Learn about Hairy inflection point problem Video

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Learn about Hairy inflection point problem

Learn about Hairy inflection point problem Welcome while I was just sent this problem and I figure you can never get enough practice getting the intuition behind intervals of increase decrease in inflection points and minimum and maxima so I’ll do another problem and this was pretty interesting because I have natural log here so I thought I would do this one. So let’s see it says F(x) with the. I like it very much lower case but F of x is equal to the natural log of x ^4 + 27 and find the intervals of increase decrease. Find intervals of concativity in the reflection point. So before I dealt on this problem let’s just make sure we understand what are we going to do. So find intervals of increase or decrease. So what are intervals of increase that means the slope is positive right. If slope is positive what is that mean? Increase, so intervals of increase means that the slope is positive that means that the derivative is positive. So that means where does f(x) where is that greater than zero right. Increase intervals and increases just positive slope. Similar intervals of decrease are just where f(x) is less than zero. The slope is negative that’s when the function is going to decrease not too sophisticated there. Now let’s see find the intervals of concativity and the inflection points. So what are those? Intervals of connectivity, so just a review of what we did before, where is the graph concave upwards? So I’ll write that as concave con up. Where is the graph concave up and what that it looks like? Well that means that the graph kind of looks like that. Well that happens whenever, whether we know it’d be true. The slope itself is increasing not just the graph is increasing. The slope itself is increasing right because the slope is negative here. It gets less negative then it goes to zero and then it get some more positive and it even more positive. So concave upwards means that the slope itself is increasing which means the slope of the slope is positive and what is the slope of the slope? Well that’s the second derivative. So that means that the second derivative of x is greater than zero and then summarily concave downwards it’s going to look some thing like that, guys is going to look like that and that’s when, that’s period of the curve where the slope itself is decreasing right if a high slope here think it’s a little bit flatter and goes to zero and it’s a little bit more negative. So that means that the slope of the slope or the second derivative is decreasing and what are inflection points? Inflection points are points where you go either from concave downward to concave upwards or concave upwards to concave downward and they look like this. So if was going from concave upward to concave downward it will look like this. The graph would look like this, I’d be concave up and then I go concave down. And this point right here is where you switch and that’s an inflection point or if you’re going–let’s say if I were to continue this graph on concave down then I would switch back to concave up, this point here will also be inflection point. And at this point, since we’re switching between concave up and concave down or vice versa the second derivative is zero so I’ll write that inflection points. Inflection points, the second derivative is equal to zero. So let’s use this bit of review to solve this problem. So intervals of increase decrease. So let’s take the first and the second derivative that’s always the useful thing to do. So let’s see f(x) sometime write f(x) and right on my hand writing to give me a little comfort is equal to and maybe give you comfort as well. The natural log of x^4 + 27, what is the first derivative? F(x) is equal to, let’s do this chain rules. Some people go outside in, I do the inner expression I take the derivative of the whole outside. So what’s the derivative of this inside? X^4 +27 well that’s just for x^3, right the derivative of 27 is nothing x the derivative ln of this with respect to this whole thing. So what’s the derivative of ln of x its one over x right? So the derivative of ln of this whole thing with respect to this whole thing is going to be equal to 1 over x^4 + 27 or you could say, you know that’s the same thing as—and I’m going to write it this way because I find this easier in my head to graph a lift for x^3 x x^4 + 27 to the -1 power. I like doing it that because we’re going to the second derivative now and take the second derivative of this we would have to know that the quotient rule but now we just have to know the product rule and I know the product rule I was to forget the quotient rule. So let’ do that. Let’s take the second derivative. So f(x)(x) this might get a little hairy. The second derivative and I will arbitrarily switch colors. So let’s do the product rule. So we take the derivative of the first expression, the first function so that’s 12x^2 x just the second function x^4 plus 27 to the -1 + just the first expression right we took the derivative here so now were just take it without the derivative here times the derivative of the second thing and here we’re going to have to use a chain rule to take the derivative of the second things let’s take the derivative of the inside that’s f4x^3 times the derivative of this whole thing so lets see that’s -1 x -1x^4 + 27 to the -2 power so that looks really really hairy but let’s see if we can simplify this a little bit. So this is a hairy problem clearly. So this is equal too and now since we have taken our derivatives we can not write as a negative exponent. I think that will make this a little easier so that equals 12x^2 over x^4 + 27 – 16x^6 over x^4 + 27 squared and now let see what we can do. Well I hopefully this—it would be very cool to send your problem that doesn’t have a clean answer but let’s keep checking for I think this might end up clean and what is this equal, Let’s a find a common denominator, well this is x^4 + 27 this is that squared so the common denominator is just that squared. The common denominator x^4 plus twice that and just see now its