Calculus: Where Functions Increase, Decrease

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Curve Sketching
Critical Points
Regions Where a Function Increases or Decreases Page [1 of 1]
Well, now we are world experts on the critical points. We know how to find them. They're actually not that hard to find. The critical points are defined to be those points where the derivative equals zero or those points on the function for which the derivative doesn't exist.
And here's the list of examples we looked at. Here on this one very complicated function and you can see these are the places, these critical points, the one's - the x's marked in green are the places where the derivative equals zero.
Also, please let me remind you that just because the derivative equals zero doesn't mean we have a max or a min. This is a place where the derivative is zero because the tangent line, even though it goes through the curve, it is a tangent line and it's horizontal. So this is neither a max nor a min because you can see the function is climbing all the way through there. But the climbing sort of looks funny, right? It's climbing, climbing, climbing, climbing, it's slowing down. It's climbing, then it's zero, but then it starts speeding up. It's climbing. So that's how the climb goes.
And then we have these places where x is - where the function's defined - notice there is a dot there on the function. But the derivative is undefined; the slope there is undefined. So, in fact, we see that critical points come in two flavors. But they do have the power to partition the world up for us into little pieces where we actually can discover the regions where a function is increasing and where a function is decreasing. Notice, for example, in this region right here, the function's increasing. It's climbing, because the derivative is positive. Remember, a function is increasing wherever it's derivative is positive because that means the slope or the tangent is positive which means that the line is going up there. The tangent therefore, is going up; therefore, the curve is rising. In this region, the curve is falling, the curve is decreasing. The function is decreasing - the derivative in this region must be negative. Here, the derivative in this region must be positive; here negative, here positive and here positive. So, actually, if we find the critical points that will cut up the x-axis into a whole bunch of places and now we can ask is the function rising or falling on this region. In this example, for example, we see that the function is rising here, falling in this region, rising in this region, falling in this region, rising in this region, and then continually rising in that region. So you can see that - what I would like to now have us think about in order for us to discover where a function is increasing and where it's decreasing, we're going to actually make up a little sort of an x-axis and look at a sign chart, just like we did in those max/min problems to see if we already had a max or a min. We made a little sign chart. Well, this sign chart will easily allow us to figure out if a function is increasing or decreasing.
Okay, let me show you how this works in practice. Okay, so let's look at an example. In fact, let's return to the examples that we looked at in our previous discussion. The first example we looked at was f of x equals four x squared plus two x minus two. And we saw the derivative equaled eight x plus two and when we set that equal to zero, we found the critical point. And we discovered there's only one critical point. There was a critical point at x equals and I believe the answer was minus one fourth.
Okay, so there's only one critical point, which means if I want to find out where this function is increasing and where the function is decreasing, I can set up a little sign chart where I plot the sign of the derivative, not of the function, but of the derivative. And all I'm going to mark on this chart are places where the critical points are.
Well, here we only have one critical point, so let me mark that one point here. Now you could mark it, you can put in zero. In fact, by the way, if you're thinking of this as a number line, you might really want to mark this zero, one, two or negative one, negative two, and then plot exactly where the negative one fourth would go. Let me invite you not to do that. Instead, just plot the critical points and no other points, because those other points will just confuse you.
Watch how I do this problem and see that I think it actually makes the process reasonably easy. What I do is there's only one so I'll just put it right here. And I'll say that is minus one fourth. So, technically, I guess zero would be over here, minus one might be way over here, one would be all the way over here, but I'm not going to plot those points. I know that here the derivative is zero. So, in fact, I'll put a little zero here to remind me that the derivative there is zero.
Now what happens here? Well, to find out what happens here all I have to do is pick a random point that's smaller than minus a fourth and plug it into the derivative and see what the sign of that is. Now it could be any point, it could be as easy or complicated, as you want to make it. I want to make it pretty easy, so I'll pick a point to the left of minus a fourth. I'll pick minus one. So if I take minus one and plug it in here, what do I see? Well, I see that f prime at minus one equals minus eight plus two, which is minus six. I don't care about the actual number; all I care about is its sign. It's negative. What does that tell me? You tell me. Take a guess.
