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Current time:0:00Total duration:20:31

Curve sketching with calculus: polynomial

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.10 (EK)
,
FUN‑4.A.9 (EK)

Video transcript

let's see if we can use everything we know about differentiation and connectivity and maximum and minimum points and inflection points to actually graph a function without using a graphing calculator so let's say our function let's say that f of X is equal to 3x to the fourth minus minus 4x to the 3rd plus 2 and of course you could always graph a function just by trying out a bunch of points but we want to really focus on the the points that are interesting to us and then just to get the general shape of the function especially we want to focus on the things that we can take out from this function using our calculus toolkit or our derivative toolkit so the first thing we probably want to do is figure out the critical points we want to figure out I'll write it here critical critical points and just as a refresher of what critical point means it's the point where F where the derivative of f of X is 0 so critical points are f prime of X is either equal to 0 or it's undefined or if undefined this function looks differentiable everywhere so the critical points that we worried about are probably well I can tell you they're definitely just the points where F prime of X are going to be equal to 0 this this this derivative F prime of X is going to actually be defined over the entire domain so let's actually write down the derivative right now so the derivative of this F prime F prime of X this is pretty straightforward the derivative of 3x to the 4th 4 times 3 is 12 12x to the it will just decrement the 4 by 1/3 right you just multiply times the exponent and then decrease the new exponent by 1 minus 3 times 4 is 12 times X to the 1 less than 3 is 2 and then the derivative of a constant the slope of a constant you can almost imagine is zero it's not changing a constant by definition isn't changing so that's F prime of X so let's figure out the critical points the critical points are where this thing is either going to be equal to zero or it's undefined now I could look over the entire domain of real numbers and this thing is defined pretty much anywhere I could put any number here and it's not going to blow up it's going to give me an answer to what the function is so that it's defined everywhere so let's just figure out where it's equal to zero so f prime of x is equal to zero so let's solve which X is so let's solve well I don't have to rewrite that I just wrote that let's solve for whether this is equal to zero and I'll do it in the same color so 12 X to the third minus 12 x squared is equal is equal to zero and so let's see what we can do what can we do to solve this well we could factor out a 12x so if we factor out a 12x then this term becomes just X and then actually let's look factor out a 12x squared we factor out a 12x squared if we divide both of these by 12x squared this term just becomes an X and then minus 12x squared divided by 12x squared is just 1 is equal to 0 I just rewrote this top thing like this you could go the other way if I multiply distributed this 12x squared times this entire quantity you would get my derivative right there so the reason why I did that is because to solve for 0 or if I want all of the X's that that make this equation equal to 0 we now have I now have written it in a form where I'm multiplying one thing by another thing and in order for this to be zero one or both of these things must be equal to 0 so 12x squared or equal to 0 which means that X is equal to 0 will make this quantity equal 0 and then the other thing that would make this quantity 0 is if X minus 1 is equal to 0 so X minus 1 is equal to 0 when X is equal to 1 so these are our two critical points our two critical points or X is equal to 0 and X is equal to 1 and remember those are just the points where our first derivative is equal to 0 where the where the slope is 0 they might be maximum points they might be minimum points they might be in flat points we don't know they might be you know this was a constant function they could just be anything so we really can't say a lot about them just yet there but they are points of interest I guess that's all we can say that they are definitely points of interest but let's keep going and let's try to understand the concavity and maybe we can get a better sense of this graph so let's let's figure out the second derivative so the second I'll do that in let me do it in this orange color so the second derivative of my function f let's see 3 times 12 is 36 x squared minus 24x so let's see well there's a couple of things we can do now that we know the second derivative we can answer the question is my graph concave upwards or downwards at either of these points so let's figure out what at either of these critical points and it'll all fit together remember if it's concave upwards then we're kind of in a u-shape if it's concave downwards then we're in a in a kind of upside-down U shape so F prime prime our second derivative at X is equal to 0 is equal to what it's equal to 36 0 squared minus 24 times 0 so that's just 0 so f prime prime so it's just equal to 0 so we're in either concave upwards nor concave downwards here it might be a transition point it may not if it is a transition point then we're dealing with an inflection point we're not sure yet now let's see what F prime prime or a second derivative evaluated it one is so that's 36 times 1 let me write it down that's equal to 36 times 1 squared which is just 36 minus 24 minus 24 times 1 so 36 minus 24 so it's equal to 12 so this is positive our second derivative is positive here it's equal to 12 which means our first driven creasing the rate of change of our slope is positive here so this at this point right here we are concave upwards upwards which tells me that this is probably a minimum point right the slope is 0 here but we are concave upwards at the points oh that's interesting so let's see if there are any other potential inflection points here we already know that this is this