only teachers or that they will actually give you this as homework you seldom see something this hairy but you’ll never know and you definitely want to see on the AP exam. They are usually testing concepts not you ability to jump to a hairy problem but let’s do it anyway, never heard. So that’s our common denominator. So this one is going to have to be that the 12^2 x x^4 + 27.I hope that make sense to you right because think about it, if we just cancelled this with the squared you will end up with just this term right there. So I do is I multiply the numerator and the denominator by x^4 + 27 and then of course the second term doesn’t change because its denominator was already this so minus 16x^6. Now let me just erase some of this stuff that are all appeared because I can't erase this because we’re going to use this later. So but I think were all almost there. And I think you remember all of this stuff hopefully. Remember just say you not—I mean often times when you’re checking to the math you lose sight of what were even working on but were just trying to figure out the second derivative and I’m just simplifying this big hairy equation and that’s actually what this seems to be an exercise in simplification but let’s—so I’ll draw a line here and let me just remind you that I’m working on the second derivative of f and that equals just to simplify so 12x^2 x x^4 = 12x^6 + whatever 12 x 27 is. So I’ll just write that down. So 12 x 27x^2 – 16x^6 all of that over x^4 + 27, well anyway what can we do now? Well we can 12x^6 – 16x^6 - 4x^6 + 324x^2 over x^4 + 27. Now what was the whole point of taking the second derivative? Well let’s answer different part of the questions because I think we finally got the—let me erase of all this here so we have some room to work because now that we’ve done kind of our mechanical part of the work we can actually think about what we’ve just done. This was just our getting to where we are now. That is just how we got to where we are now and I actually I can even erase this. This is the second derivative. So I can erase all of that work and I hope you understand how we got there and I can even write it. So this is f(x). So lets to the first part. Find the intervals of increase and decrease. So essentially where is the first derivative positive and when is it negative? So this is the first derivative right. So when is this terms, when is this expression positive or negative? When is this express so we can also written this as 4x^3 over x^4 + 27. When is this positive and when is this negative? Well first of all the denominator is always going to be positive right. You take anything at the fourth power, its automatically going to be positive at least its not going to be negative right if we put zero there you just get zero but any other number there is going to be positive just let me sound imaginary. So the denominator is always going to be positive. So if you want to know when is this positive or negative the question only wills down to what is 4x^3 positive or negative. And so if you want to say when is it positive, well for—actually its kind of you know 4x^3 is positive when x^3 is positive and when is x^3 positive? Well when x is greater than zero right. If you put in any number greater than zero here this term will be positive and likewise when is this term negative? Well you put a negative number you take the negative to an odd number exponent. You’re going to get a negative number. So let’s see the enables of increase. So, f(x) increases when x is greater than zero and how do we know that we just look at the first derivative and we said when is this greater than zero. The denominator doesn’t matter because it’s always positive. So when is this greater than zero, when x^3 is greater than zero. When is x^3 greater than zero well sure when x is greater than zero write only a positive number to the third power is positive and similarly and I’ll write in different color f(x) decreases when x is less than zero because that’s when x^3 is less than zero which will make four actually less in zero which will make this whole thing less than zero. And it didn’t ask for because for a minimum or maximum point but we do have a critical point exactly when x = 0 right because when x = 0, we have the first derivative is equal to zero and we have zero slope and do we know whether that’s a minimum or maximum point? Well let’s see, f(x) increases when x is greater than zero so the slope is upward after zero and f(x) decreases when x is less than zero so slopes is negative. So write at x = 0, we must have a minimum point they didn’t even ask of that but anyway so let’s figure out the intervals of concativity and the inflection points now. So we know x = 0 is a minimum point. We know when x is greater than zero we’re increasing. When x is less than zero we’re decreasing and now lets figure out the calculative inflection and that’s were going to use the second derivative and let me erase all of this just so we have some space to work with. Let’s see okay. So it was a hairy, hairy problem but in the end all you have to do is check through the algebra and has faith and in the end something bagley usable will come out. So fine the enables of concativity. So what we want to figure out is when is the second derivative positive when is it negative and when is it zero. Zero is inflection point. Positive it means concave upwards, negative means its concave downwards and watch the previous three or four videos I’ve done on intuition of concativity because I don’t want you just to memorize. I want you to always you know when you’re 50 years old or 40 years old and your kids learning this I don’t want you to have go back to the book. I want it to be in grin Oh concave up versus means it’s the slope is increasing so that means that second derivative must be positive right. You don’t have to memorize these things. But anyway, so let’s figure this out. When is this positive negative or zero? Well we can use the same logic. This denominator is always going to be positive right because no matter what number you put in here, you can either positive or zero and you get left with 27 squared you get a positive number and