much sharper falling and rising lines, so my graph is falling much more rapidly as I get close to 4 and rising much more rapidly as we move away from x equals 4, so there's our stick figure sketch of a graph with derivative x plus 3 squared over x minus 4.