Curve sketching with calculus: logarithm
Sal sketches a graph of f(x)=ln(x⁴+27) including extremum and inflection points. Created by Sal Khan.
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- Instead of saying ". . just less/greater than. . ."
Wouldn't it just be "just in that sign direction until the critical value?"
So at13:24, Instead of saying, "So when x is slightly less than zero. . " Would it be the same as saying, "When x is between 0 and -3. . . "?
(If there are no more critical points between the value and +/- infinity, then you would just say, "When x is between 0 and negative infinity. . "_
If not, I would like to know why.
- You are correct. I think Sal was just trying to stress the meaning of the inequality.
In general there are many ways to say the same thing. Some ways are more formal, some are more standard, some are easier to understand. A good mathematician is always thinking about the best way to say something just like you are.(11 votes)
- I don't completely understand why x = 0 is a minimum point. Sal's logic does make sense (~15:14) that since f"x is positive at both sides of 0, it is concave up all around and x = 0 is a minimum, but in the last video Sal said that in order to verify that a critical point is a min/max you should take the second derivative at that point. If the second derivative is positive it's a min, if it's negative it's a maximum and if it's zero it is neither but may be a potential inflection point. But, if you take the second derivative of zero, in this case, it is zero, so with that logic shouldn't it not be a minimum?(9 votes)
- The local max / min (also called "extrema") occur at the bottom or top of a curve. At a max, the curve stops increasing and changes direction to start decreasing (the slope switches from positive to negative). Thus the slope at that exact point where the change occurs is zero -- this is as high as the curve gets in that local area. For a min, it is the other way around, the curve stops decreasing, turns around and starts increasing (the slope switches from negative to positive). Thus the slope at that exact point is zero.
However, you can have a place where the slope is zero, but the curve does not turn around.
So, anywhere the slope is zero is a possible (but not certain) max or min. You have to check it to make sure. You can check by graphing, or you can check using the second derivative.(0 votes)
- If you found the third derivative of a function and you plug in your inflection point candidates into it and you get 0 for one of them, does that immediately tell you that that specific candidate is NOT an inflection point?(4 votes)
- No. If the third derivative is 0, the test in inconclusive. But if the third derivative exists and is non-zero at a point where the second-derivative is zero, then that point is an inflection point. If the third derivative is 0, you cannot tell one way or the other.(5 votes)
- If you have a set of possible inflection points, why pick numbers slightly above/below each, instead of any point in between? For example, if you have possible inflections at 0 and 3, why check for positive/negative at both 0 plus a little and 3 minus a little, instead of just checking f''(1) and using it for both?(3 votes)
- You do only need to check f ''(1). (That is, as long as your second derivative is continuous; if you have a discontinuity somewhere you have to consider that as a point where the concavity could change)(2 votes)
- I'm confused as to why when using the product rule [f'(x)g(x) + f(x)g'(x)] to calculate the second derivative. You first define the f(x) to be 4x^3 and g(x) to be (x^4 + 27)^-1. Where then, when calculating the second part of the product rule equation [f(x)g'(x)], does he get the second 4x^3? If f(x)=4x^3 and g'(x)= -1(x^4+27)^-2. Why does he multiply 4x^3 twice and then g'(x). Wouldn't that be f(x)f(x)g'(x)?(2 votes)
- Pay attention to the colours used. You get the second 4x^3 from the derivative of the inside of (x^4+27)^-1. g'(x) has to be -1(x^4+27)^-2*4x^3. Say, you have a function h(x)^-1. The derivative is -1*h(x)^-2*h'(x).(2 votes)
- I'm so confused.. at about11:00Sal decided to factor out 27·12x^2 - 4x^6 = 0 to 4x^2(27·3-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important?(2 votes)
- You forgot to divide the 324 in your final equation by 4. :-)(1 vote)
- Is it possible to figure out if 0 is an inflection point by writting a limit instead of just taking really close to zero numbers?(2 votes)
- No. That won't work for an inflection point. There are three requirements for an inflection point.
There is an inflection point at x=c if and only if:
1. f(x) is continuous at x=c
2. f''(c) = 0
3. f''(c-ε) has a different sign from f''(c+ε)
This test always works.
A few (not most) inflection points can be found by the first derivative test:
If f(x) is continuous at x=c ,
If f'(c) = 0
If f'(c-ε) has the same sign as f'(c+ε)
f(x) has an inflection point at x=c
In other words, if the first derivative is 0 at some point but it is not a max or min, then it is an inflection point. However, most inflection points cannot be found this way because they will be located at some point where the first derivative is not 0.(1 vote)
- Instead of calculating the double derivative for values just less than and just greater than some value, can't we find the triple derivative(the derivative of double derivative ) of that value to indicate if the concavity is going from upwards to downwards(if the triple derivative is negative) or from downwards to upwards(if the triple derivative is positive)?(2 votes)
- Yes that would be possible, but if you keep differentiating, you might end up with a function that is not continuously differentiable, and you could end up with a concavity of zero, which is not helpful. So sometimes it can help, but not always.(1 vote)
- We didn't actually have enough info in this video to estimate the graph. We can't be sure what happens on the ends without knowing more about X intercepts and Limits as x approaches positive and negative infinity.(2 votes)
- We need to find:
lim as x-> negative infinity (ln(x^4+27))
lim as x-> positive infinity (ln(x^4+27))
Since x^4 is always going to be a positive number, we can simplify:
lim as |x|-> infinity (ln(x^4+27))
We can also write this as:
lim as |x|-> infinity (ln(x^4+27)) = ln (lim as |x|-> infinity (x^4+27))
As |x| approaches infinity, the 27 becomes insignificant, so:
lim as |x|-> infinity (ln(x^4+27)) = ln (lim as |x|-> infinity (x^4))
So now we are calculating the natural logarithm values closer and closer to infinity. If you calculate the natural logarithm of greater and greater input values in a calculator, you'll see that the output values will also increase.