is a potential inflection point let me circle it in red it's a potential inflection point we don't know whether our function actually transitions at that point we'll have to experiment a little bit to see if that's really the case but let's see if there are any other inflection points so to figure that or potential inflection points so let's see if this equals zero anywhere else so 36 x squared minus 24 X is equal to zero let's solve for X let us factor out a well you can factor our 12 X 12 X times 3x right 3x times 12x is 36 x squared minus 2 is equal to 0 so this is these two are equivalent expressions if you multiply this out you'll get this thing up here so this thing is going to be equal to 0 either of 12x is equal to 0 so 12 X is equal to 0 that gives us X is equal to 0 so x equals 0 this thing equals 0 so the second the second derivative is 0 then we already knew that because we tested that number out or this thing if this expression was 0 then the entire second derivative would also be 0 so let's write that so 3x minus 2 is equal to 0 3x is equal to 2 just adding 2 to both sides 3x is equal to 2/3 so this is another interesting point that we haven't really hit upon before that might be an inflection point the reason why I say it might be is because the second derivative is definitely 0 here you put 2/3 here you're going to get 0 and what we have to do is see whether the second derivative is positive or negative on either sides of of 2/3 and we already have a sense of that I mean we could try out a couple of numbers we know that you know if we take X's if we say that X is greater than 2/3 let me scroll down a little bit so so if we have some space so let's see what happens when X is greater than 2/3 X is greater than 2/3 what is f prime of X so let's see all right sorry what is f prime prime what is the second derivative so let's try out a value that's pretty close just to get a sense of things just to get a sense of things so let me rewrite it F prime prime of X is equal to let me write it like this I mean I could write it like this but this might be easier to deal with it's equal to 12x times 3x minus 2 so if X is greater than two-thirds this term right here is going to be positive that's definitely you know any it's any positive number times 12 is going to be positive but what about this term right here three times two-thirds minus two is exactly zero right that's two minus two but anything larger than that three times you know if I had to point one thirds right so if here this is going to be a positive quantity any value of x greater than two-thirds will make this thing right here positive right this thing is also going to be positive so that means that when X is greater than two-thirds that tells us that the second derivative is positive it is greater than zero so in our domain as long as X is larger than two-thirds we are concave upwards and we saw that here at F at X is equal to one we were concave upwards but what about X being less than two-thirds so when X is less than 2/3 let me write it let me scroll down a little bit when X is less than 2/3 what's going on I'll rewrite it F prime prime of X second derivative 12x times 3x minus 2 well if we go you know really far less we're going to get a negative number here and this might be neighbor but let's see if we just could really like you know right below two-thirds when we're still in the positive domain so you know if this was like 1.9 thirds which is you know mixture of a decimal in a fraction or even one-third this thing is still going to be positive right below two-thirds this thing is still going to be positive we're going to be multiplying 12 by a positive number but what's going to be going on what's going on right here what's going on right here at two-thirds we're exactly zero but as you go is anything less than two-thirds three times one-third is only one one minus two you're going to get negative numbers so when X is less than 2/3 this thing right here is going to be negative so the second derivative if X is less than two-thirds the second derivative right right to the left or right when you go less than two-thirds the second derivative of X is less than zero now the fact that we have this transition from when we're less than two-thirds we have a negative second derivative and when we're greater than two-thirds we have a positive second derivative that tells us that this indeed is an inflection point that X is equal to two-thirds is definitely an inflection point for our regional function up here now we have one more candidate inflection point and then we're ready to graph then you know once you know all the inflection points in the maximum minimum you are ready to graph the function so let's see if X is equal to zero is an inflection point we know that the second derivative is zero at zero but what happens above and below the second derivative so let's let me do our little test here so when X is let me draw a line so I don't so we don't get confused with all of the stuff that I wrote here so when X is greater than zero what's happening the second derivative remember the second derivative was equal to 12x times 3x minus two I like writing it this way because you've kind of decomposed it into two linear expressions and you can see whether each of them are positive or negative so if X is greater than zero this thing right here is definitely going to be positive and then this thing right here right when you go right above X is greater than zero so we have to make sure to be very close to this number right so this number is like point let's say it's point one right so you know you're right above zero so this is going to be true for all of X greater than zero we just want to test exactly what happens right when we go right above zero so this was point one you would have 0.3 0.3 minus 2 that would be a negative number right so right as X goes right above zero this thing right here is negative so as X is greater than 0 you will have your second derivative is going to be less than 0 you're concave downwards which makes sense because we're at some point we're going to be hitting a train remember where we were concave downwards