denominator, so f(x) (x) as second derivative is zero when this thing is equal to zero, when the numerator is equal to zero. Its positive when the numerator is positive and its negative when the numerator is negative right because this is just you knows this is always going to be positive down here. So let’s figure out those intervals. So let’s just figure out first of all when this equals zero. So minus four x^6 plus 324x squared. Let’s figure out when that equals zero. Well the first thing I would do if we just—let’s just divide both sides by x squared just to get some you know get this squared another way so I get minus four x^4 + 324 = 0. Let’s take that of course the force on that side of equation so we then now switch swap sides of the equation again so I get 4x^4 = 324. Well I just move it on that side and then switched and then I get x^4 = 324 divided 4 is equal to 81 that’s a good number and what to the fourth power is equal to 81. Well three right because 3 squared is 9, 9 squared is 81 so x—and this is interesting though x is what? X is equal to plus or minus three right because minus three to the fourth power is also going to be—so x = +/- 3. So these are the inflection points. X = +3 and x = -3 and that’s actually just the x. If you want the actual inflection point you would take f of both of those points. So let’s see. So the point if you actually want the actual core and points for the inflection points you would say the inflection points and I’ll do them in green green for inflection points. Positive three and you put three back in here and the natural log of 3^4 which is 81 + 27 = Right that’s one of the inflection points if you were to graph that point you could evaluate this in you calculator and you get the exact y value and the other point is going to be -3 and could have been the same value right because -3^4 still 81 so its also the natural log of 108. So if you were to graph this on a graphing calculator and I encourage you to do so if you get the intuition. These would be two of the inflection point and you would see it because we’ll be switching between our concativity at. Fair enough, so between these points we must be one type of concativity and then after this point we must switch. So let’s figure out what type of concativity we are between these points. So what’s—oh actually no, I just realize this up here I shouldn’t have just divide both sides by x squared. I should have just factor out an x squared. When this thing equal zero is actually, sorry my mistake. I should’ve written—I should’ve factored out I should said minus four x squared. I would redo the view but I’ve gone so far into it that I’ll just carry forward and you’ll see that sometimes even I do simple algebraic errors. So if I were to factor this out minus 4x square times x^4 minus81 = 0. So we solve when this equal zero right, this equal zero when x is plus or minus three and then when else could this expression equal zero. Well if this factor equal zero. So if x is equal to zero, so the other inflection point we could say is also zero right because if you put x = 0 the second derivative is also going to be = 0. Sorry I should know just divide it vote sides by x squared. In fact in order to do that you have to assume that x is not equal to zero so anyway I shouldn’t have just factored it out. And then you say well you know this is zero or this is zero. So the other inflection point is the point zero, what? Well if you put zero into the original equation you get the natural log of 27. So these are the three inflection points and so let’s see I think I’m pushing 21 minutes. So those are the inflection points for this and this was an extremely, extremely hairy problem. So what are the concavities? So all you have to do to figure out the concativities is try x values in between this points—the value at the second derivative, an x values in between those points and if the second derivative is positive you its concave upwards from you know let me draw a number line. This might be the longest video I ever made. So let’s see, we get an inflection at zero, positive three and negative three and we are here to show what these points are. We could graph them but that’s too much time. But what happen is let’s pick some points. Are we concave upward or downward here? Actually what we figure out once we know that we keep switching at every inflection point so that is all we know. So let’s try x = 1 so x = 1 what’s f(1)? F the second derivative of 1 =? It’s equal to minus 4 plus 324 = 320. We didn’t even care with the number is all we care is that number is positive right. So if the number is positive at one that means from zero to three, from x = 0 to 3 we are concave upwards. And so every inflection points were switching. So from greater than three we are concave downwards and then we also concave downwards over this interval and then since we switch with minus three were concave upwards here. Well anyway, this ones a very the nine looking problem but it end up being much hairy than I thought and I even made the simple mistake here but hopefully that didn’t throw off to much. The simple idea is intervals are increasing decreasing when the slope increase, when the slope decrease that’s the thing when there’s a first derivative increase when does the first of your decrease. Concativity, well first thing you’d find is an inflection point that’s what I would do. That’s where the second derivative is = 0 and this was extremely hairy but we found those points and then the actual points on the graph are these three with a natural log etcetera and then at these points you’re switching between concativity upwards and downward so all you have to do is find the concativity at one of the points in the interval and then at each of those points they will switch and I encourage you we don’t have a time of this. I’m pushing 24 25 minutes but I encourage you to graph that and verify and see for yourself that the graph really are concave upwards for all x lesson three in concave downward between negative three and zero and see what an inflection point looks like with this points look like. Anyway, I’ll leave it there and I will see you in the next video.