Well, this tells me that since the derivative at minus one is negative, the function must be falling there. So, in fact, I'm going to put a negative sign over here and, in fact, I'm going to put negative signs everywhere. How do I know that in this whole region, the function must be falling? Well, for - if the function weren't falling somewhere in here that would mean it would have to be rising. But if it were rising, and then at some point falling, there has to be a place where you go from rising to falling. So one of two things has to happen. Either you gently taper off, which means the derivative is zero, but that would mean another critical point, or you get one of those wave-like cusp forms, where you're rising and then falling like this. But then you get to a place where the derivative doesn't exist. But that would also be a critical point. So bottom line is, if these weren't all negative in this whole region, then there would have to be some other place where we'd have a critical point, so we only have one. So, in fact, every one here is negative.
Similarly, if we take any point here, I'll pick zero, which is to the right of negative four, plug in zero you see positive, so f prime of zero equals two. I don't care the value, just the sign, and it's positive, and it's got to be positive on the entire region for there are no other critical points.
So what have I just discovered? Well, I've just discovered that before this point, the function is falling. The derivatives are negative, so I denote that by putting arrows going down like this. The function is decreasing on this region. Here, the function is increasing. So I just discovered that f of x is increasing for all x values that are less than minus a fourth and increasing, and decrea... oh I'm terribly sorry. It's not increasing for these values, it's decreasing. This is a typographical error, I'm very sorry. It should be decreasing. Decreasing because the function is falling for all these values and it's increasing - now I'm going to tell you when it's increasing. It's increasing for all the values x that are greater than minus a fourth. So now, I see on what region is the function increasing, decreasing and what region is the function increasing? By the way, using this information, what can you conclude about this point minus a fourth? Is that located at a max, a min or neither? This has to be a min because we're falling and then going up. So, in fact, you also see for free and minus a fourth is at a min. So in fact, using the sign chart, we also find out where the minima are and where the maxima are.
Okay, let's take a look at another example. Let's take the second example we looked at, in fact. So here's the second example b, g of x equals x cubed minus three x squared minus nine x plus one. We took the derivative before, and I'll remind you of that. We got three x squared minus six x minus nine. When we set that equal to zero and solved, what we saw was we got two answers, so the critical points are at x equals three and x equals minus one.
Let's now find in what region is the function increasing and what region is the function decreasing. How do I proceed? What I'm going to do is I'm going to create a sign chart for the derivative. So I create a sign chart for the derivative g prime of x. Where do I put mesh marks down? Where all of the critical points are. I see a critical point at minus one and a critical point at three. So I'm going to put the minus one over to the left and the three here. Now, remember I don't like to put in any other hash marks because it gets me confused. I just put in the ones that I need and understand that there's zero and one and two here and then four here and minus two here. But I don't write those in because they confuse me. These are places where the derivative equals zero, so you might want to put in zeroes here just to remind you that the derivative is zero there. And now I want to see what goes over here. All I need to do, by the way, is just pick a point, any point in this region, and whatever the sign of the derivative is there, I know that's the sign of the derivative everywhere in that region - similarly in between here and similarly out here. So there's three regions we have to check.
Okay, so now what do I do? Well, I'll pick a point here, you can pick any point you want, and I'll just pick let's say negative two. So let's take a look at g of negative two - and I care about the sign. Now, actually, I deliberately wrote something that's wrong. I wrote g of negative two. Is that really what I want to look at? Well, no, because what is g of negative two? That just tells me where the dot goes on the graph, but I'm not interested in what the dot is doing. I want to know how the function is moving through that dot. Is it going up, is it going down? So to see if things are increasing or decreasing, I must look at the derivatives, which represent the slope of the tangent. So, don't make the kind of mistake of plugging into the original function. Understand this function is a machine that spits out where the points are located. This function is a machine that spits out the slope of the tangent. Okay, if we plug in minus two into the derivative, so we have to go back to here, what do we see? Well, I see minus two squared, which gives me a four. Four times three equals twelve. And then I see a twelve and then I see six times negative two, which gives me a negative twelve. But I have a negative sign in front of it, so this gives me another twelve, that's twenty-four. Twenty-four minus nine is positive. That's all I care about. I don't care exactly what it equals; all I care about is that it's positive. So if it's positive that means that everybody here must positive. And if the derivative is positive, is the function increasing or decreasing? Increasing. Slopes are positive, the function's on the rise.
