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# Curve sketching with calculus: logarithm

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

## Video transcript

let's say we've got the function f of X is equal to the natural log of X to the fourth plus 27 and what we want to do is take its first and second derivatives and use as much as we of our techniques as we have at our disposal to attempt to graph it without a graphing calculator if we have time I'll take out the graphing calculator and see if it if our answer matches up so a good place to start is to take the first derivative of this let me do that over here so the derivative of F well you take the derivative of the inside so take the derivative of that right there which is 4 X to the third and then multiply it times the derivative of the outside with respect to the inside so the derivative of natural log of X is 1 over X so the derivative of this whole thing with respect to this inside expression is going to be so times 1 over X to the fourth plus 27 if you found that confusing you might want to re-watch the chain rule videos but that's the first derivative of our function and I could rewrite this this is equal to 4 X to the third over X to the fourth plus 27 or I could write it as 4 X to the third times X to the fourth plus 27 to the negative 1 all three of these expressions are equivalent I'm just writing I multiplied it out or I could write this as a negative exponent or I could write this as a fraction with this in the denominator they're all equivalent so that's our first derivative let's do our second derivative our secondary this this looks like it'll get a little bit this will get a little bit hairier so our second derivative is the derivative of this so it's equal to we can now use the product rule it's the derivative of this first expression times the second expression so the derivative of this first expression 3 times 4 is 12 12 x squared right we just decrement the 3 by 1 times the second expression times X to the fourth plus 27 to the minus 1 and then to that we want to add we want to add just the first expression not as derivative so four X to the third time's the derivative of the second expression and it Rizza the second expression we could take the derivative of the inside which is just four X to the third this derivative of 27 is just zero so this so times four X to the third times the derivative of this whole thing with respect to the inside so times so you get take this exponent put it out front so x minus one times this whole thing X to the fourth plus 27 to the we decrement this by one more so minus two so let's see if I can simplify this expression a little bit so this is equal to so this right here is equal to 12x squared over this thing X to the fourth plus 27 and then let's see if we multiple gonna have a minus here so it's minus you multiply these two guys four times four is 16 16 X to the third times X to the third is X to the sixth over this thing squared over X to the fourth plus 27 squared that's just another way to rewrite that expression right there right to the minus two you could just put in the denominator and make it to a positive two in the denominator same thing now if you you've seen these problems in the past we always want to set these things equal to zero we want to solve for x equals zero so it'll be useful to have this expressed as just one fraction instead of the difference or the sum of two fractions so what we could do is to have we could have a common denominator so we could multiply both the numerator denominator of this expression by X to the fourth plus 27 and what do we get so this is equal to so if we multiply this first expression times X to the fourth plus 27 we get 12x squared times X to the fourth plus 27 and then in the denominator you have X to the fourth plus 27 squared all I did I multiply this numerator and this denominator by X to the fourth plus 27 I didn't change it and then we have that second term minus 16x to the sixth over X the fourth plus 27 squared the whole reason why I did that now I have a common denominator now I can just add the numerators so this is going to be equal to this is going to be equal to let's see let's melt well the denominator we know what the denominator is it is X to the fourth plus 27 squared that's our denominator and then we can multiply this out this is 12x squared times X to the fourth so that's 12 X to the sixth plus 27 times 12 I don't even feel like multiplying 27 times 12 so I'll just write that out so plus 27 times 12x squared I just multiply the 12x squared times this 27 and then minus 16x to the sixth minus 16x to the sixth and this simplifies to let's see if I can simplify this even further so I have an X to the sixth here X to the sixth here so this is equal to do this in pink this is equal to the 27 times 12x squared I don't feel like figuring that out right now times 12x squared and then you have minus 16x to the 6 and plus 12x to the 6 so you add those two you get minus 4 12 minus 16 is minus 4x to the sixth all of that over X to the fourth plus 27 plus 27 squared and that is our second derivative now we've done all of the derivatives and this was actually a pretty hairy problem and now we can set we can solve for when the first and the second derivative equals zero and we'll have our candidate well we'll know our critical points and then we'll have our candidate inflection points and see if we can make any headway from there so first let's see where where our first derivative is equal to zero and get our critical points or at least maybe also maybe we're there where it's undefined so this is equal to zero if we want to set if this the only place that this can