Function Graphs 1,474 views
Increasing and decreasing functions are a critical topic on the AP test. It's practically guaranteed that you're going to have to find some minimums, some maximums, probably some increasing and decreasing intervals. You're probably going to have to find some concavity, inflection points, the whole bit. This time we're going to be doing basically just one problem, and gradually developing all those aspects on a single problem.
I hate to be an extremist about this but, I'm absolutely certain you're going to face some of these on the AP test. Relative and absolute extrema are the highest and lowest points that a graph reaches. Either absolute is highest and lowest in the entire region or relative is highest and lowest limit within a particular section. Let's have a look.
So we have a problem here where we have to look at this graph on the closed interval from -4 to +4. Because it's a closed interval, you can have an absolute maximum or minimum. If it's not a closed interval, you might not have an absolute maximum or minimum. Looking at our graph it appears that right above 4 is this spot here and it's a closed interval so that seems to be our absolute maximum.
Remember, if that was an open circle, in other words, if this was an open parenthesis, it would be an open interval and there really would be no absolute maximum. This one, that's a minimum. I don't know if it's a absolute or relative minimum right now, we'll find out in a minute. This one sure looks like it's a relative maximum. It's the highest spot within its region. Be careful with this graph.
There is a little problem with it, and if you see these on your calculator, it might fool you. The lowest this interval goes is to -4. Straight above -4 is about right there. So we shouldn't even be looking at that region on the graph, so just ignore it. This looks like it's probably a relative minimum and this is probably an absolute minimum. Let's find out.
We're going to have four places to test in this. Both end points in these two places. A lot of times students will go and get involved in all the steps this and say, I've got to find the place where the place where the slope is 0, the place where the slope is 0, and then they forget to test the end points. But the end points, they are important. But let's go and find this.
Remember last episode, what we did was we found the places where the slope was 0, in other words where the tangent line was horizontal. That gives you relative maximums and minimums. Take the derivative. The derivative of this is 3 times 1/3. One power less than 3 + 2x - 3 and then said it equals 0. 0 equals xÂ² + 2x - 3. That's a really one to factor and solve, I'm just going to cut to what the solutions are. The x values are at -3 and +1. The next thing we have to do is to test all of these locations so I have to put -4 into the original function, -3, +1 and +4 into it, to prove which ones are the minimums and maximums.
If you see one of these on the free response section, you're going to be expected to show the proof that it's a minimum or a maximum. Let's do that. Proof.
If I put -4 into the original function, f(-4) turns out to be 5/3. Now we'll put -3 into the function. f(-3) substituting into there, is going to turn out to be 4. I'm going to write the other two right on the graph. If you put 1 to the formula, you will get a result of exactly -6 2/3. And if you put +4 into it, you'll get a result of 20 1/3.
So, let's analyze these all now. It looks like the smallest it ever goes is a -6 2/3. That's an absolute minimum in the interval. 20 1/3, highest it ever goes, that's an absolute maximum. This is a relative minimum here, it's not the lowest it ever goes but it's the lowest within that section. And the relative maximum, looks like it's 4. There you have it.
Increasing and decreasing intervals. Those are the sections of a graph where the result of the function is increasing or decreasing. Here is what it sounds like. We are going to continue on with the problem we were just doing, adding this as an extra step.
You remember last time we took the derivative and we used the derivatives to find the minimums and maximums. Looking at this graph, you can see this looks like this is an increasing interval from here to here. Remember that section of the graph doesn't count for this problem. Another increasing interval looks like it's from here to here, and the decreasing section is from there to there. You might not see a graph on the AP test, you might have to do this in the section where you're not even allowed your graph and calculator. So you need to know how to prove it manually.
Well, if it's increasing, what has to be true about the graph? It's going bigger and bigger and that means that the slope is a positive. So all you really need to do is find out if the slope is positive or negative within a section and tell if it's increasing or decreasing. You have to set up intervals. So our intervals on this are between -4 and -3. We have an interval between -3 and +1 and an interval between +1 and +4. You need to pick test numbers for that to put into the formula. So I've picked out a few in advance. I picked out -3.5, that's an x value. I picked out 0 and I picked out 2. -3.5 is in interval 1, 0 is in interval 2 and the number 2 is in the third interval.