Okay, what about here? What point would you pick? Well, I would pick zero, which is in between minus one and three, that's an easy point. Plug in zero here. I see negative nine that plainly is negative. So g prime at zero is negative and so I put in negative signs everywhere here. You see why there can't be any change in sign? If there were a change in signs, either it would have to be a place where we taper off like this, which means derivative is zero, or taper off like this which means derivative is undefined. But there are no such points there because we found all of the critical points. And what about here? Well, you could try something exotic, like maybe, uh, four squared is sixteen, sixteen times three and then compare that to minus twenty-four and subtract nine. You can actually work that out and see it's actually positive. So, in fact, this is positive.
So what have I discovered through my little sign chart? Well, I can actually make a lot of discoveries here. What's the function doing? Well, here the derivative is positive, so the function's on the rise; it's rising. Here, between minus one and three the function's falling and then we rise again. So I see that the function g of x is increasing. It's on the rise for all x, let's see, all x that are less than minus one. You see all the x's that are smaller than negative one, the function's increasing. But it's also increasing here and for all x that are greater than three. So all x smaller than negative one and x greater than three, the function is rising.
The function is decreasing. Where is it decreasing? Well, it's decreasing in this region for all x that are between minus one, so x has to be bigger than minus one, but less than three. This just means that all of the x's that are bigger than minus one and less than three - so it's that region right there, the function is decreasing. So now, I've found that all of the regions where the function is decreasing and all the regions where the function is increasing. And for free - this is my favorite part by the way - for free, we also know what happens at these critical points. Is this a max or a min? It's a max because I'm rising, leveling off and then falling. That's a high point. So we also see x equals minus one, max - the function's a maximum there, and what about three? Well, we're decreasing, we level off to a derivative of zero, slope zero, and then we go up, it must be a min. So and x equals three is a min. So for free, we actually find all of the maxima and minima.
Notice what we're doing here. We're not only finding where things are increasing or decreasing, but we also find out maxima and minima for free using this little sign chart. This sign chart, I think, is really easy, really nice and gives you a lot of information.
Let's quickly try that very last example of ours, that was example c. H of x, that was this tricky function, which equals x to the one third power. You take the derivative, which we did. What did we see? We saw one over three x to the two-thirds power. And we saw a critical point at x equals zero, because the derivative was undefined but the function was defined. So there was only one critical point.
So let's take a look and see where we have an increasing and a decreasing region. So this is h prime of x. I plug in the critical point zero - you might want to put in not zero but you might want to put undefined or does not exist because the derivative is undefined there. Now pick a point to the left like - let's say, negative one and what do we get? If I plug in negative one into here, what would I see? Well, I'd say negative one and then I've got to take the cube root of it, and the cube root of negative one is negative one. If you don't believe me, take negative one and multiply it by itself three times. Negative one times negative one times negative one gives you negative one.
Then I square it and what do I see? Well, I see positive one and then I have that times three, so I see this whole thing becomes positive. So this is positive. So, in fact, everywhere here, we must have positive. Let's pick a point over here -- h prime at one, plug in one. Well, the cube root of one is one, one squared is one; this whole thing is still positive. So what have I discovered? Well, I discover that here the function is increasing. The derivatives are positive, slopes are positive and also over here. And so what's the moral? The moral is h of x is increasing for all x except that one point where it's undefined. For all x except x equals zero. And it's never decreasing -- h never decreases. So now I ask the last question here, is zero a max or a min? In fact, why don't you take a guess right now? Is there a maximum or a minimum at this point? Try it.
Well, let's think. If it's a maximum, the curve has to go up and then come down. Well, that's not what's happening. If it's a minimum, the curve has to go down and then up, but that's not happening. So what do you conclude about this critical point? It's neither a max nor a min. And so a big surprise ending, x equals zero is neither a max nor a min. Here's an example where we see a critical point that's neither a max nor a min. What happens? We see that the function is increasing and then it sort of levels off and then keeps increasing. It's like that very last example in that picture I drew where, in fact, the tangent becomes horizontal but in fact, the function keeps going up -- just a little wiggle in it. So, in fact, this has no max and no min.
All right, well good for us. Now we know how to find max and min and check them. We take all the critical points and we put them on a sign chart. Take a look at the signs of the function. You can then immediately get where the function's increasing/ decreasing, find out where we have all the max and all the minima. Great! Let's see what's next. See you soon.