equal to zero is if this numerator is equal to zero this denominator actually if we are assuming we're dealing with real numbers this term right here is always going to be greater than or equal to 0 for any value of x cuz it's an even exponent so this thing can never equal zero right because you're adding 27 to something that's non-negative so this will never equal zero so this will also never be undefined so there's no there's no undefined critical points here but we could set the numerator equal to zero pretty easily if we wanted to set this equal to zero we just say 4x to the third is equal to zero and we know what x value will make that equal to zero X has to be equal to zero four times something 2/3 is equal to zero that something has to be zero X to the third has to be zero X has to be 0 so we can write F prime of zero is equal to zero so zero is a critical point zero is a critical point create a goal point the slope at zero is zero we don't know if it's a maximum or a minimum or an inflection point yet or it I mean it could be you know though well well we'll explore it a little bit more and actually just so we get the coordinate what's the coordinate the coordinate X is zero and then Y is the natural log of X is zero this is just turns out so so natural log of 27 so it's the natural log of let me to figure out what that isn't get the calculator out I said I wouldn't use a graphing calculator but I could use a regular calculator so 27 if I were to take the natural log of that it's like three point well for our purpose this is call it three point three we're just trying to get the general shape of the graph so three point three three point well we could just say two nine and it keep it kept going so this is a critical point right here the slope is zero here slope is equal to zero at X is equal to zero so this is one thing we want to block off and let's see if we can find any candidate inflection points and remember candidate and inflection points are where the second derivative now the second derivative equals zero that doesn't tell us that those are definitely inflection points let me make this very clear if let me write this let me do it in a new color if X is inflection inflection then then the second derivative at X is going to be equal to zero because you're having a chain the concavity I always say conductivity but it's the concavity you have a change in the slope goes from either increasing the decreasing or from decreasing to increasing but if the if the derivative is equal to zero you cannot the second derivative is equal to zero you cannot assume that is an inflection point so what we're going to do is we're going to find all of the points at which this is true and which this is true and then see if we actually do have a sign change in the second derivative at that point and only if you have a sign change then you can say it's an inflection point so let's see if we can do that so just because the second derivative zero that by itself does not tell you it's an inflection point it has to have a second derivative zero and when you go above or below that X you have to the the second derivative has to actually change signs only then so we can say if f prime changes changes signs around around X then we can say that X is an inflection X is inflection and if there's changing signs around X then it's definitely going to be zero right at X but you have to actually see that if it's negative before X has to be positive after X or it's positive before X has to be negative after X so let's test that out so the first thing we need to do is find these candidate points remember the candidate points are where the second derivative is equal to zero we're gonna find those points and then see if this is true that the sign actually changes so we want to find where this thing over here is equal to zero and once again to for this to be equal to zero the numerator has to be equal to zero this denominator can never be equal to zero if we're dealing with real numbers which i think is a fair assumption so let's see where this where this where the where the our numerator can be equal zero for the second derivative so let's set the second derivative the numerator of the second derivative twenty-seven times twelve x squared minus four X to the sixth is equal to zero remember that's just a numerator of our second derivative any X that makes the numerator zero is making the second derivative zero so let's factor out a let's factor out a four x squared so four x squared now we'll have 27 times if we factor 4 out of the 12 we will just get a 3 and we factored out the x squared minus we factored out the 4 we factor out an x squared so we have X to the fourth is equal to 0 so the X's that will make this equal 0 either well will satisfy either I'll switch colors either 4 x squared is equal to 0 or now 27 times 3 I can do that in my head that's 81 20 times 3 is 60 7 times 3 is 21 60 plus 21 is 81 or 81 - minus X to the fourth is equal to 0 any X that satisfies either of these will make this entire expression equal to 0 because if this thing is 0 the whole thing is going to be equal to 0 if this thing is 0 the whole thing is going to be equal to 0 let me be clear this is 81 right there let's solve this this is going to be 0 when X is equal when X is equal to 0 itself this is going to be equal to 0 when X let's see if we add X - 4 at the both sides you get X to the fourth is equal to 81 if we take the square root of both sides of this you get x squared is equal to 9 or so you get X is plus or minus 3 X is equal to X is equal to plus or minus 3 so these are our candidate inflection points X is equal to 0 X is equal to +3 or X is equal to minus 3 so what we have to do now