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Learn about Hairy inflection point problem

Welcome while I was just sent this problem and I figure you can never get enough practice getting the intuition behind intervals of increase decrease in inflection points and minimum and maxima so I’ll do another problem and this was pretty interesting because I have natural log here so I thought I would do this one. So let’s see it says F(x) with the. I like it very much lower case but F of x is equal to the natural log of x ^4 + 27 and find the intervals of increase decrease. Find intervals of concativity in the reflection point. So before I dealt on this problem let’s just make sure we understand what are we going to do.

So find intervals of increase or decrease. So what are intervals of increase that means the slope is positive right. If slope is positive what is that mean? Increase, so intervals of increase means that the slope is positive that means that the derivative is positive. So that means where does f(x) where is that greater than zero right. Increase intervals and increases just positive slope. Similar intervals of decrease are just where f(x) is less than zero. The slope is negative that’s when the function is going to decrease not too sophisticated there. Now let’s see find the intervals of concativity and the inflection points. So what are those? Intervals of connectivity, so just a review of what we did before, where is the graph concave upwards? So I’ll write that as concave con up. Where is the graph concave up and what that it looks like? Well that means that the graph kind of looks like that.

Well that happens whenever, whether we know it’d be true. The slope itself is increasing not just the graph is increasing. The slope itself is increasing right because the slope is negative here. It gets less negative then it goes to zero and then it get some more positive and it even more positive. So concave upwards means that the slope itself is increasing which means the slope of the slope is positive and what is the slope of the slope? Well that’s the second derivative. So that means that the second derivative of x is greater than zero and then summarily concave downwards it’s going to look some thing like that, guys is going to look like that and that’s when, that’s period of the curve where the slope itself is decreasing right if a high slope here think it’s a little bit flatter and goes to zero and it’s a little bit more negative.