before we got to two-thirds right so this is consistent from zero to two-thirds we are concave downwards and then at two-thirds we become concave upwards now let's see what happens when X what X is right less than when X is just barely just barely less than zero so once again F Prime the second derivative of X is equal to 12x times 3x minus 2 well right if you know if X was minus 0.1 or minus 0.0001 no matter what this thing is going to be negative this expression right here is going to be negative the 12x right you have some negative value here times 12 it's going to be negative and then what's this going to be well three times minus 0.1 is going to be minus 0.3 minus 2 is minus two point three you're definitely going to have a negative this value right here is going to be negative and then when you subtract from a negative it's definitely going to be negative so that is also going to be negative but if you multiply a negative times a negative you're gonna get a positive so actually right below X is less than 0 the second derivative is positive the second derivative is positive now this one might have been a little bit confusing but we've we should now have the payoff we now have the payoff we have all of the interesting things going on we know that it X is equal to 1 we know that X is equal to 1 let me write it over here and that way we figured out at X is equal to 1 the slope is 0 so f prime prime is sorry let me write it this way I should have said we know that the slope is zero slope is equal to 0 and we figured that out because the first derivative was 0 this was a critical point and we know that we're dealing with the the function is concave upwards at this point and that tells us that this is going to be a minimum point so what is f of and we should actually get the coordinates so we can actually graph it that was the whole point of this video so and F of 1 is equal to what F of 1 let's go back to our original function is 3 times 1 right one to the fourth is just 1 3 times 1 minus 4 plus 2 right so 4 times so it's it's it's well 3 times 1 minus 4 times 1 which is minus 1 plus 2 well that's just a positive 1 so f of 1 is 1 and then we know that X is equal to 0 we also figured out that the slope is equal to 0 but we figured out that this was an inflection point right the the connectivity switches before and after so this is an inflection point inflection point and we are concave below 0 below 0 so when X is less than 0 we are upwards we are upwards our second derivative is positive and when X is greater than 0 we are downwards we are concave downwards right above not for all of the domain X and 0 just right above 0 downwards and then what is f of 0 just so we know we want to graph that point F of 0 let's see F of 0 this is easy 3 times 0 minus 4 times 0 plus 2 well that's just 2 f of 0 is 2 and then finally we got the point X is equal to 2/3 X is equal to let me do that in another color we had the point X is equal to 2/3 we figured out that this was a inflection point the slope isn't necessarily lower the slope definitely isn't 0 there because it wasn't one of the critical points and we know that we are downwards we know that when X is less than 2/3 or right lessons 2/3 we are concave downwards and when X is greater than 2/3 we saw it up here when X was greater than 2/3 right up here we were concave upwards the second derivative was positive we were upwards and we could actually figure out what's F of 2/3 that's actually a little bit complicated we don't even have to figure that I don't think to graph I think we could do a pretty good job of graphing it just with what we know right now so that's our takeaways let me do a rough graph let's see so let me do my axes just like that so let's see we're going to want to graph the point zero two so let's say that the point zero two so this is X is equal to zero and we go up one two so this is the point zero two maybe I'll do it in that color the color I was using so that's this color so that's that point right there then we have the point X we have F of one which is the point 1 1 right so this point right here is will go up 1 so that's the point 1 1 this was the point 0 2 and then we have the X is equal to 2/3 which is our inflection point so when X is 2/3 we don't know exactly what number F of 2/3 is might be here someplace let's say F of 2/3 is right there so that's the point 2/3 and then whatever F of 2/3 is it looks like it's going to be 1 point something F of 2/3 you could calculate it if you like you just have to substitute back in the function but we're ready to graph this we're ready to graph this thing so we know that it X is equal to 1 the slope is 0 we know that the slope is 0 it's flat here and we know it's concave upwards so we're dealing it looks like this it looks like that over that interval we're concave upwards and we know we're concave upwards from X is equal to 2/3 and on right front let me do it in that color we knew X you 2/3 and on we're concave upwards and so that's why I was able to draw this u-shape now we know that when X is less than 2/3 and greater than 0 we're concave downwards so the graph would look something like this over this interval will be concave downwards let me draw it nicely over this interval the slope is decreasing and you could see it if you keep drawing tangent lines it's flattish there it gets negative more negative more negative more in it until the inflection point and then it starts increasing again because we go back to concave upwards and then finally the last interval is below 0 and we know below 0 when X is less than 0 we're concave upwards so the graph looks like this well graph looks like that and we also know that X is equal to zero was a critical point the slope was zero so this graph is actually flat right there - so this is an inflection point where the slope was also zero so this is our final graph we're done after all that work we were able to use our calculus skills and our knowledge of inflection points and concavity and transitions and concavity to actually graph this fairly fairly hairy looking graph but this this should be kind of what it looks like if you graph it on your calculator