is to see whether the second derivative changes signs around these points in order to be able to lay them they label them inflection points so what happens when X is slightly below 0 so let's say let's take the situation let's do all the scenarios what happens when X is slightly below 0 not all of them necessary but if you're like X is like 0.1 what is the second derivative going to be doing if X is point 1 this or if X is minus 0.1 this term right here is going to be positive and then this is going to be 81 minus 0.1 to the fourth so that's going to be a very small number right so it's going to be some positive number times 81 minus a small number so it's going to be a positive number so when X is less than zero or just slightly less than zero our second derivative our second derivative is positive now what happens when X is slightly larger this isn't when I write this notation I want to be careful I mean really just write below zero now when X is right above zero what happens let's say X was point O one or point one positive point one what's going to be the same thing because in both cases were squaring and we're taking the fourth so you're kind of losing your sign information so if X is point one this thing is going to be a small positive number you're going to be subtracting a very small positive number from 81 but 81 minus a small number is still going to be positive so you're gonna have a positive times a positive so your second derivative is still going to be greater than zero so something interesting here at your second derivative is zero when X is equal to zero but it is not an inflection point because notice the concavity or sorry the concavity did not change around zero our second derivative is positive as we approach zero from the left and it's positive as we approach zero from the right so in general at zero we're always as we as we need as we're near zero from either direction we're going to be concave upwards so the fact that the fact that 0 is a critical point and that we're always concave upward as we approach zero from either side this tells us that this is a minimum point minimum minimum point right there because we're concave upwards all around zero so that's our so zero is not an inflection point let's see if positive and negative 3 are inflection points and if you study this equation let me let me write R and actually I just want to be clear I've just been using the numerator of the second derivative the whole second derivative is is this thing right here but I've been ignoring the denominator because the denominator is always positive so if we're trying to understand whether things are positive or negative we just really have to determine whether the numerator is positive or negative because this expression right here is always positive it's something to the second power so let's test whether we have a change in concavity around X is equal to positive or negative three so remember the numerator of our let me just rewrite our second derivative just so you see it here F prime prime of X the numerator is this thing right here it's 4x squared times 81 minus X to the fourth and the denominator was up here X to the fourth plus 27 squared X to the fourth plus 27 squared that was our second derivative let's see if if this changes signs around positive or negative three and actually we should get the same answer because regardless of whether we put positive or negative three here you lose all your sign information because you're taking it to the fourth power you're taking to the second power and obviously anything to the fourth power is always going to be positive anything to the second power it's always going to be negative so kind of when we do our test if it's true for positive three is probably going to be true for negative three as well but let's just try it out so when X is just a little bit less than positive three what's the sign of F prime prime of X so it's going to be 4 x times 9 so that you know it's going to be 4 times a positive numbers it might be like two point nine nine nine but this is still going to be positive so this is going to be positive when X is approaching three and then this is going to be well if X is 3 this is zero so if X is a little bit less than three if X is a little bit less than three if it's like two point nine nine nine nine this number is going to be less than 81 so this is also going to be positive and of course the denominator is always positive so as X is less than three is approaching from the left we are concave upwards this thing is going to be a positive then F prime prime is greater than 0 we are upwards concave upwards when X is just larger than three what's going to happen well this first term is still going to be positive but if X is just larger than three X to the fourth is going to be just larger than 81 and then so this second term is going to be negative in that situation it's going to be negative let me do it a new color it's going to be negative when X is larger than three because this is going to be larger than 81 so if this is negative and this is positive then the whole thing is going to be negative because this denominator is still going to be positive so then F prime prime is going to be less than zero so we're going to be concave downwards one last one what happens when X is just greater than -3 so just being greater than -3 that's like minus 2 point 9 9 9 9 9 so when you take minus 2 point 9 9 and square it you're gonna get a positive number so this is going to be positive and if you take minus 2 point 9 9 to the fourth that's going to be a little bit less than 81 right cuz 2.99 to the fourth is a little bit less than any one so this is still going to be positive so