So that means that the slope of the slope or the second derivative is decreasing and what are inflection points? Inflection points are points where you go either from concave downward to concave upwards or concave upwards to concave downward and they look like this. So if was going from concave upward to concave downward it will look like this. The graph would look like this, I’d be concave up and then I go concave down. And this point right here is where you switch and that’s an inflection point or if you’re going–let’s say if I were to continue this graph on concave down then I would switch back to concave up, this point here will also be inflection point. And at this point, since we’re switching between concave up and concave down or vice versa the second derivative is zero so I’ll write that inflection points.

Inflection points, the second derivative is equal to zero. So let’s use this bit of review to solve this problem. So intervals of increase decrease. So let’s take the first and the second derivative that’s always the useful thing to do. So let’s see f(x) sometime write f(x) and right on my hand writing to give me a little comfort is equal to and maybe give you comfort as well. The natural log of x^4 + 27, what is the first derivative? F(x) is equal to, let’s do this chain rules. Some people go outside in, I do the inner expression I take the derivative of the whole outside. So what’s the derivative of this inside? X^4 +27 well that’s just for x^3, right the derivative of 27 is nothing x the derivative ln of this with respect to this whole thing. So what’s the derivative of ln of x its one over x right? So the derivative of ln of this whole thing with respect to this whole thing is going to be equal to 1 over x^4 + 27 or you could say, you know that’s the same thing as—and I’m going to write it this way because I find this easier in my head to graph a lift for x^3 x x^4 + 27 to the -1 power. I like doing it that because we’re going to the second derivative now and take the second derivative of this we would have to know that the quotient rule but now we just have to know the product rule and I know the product rule I was to forget the quotient rule. So let’ do that.

Let’s take the second derivative. So f(x)(x) this might get a little hairy. The second derivative and I will arbitrarily switch colors. So let’s do the product rule. So we take the derivative of the first expression, the first function so that’s 12x^2 x just the second function x^4 plus 27 to the -1 + just the first expression right we took the derivative here so now were just take it without the derivative here times the derivative of the second thing and here we’re going to have to use a chain rule to take the derivative of the second things let’s take the derivative of the inside that’s f4x^3 times the derivative of this whole thing so lets see that’s -1 x -1x^4 + 27 to the -2 power so that looks really really hairy but let’s see if we can simplify this a little bit. So this is a hairy problem clearly. So this is equal too and now since we have taken our derivatives we can not write as a negative exponent. I think that will make this a little easier so that equals 12x^2 over x^4 + 27 – 16x^6 over x^4 + 27 squared and now let see what we can do.

Well I hopefully this—it would be very cool to send your problem that doesn’t have a clean answer but let’s keep checking for I think this might end up clean and what is this equal, Let’s a find a common denominator, well this is x^4 + 27 this is that squared so the common denominator is just that squared. The common denominator x^4 plus twice that and just see now its only teachers or that they will actually give you this as homework you seldom see something this hairy but you’ll never know and you definitely want to see on the AP exam. They are usually testing concepts not you ability to jump to a hairy problem but let’s do it anyway, never heard. So that’s our common denominator.

So this one is going to have to be that the 12^2 x x^4 + 27.I hope that make sense to you right because think about it, if we just cancelled this with the squared you will end up with just this term right there. So I do is I multiply the numerator and the denominator by x^4 + 27 and then of course the second term doesn’t change because its denominator was already this so minus 16x^6. Now let me just erase some of this stuff that are all appeared because I can't erase this because we’re going to use this later. So but I think were all almost there. And I think you remember all of this stuff hopefully. Remember just say you not—I mean often times when you’re checking to the math you lose sight of what were even working on but were just trying to figure out the second derivative and I’m just simplifying this big hairy equation and that’s actually what this seems to be an exercise in simplification but let’s—so I’ll draw a line here and let me just remind you that I’m working on the second derivative of f and that equals just to simplify so 12x^2 x x^4 = 12x^6 + whatever 12 x 27 is. So I’ll just write that down.