Well, how do you find out if the slope is positive or negative? Use your slope formula. I'm going to substitute these in. Now, again, on the AP test you are going to be required to show some proof if it's in the free response. But the proof doesn't have to be anything more than a little chart showing what you got for the result. So I'll make this into a little chart with x values and f'(x). So if you put -3.5 into the slope formula, the result you get is +2.25. The exact number necessarily doesn't matter. All you care about is if it's positive or negative.
So, as proof that you know what you are talking about, just write down that it's greater than 0. And that's enough proof that you can go and say, the slope is greater than 0 so that's an increasing interval. This increasing interval starts at -4 and it goes to but it doesn't include -3. It can't include -3 because it stopped increasing there. You have to write it like this. Closed interval on the left, open interval on the right. Make sure you get the notation right, you'll be graded on that on the AP test.
Next section, if I take and substitute this 0 into this, f'(0) is, that's easy, -3. That is less than 0, that means the slope is negative, which means it's getting lower and lower and that's our decreasing interval. That decreasing interval goes between the end points -3 and +1. Again, it doesn't include those because those are the places where the slope is 0 and it's not increasing or decreasing at the moment that it's 0. You have to write it like this. Open parenthesis (-3, 1)
Last section, if you substitute 2 in, the result of substituting 2 into the slope formula is 5. You don't care what the number is, you just care it's a positive. It is a positive and since it's a positive, the slope is going up, it's an increasing interval. Increasing interval starts at 1, goes up to 4. Doesn't include 1, but since it's closed, it does include 4.
Inflection points and concavity. This is a little bit harder than what you just did. It's really pretty obvious to see when you have an increasing or a decreasing slope. Inflection points are tougher to see. They are the places where the slope stops increasing or decreasing and reverses. For example, you're going along on a curve, it's getting steeper and steeper. The inflection point is where it's still getting steeper but not as rapidly. That's kind of tough to see. Back to the same problem.
Inflection points happen when the rate of the rate of change hits 0. Well, that's the second derivative. So we're going back to this original function with the first derivative. The second derivative for this is just 2x + 2. If you set this equal to 0, that's where you're going to find your inflection points. I'm going to show them to you on the graph. Inflection points may or may not be critical points. Critical points remember, are the places where the slope actually equals 0. This one is a critical point. The slope is 0 but it's not a place where you call it an inflection point.
This one though, that's an inflection point, because this is the spot where the slope stops getting more negative and it starts getting less and less negative. This one is a critical point but not an inflection point. And here the slope seems to be getting steeper and steeper, so there is probably no other inflection points. Let's find out for sure.
You know that an inflection point happen0s when the slope of the slope, in other words the second derivative, equals 0. So 0 equals 2x + 2, that happens, it's easy enough, 1x equals -1. That's our only inflection point. So we know the inflection points happens at x equals -1. And the y coordinate of that, I'm just guessing right here. It just looks like it's about -1. something.
Sometimes you'll be asked to give the whole point sometimes just the x coordinate. I'm just going to leave a question mark with that for now. Concavity. Concavity refers to whether or not it's going downwards with a negative concavity or blowing upwards, positive concavity. Concavity can be tested by using the second derivative. If substituting a number into the second derivative gives you a negative result, you've got negative concavity. This looks like a section with negative concavity, this one looks like one with positive concavity. But we'll do a couple of test numbers to find out. I'm looking at this, well -3. That's a nice neat spot that's an interval between -1 and -4. Test it out.
Again, you have to show proof on these things. So I'm going to make a chart like I did in the last episode. One of the intervals has -3 as a test number. I could have picked -2 or -3.5 or something like that. In this next interval, 0 is nice and neat. We'll use that as a test number. This chart should say that you have the double derivative. And if you substitute these in, y''(-3) turns out to be -4. You don't care about the actual number again, you just care about positive or negative. That is less than 0, and that tells us that at that spot it's concave down. It's going to be concave down all the way between the end point and the inflection point. Concave down between -4, includes the -4 and -1.
But it doesn't include the -1 because that's a spot where the concavity is momentarily 0, in other words an inflection point. Test the next region. If I put 0 into the double derivative formula, what I'm going to get is, easy enough and that double derivative result is greater than 0. So that tells us that this region has positive concavity. So it's concave up between -1, not including -1, and +4. It doesn't include the +4 because that's a closed interval. I'm going to do a quick side note here.