you have positive times a positive divided by a positive so you're going to be concave upwards because your second derivative is going to be greater than zero concave upwards and then finally when X is just just less than negative 3 remember when I write this sum I don't mean for all X's a larger than negative 3 are all X is smaller than negative 3 I'm really I there's actually no well I can't think of the notation that would say just as we just approached 3 in this case from the left but what happens if we just go to minus 3 point 1 1 or you know 3.01 I guess is a better one or 3 point 1 well this term right here is going to be positive but if we take minus 3 point 1 to the fourth that's going to be a that's going to be larger than positive 81 right the sign will become positive it'll be larger than 81 so this will become negative so in that case as well we'll have a positive times a negative divided by a positive so then our second derivative is going to be negative and so we're going to be downwards downwards so I think we're ready to plot so first of all our is X plus or minus 3 inflection points sure as we approach X is equal to 3 from the left we are concave upwards and then as we cross 3 we the second derivative is 0 the second derivative 0 I lost it up here the second derivative is 0 and then as we go to the right of 3 we become concave downwards so we got our sign the second derivative so X is equal to 3 so this 3 is definitely inflection point and the same argument could be made for negative 3 we switch signs as we cross 3 so these definitely are inflection points inflection just so we get the exact coordinates let's figure out what F of 3 is or F of positive and negative 3 and then we're ready to graph so first of all we know that F we know that the point 0 comma 3 point 2 9 that this was a minimum because 0 was a critical point the slope is 0 there and because it's concave upwards all around zero so the zero is definitely not an inflection point and then we know that the points positive 3 and minus 3 are inflection points and in order to figure out their y-coordinate so we can just evaluate them so they're actually gonna have the same y-coordinates because if you put a minus 3 or positive 3 and take it to the fourth power you're gonna get the same thing let's figure out what they are so if we take 3 to the fourth power that's what 81 81 plus 27 is equal to 108 and then we won't take the natural log of it I'm gonna take the natural log so it's like 4 point let's just say four point seven just to get a rough idea four point seven and that's true hardly we do positive or negative 3 because we took to the fourth power so it's 4.7 4.7 these are both inflection points and we should be ready to graph it let's graph it alright let me draw my axis let me draw my axis just like that then this is my y-axis so this is my x axis this is why you can even call it the f of X axis if you like this is X and so the point zero 3 point 2 9 so let's say this is 1 2 3 4 5 so the point 0 3 point 2 9 so it's 0 1 2 3 a little bit above 3 is right there that's the minimum point and then we're concave the slope is 0 right there we figured that out because it was a the first derivative was 0 there so it's a critical point and it's concave upwards around there so that told us we're at a minimum point right there and then at positive 3 so 1 2 3 at positive 3 4 point 7 so four point seven will look something like that we have an inflection point before that before that we're concave upwards and then after that were concave downward so it'll look something like this so we're concave upwards up to that point maybe actually you should let me ignore that yellow thing I drew before that let me get rid of like that so let me draw it like one two three three four point seven looks like that and minus three four point seven one two three four point seven looks like that so we know it's zero we are we have a slope of zero and we're concave upwards so we look like this we're concave upwards until X is equal to three and then at x equals three we become concave downwards we've become concave downwards and we go let me drum try my best to draw it well and we go off like that and then we're concave upwards around zero until we get we're concave upwards as long as X is greater than minus three and then at minus three we become concave downwards again maybe I should do it in that color this concave downwards right here that's this right here that's that right there and this concave downwards right here sorry I meant to do it in that red color this concave downwards right here is this right there and then the concave upwards around zero is right there or you could even imagine this concave upwards that we measured that's this concave upwards and did this concave upwards is that and then around zero we're always upwards so this is my sense of what the graph will look like and maybe it'll it'll just you know it turns into well you could think about what it does is X approaches positive or negative infinity some of the terms well I won't go into that but let's test whether we've graphed it correctly using a graphing calculator so let me get out my my ti-85 trusty ti-85 and let's graph this sucker all right press graph y equals the natural log of X to the fourth plus 27 all right I want to hit that graph there so I do second graph and let's cross our fingers it looks pretty good it looks almost exactly like what we drew so I think our I think our our mathematics was correct so this was actually very satisfying so hopefully under you appreciate the usefulness of inflection points and second derivatives and first derivatives when you in graphing some of these some of these functions