So 12 x 27x^2 – 16x^6 all of that over x^4 + 27, well anyway what can we do now? Well we can 12x^6 – 16x^6 - 4x^6 + 324x^2 over x^4 + 27. Now what was the whole point of taking the second derivative? Well let’s answer different part of the questions because I think we finally got the—let me erase of all this here so we have some room to work because now that we’ve done kind of our mechanical part of the work we can actually think about what we’ve just done. This was just our getting to where we are now. That is just how we got to where we are now and I actually I can even erase this. This is the second derivative. So I can erase all of that work and I hope you understand how we got there and I can even write it. So this is f(x).

So lets to the first part. Find the intervals of increase and decrease. So essentially where is the first derivative positive and when is it negative? So this is the first derivative right. So when is this terms, when is this expression positive or negative? When is this express so we can also written this as 4x^3 over x^4 + 27. When is this positive and when is this negative? Well first of all the denominator is always going to be positive right. You take anything at the fourth power, its automatically going to be positive at least its not going to be negative right if we put zero there you just get zero but any other number there is going to be positive just let me sound imaginary.

So the denominator is always going to be positive. So if you want to know when is this positive or negative the question only wills down to what is 4x^3 positive or negative. And so if you want to say when is it positive, well for—actually its kind of you know 4x^3 is positive when x^3 is positive and when is x^3 positive? Well when x is greater than zero right. If you put in any number greater than zero here this term will be positive and likewise when is this term negative? Well you put a negative number you take the negative to an odd number exponent. You’re going to get a negative number. So let’s see the enables of increase. So, f(x) increases when x is greater than zero and how do we know that we just look at the first derivative and we said when is this greater than zero. The denominator doesn’t matter because it’s always positive. So when is this greater than zero, when x^3 is greater than zero. When is x^3 greater than zero well sure when x is greater than zero write only a positive number to the third power is positive and similarly and I’ll write in different color f(x) decreases when x is less than zero because that’s when x^3 is less than zero which will make four actually less in zero which will make this whole thing less than zero. And it didn’t ask for because for a minimum or maximum point but we do have a critical point exactly when x = 0 right because when x = 0, we have the first derivative is equal to zero and we have zero slope and do we know whether that’s a minimum or maximum point? Well let’s see, f(x) increases when x is greater than zero so the slope is upward after zero and f(x) decreases when x is less than zero so slopes is negative.

So write at x = 0, we must have a minimum point they didn’t even ask of that but anyway so let’s figure out the intervals of concativity and the inflection points now. So we know x = 0 is a minimum point. We know when x is greater than zero we’re increasing. When x is less than zero we’re decreasing and now lets figure out the calculative inflection and that’s were going to use the second derivative and let me erase all of this just so we have some space to work with. Let’s see okay.

So it was a hairy, hairy problem but in the end all you have to do is check through the algebra and has faith and in the end something bagley usable will come out. So fine the enables of concativity. So what we want to figure out is when is the second derivative positive when is it negative and when is it zero. Zero is inflection point. Positive it means concave upwards, negative means its concave downwards and watch the previous three or four videos I’ve done on intuition of concativity because I don’t want you just to memorize. I want you to always you know when you’re 50 years old or 40 years old and your kids learning this I don’t want you to have go back to the book. I want it to be in grin Oh concave up versus means it’s the slope is increasing so that means that second derivative must be positive right. You don’t have to memorize these things.

But anyway, so let’s figure this out. When is this positive negative or zero? Well we can use the same logic. This denominator is always going to be positive right because no matter what number you put in here, you can either positive or zero and you get left with 27 squared you get a positive number and denominator, so f(x) (x) as second derivative is zero when this thing is equal to zero, when the numerator is equal to zero. Its positive when the numerator is positive and its negative when the numerator is negative right because this is just you knows this is always going to be positive down here. So let’s figure out those intervals.

So let’s just figure out first of all when this equals zero. So minus four x^6 plus 324x squared. Let’s figure out when that equals zero. Well the first thing I would do if we just—let’s just divide both sides by x squared just to get some you know get this squared another way so I get minus four x^4 + 324 = 0. Let’s take that of course the force on that side of equation so we then now switch swap sides of the equation again so I get 4x^4 = 324. Well I just move it on that side and then switched and then I get x^4 = 324 divided 4 is equal to 81 that’s a good number and what to the fourth power is equal to 81. Well three right because 3 squared is 9, 9 squared is 81 so x—and this is interesting though x is what? X is equal to plus or minus three right because minus three to the fourth power is also going to be—so x = +/- 3. So these are the inflection points. X = +3 and x = -3 and that’s actually just the x. If you want the actual inflection point you would take f of both of those points. So let’s see.