Remember on the very beginning of this episode, we were talking about finding maximums and minimums? Well, there is one detail that I omitted at that point. Having a slope of 0 doesn't really prove whether it's a max or a minimum, there is a little more to it. Take a look at this one right here. Remember you do the first derivative to find a critical point, then you substitute in a 0 for the first derivative. We'll do this one quickly.
So g' would be 3xÂ² - 12x + 12 and then you substitute in the 0 for g'. If you do that, what you are going to find out is that, this spot right here is a place where the slope is 0. It's a critical point. But look at the graph. It's not a minimum. It's not a maximum. It turns out if you do the next derivative, the double derivative, that's also a place where the double derivative is 0. So, it's both a critical point and an inflection point. If it's a critical point and an inflection point, then you can't say it's a minimum or a maximum. We're going to go on with the first and second derivative test to really prove something isn't a minimum or maximum.
The first derivative test has a way of proving absolutely that you have a minimum or a maximum. On the free response section, it's not going to be enough to say, this is a minimum, this is a maximum. You have to prove it. But it won't take long. Looking at this little guy again, we have to prove that this is a relative minimum. You have to prove that that spot is a relative maximum.
Remember in the first part of this episode, what we did was we found out that these were the spots where the slope equalled 0. Slope equals 0 when x equals -3 and the slope equals 0 when x equals +1. Well, what we have to do with the first derivative test to prove that it's a minimum or a maximum, is to find the slopes on either side. What I'm going to do is pick some numbers within the intervals, then I'm going to find the slopes and then I'll interpret it. We'll just try it again just like the last ones.
This time we do care about the slopes but again, just whether they are positive or negative. First interval is between -4 and -3 so I'm going to pick -3.5 again. Next interval is between -3 and +1, I'm going to pick to pick my favorite number, 0. That's my favorite number because it's too easy to substitute. The last interval is between 1 and 4, I think I'll pick 2. Notice what the chart says, we're finding the slopes, we need the slope numbers.
So I'm going to substitute these into the slope formula and here are the results. The slope at -3.5 is 16.25. Pretty steep. The slope at 0 is -3 and the slope at 2 is 5. These slopes are proof of whether or not we have minimums or maximums. So, if you look over here, the slope is positive, and on this side you have a negative slope.
This is part of the proof. Well, if you have a positive slope on the left side of a critical point, and a negative slope on the right side, that proves that it has to be a relative maximum. Now, we've proven that -3 is the relative maximum. This interval right here has a negative slope, the next interval has a positive slope. If your critical point has a negative slope on the left, and a positive slope on the right, you had to have an absolute minimum, a relative minimum. So, relative minimum is at 1.
One last thing, the second derivative test. The second derivative test is another way of proving minimums or maximums. If on a problem you've already done the first derivative test to prove the minimum or maximum, don't do this, you'll just be repeating yourself. But once you've see the second derivative test, you might not ever want to do the first derivative test again. If you have a relative maximum, look at the concavity, it's pointing downwards. All you really need to do is put that location into the second derivative and if you get a negative result, it's concave down and it's a relative maximum.
That's the counter intuitive part. A negative usually gives you a maximum. If you can remember this though, it saves you a lot of time. So, what we'll do for the second derivative test is again set up our chart but this time in the chart we don't have to do three intervals. All we have to do is test the things that we suspect are minimums and maximums. Those two are at -3 and +1.
And we're going to put them in the second derivative. Substituting into the second derivative, y"(-3) is going to be -4. That -4 is a negative result, it's less than 0. That proves that you're concave down and since you're concave down, that spot has to be a maximum. Substituting the 1. If you substitute 1 in there into the second derivative you'll get +4 for the result. +4 is greater than 0, that tells you it's concave upwards at that spot. If it's concave upwards at a critical point then you know it has to be a relative minimum. Relative minimum of 1.
We've covered a ton of detail in this episode. You may not have to do all these steps on any one problem in the AP test but it's good to be prepared for all of them because there is a possibility that one of the longer free response questions will ask you to do all of this stuff. Again, if they do, remember, put in those charts. Some kind of a proof that you know that you're proving minimums, maximums, increasing and decreasing intervals, inflection points and concavity.
There is one other kind of thing you need to be aware of. there are sometimes problems where say you have an absolute value function that forms a perfectly sharp V, at the bottom of that V, you can't do the derivative. So you can't do the first and second derivative test. There are other ways that you have to deal with this.
If you look in the bonus materials there is a really good problem involving all of these steps for an absolute value function. Make sure you check those bonus materials, it's all in there including the entire solution.