So the point if you actually want the actual core and points for the inflection points you would say the inflection points and I’ll do them in green green for inflection points. Positive three and you put three back in here and the natural log of 3^4 which is 81 + 27 = Right that’s one of the inflection points if you were to graph that point you could evaluate this in you calculator and you get the exact y value and the other point is going to be -3 and could have been the same value right because -3^4 still 81 so its also the natural log of 108.

So if you were to graph this on a graphing calculator and I encourage you to do so if you get the intuition. These would be two of the inflection point and you would see it because we’ll be switching between our concativity at. Fair enough, so between these points we must be one type of concativity and then after this point we must switch. So let’s figure out what type of concativity we are between these points. So what’s—oh actually no, I just realize this up here I shouldn’t have just divide both sides by x squared. I should have just factor out an x squared. When this thing equal zero is actually, sorry my mistake. I should’ve written—I should’ve factored out I should said minus four x squared. I would redo the view but I’ve gone so far into it that I’ll just carry forward and you’ll see that sometimes even I do simple algebraic errors.

So if I were to factor this out minus 4x square times x^4 minus81 = 0. So we solve when this equal zero right, this equal zero when x is plus or minus three and then when else could this expression equal zero. Well if this factor equal zero. So if x is equal to zero, so the other inflection point we could say is also zero right because if you put x = 0 the second derivative is also going to be = 0. Sorry I should know just divide it vote sides by x squared. In fact in order to do that you have to assume that x is not equal to zero so anyway I shouldn’t have just factored it out. And then you say well you know this is zero or this is zero.

So the other inflection point is the point zero, what? Well if you put zero into the original equation you get the natural log of 27. So these are the three inflection points and so let’s see I think I’m pushing 21 minutes. So those are the inflection points for this and this was an extremely, extremely hairy problem. So what are the concavities? So all you have to do to figure out the concativities is try x values in between this points—the value at the second derivative, an x values in between those points and if the second derivative is positive you its concave upwards from you know let me draw a number line. This might be the longest video I ever made. So let’s see, we get an inflection at zero, positive three and negative three and we are here to show what these points are. We could graph them but that’s too much time.

But what happen is let’s pick some points. Are we concave upward or downward here? Actually what we figure out once we know that we keep switching at every inflection point so that is all we know. So let’s try x = 1 so x = 1 what’s f(1)? F the second derivative of 1 =? It’s equal to minus 4 plus 324 = 320. We didn’t even care with the number is all we care is that number is positive right. So if the number is positive at one that means from zero to three, from x = 0 to 3 we are concave upwards. And so every inflection points were switching. So from greater than three we are concave downwards and then we also concave downwards over this interval and then since we switch with minus three were concave upwards here.

Well anyway, this ones a very the nine looking problem but it end up being much hairy than I thought and I even made the simple mistake here but hopefully that didn’t throw off to much. The simple idea is intervals are increasing decreasing when the slope increase, when the slope decrease that’s the thing when there’s a first derivative increase when does the first of your decrease. Concativity, well first thing you’d find is an inflection point that’s what I would do. That’s where the second derivative is = 0 and this was extremely hairy but we found those points and then the actual points on the graph are these three with a natural log etcetera and then at these points you’re switching between concativity upwards and downward so all you have to do is find the concativity at one of the points in the interval and then at each of those points they will switch and I encourage you we don’t have a time of this. I’m pushing 24 25 minutes but I encourage you to graph that and verify and see for yourself that the graph really are concave upwards for all x lesson three in concave downward between negative three and zero and see what an inflection point looks like with this points look like.

Anyway, I’ll leave it there and I will see you in the next video.

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