In other words, the value of ln(x^4) will increase as the value of x^4 increases.
lim as |x|-> infinity (ln(x^4+27)) = infinity
Based on this, the graph sketch in the video seems to be a good estimate of the actual graph, since the function that Sal drew appears to increase as x becomes more and more negative, or more and more positive.
I hope my explanation was clear enough.(1 vote)
- Isn't the domain of f(x) = ln(4x + 27) be greater than 0 and not real numbers like shown here... Since, natural log of any number <= 0 is not defined....
I will greatly appreciate if someone could clear my confusion...(1 vote)
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 all we want to do is take its first and second derivatives, and use as much 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 our answer matches up. So a good place to start is to take the first derivative of this. So 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 4x to the third, and then multiply it times the derivative of the outside, with respect to the inside. So the derivative of the natural log of x is 1over 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 rewatch the chain rule videos. But that's the first derivative of our function. I could rewrite this, this is equal to 4x to the third over x to the fourth, plus 27. Or I could write it as 4x to the third times x to the fourth, plus 27 to the negative 1. All three of these expressions are equivalent. I'm just the 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 the equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative, this looks like it'll 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, is the derivative of this first expression, times the second expression. So the derivative of this first expression, 3 times 4is 12. 12x 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 just the first expression, not its derivative, so just 4x to the third, times the derivative of the second expression. And the derivative of the second expression, we could take the derivative of the inside, which is just 4x to the third, the derivative of 27 is just 0, so times 4x to the third, times the derivative of this whole thing with respect to the inside. So times, so you take this exponent, put out front, so times minus 1, 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 multiply we're going to have a minus here, so it's a minus, you multiply these two guys, 4 times 4is 16, 16x 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 2, you just put in the denominator and make it into a positive 2 in the denominator. Same thing. Now, if you've seen these problems in the past, we always want to set these things equal to 0. We want to solve for x equals 0. 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 can do, is we could have a common denominator. So we could multiply both the numerator and 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 multiplied this numerator and this denominator by x to the the fourth plus 27. I didn't change it. And then we have that second term. Minus 16 x to the sixth over x to the fourth plus 27 squared. The whole reason why did that? Now I have a common denominator, now I can just add the numerators. So this is going to be equal to, let's see. 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. That's 12x 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 multiplied the 12x squared times the 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. 7x 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 sixth and plus 12x to the sixth. So you add those two, you get minus 4. 12 minus 15is minus 4, x 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 solve for when the first and the second derivatives equal 0, 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 our first derivative is equal to 0, and get our critical points. Or at least maybe, also maybe, where it's undefined. So this is equal to 0. If we want to set, if the only place that this can equal to 0 is if this numerator is equal to 0. 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, because it's an even exponent. So this thing can never equals 0, right, because you're adding 27 to something that's non-negative. So this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we could set the numerator equal to 0 pretty easily. If we wanted to set this equal to 0, we just say 4x to the third is equal to 0, and we know what x-value will make that equal to 0, x has to be equal to 0. 4 times something to the third is equal to 0, that something has to be 0. x to the third has to be 0, x has to be 0. So we can write, f prime of 0 is equal to 0. So 0 is a critical point. 0 is a critical point. The slope at 0 is 0. We don't know if it's a maximum or a minimum, or an inflection point yet. We'll explore it a little bit more. And actually, just so we get the coordinate, what's the coordinate? The coordinate x is 0, and then y is the natural log-- if x is 0, this just turns out, it's the a natural log of 27. Let me figure out what that is, I'll get the calculator out. I said I wouldn't use a graphing calculator, but I can use a regular calculator. So27, if I were to take the natural log of that, for our purposes let's just call it 3.3. We're just trying to get the general shape of the graph. So 3.3. Well, we could just say 2.9 and it kept going. So this is a critical point right here. The slope is 0 here. Slope is equal to 0 at x is equal to 0. 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 inflection points are where the second derivative equals 0. Now if the second derivative equals 0, that doesn't tell us that those are definitely inflection points. Let me make this very clear. If, let me do it in a new color. If x is inflection, then the second derivative at x is going to be equal to 0. Because you're having a change concavity. You have a change in the slope, goes from either increasing to decreasing or from decreasing to increasing. But if the derivative is equal to 0, the second derivative is equal to 0, you cannot assume that is an inflection point. So what we're going to do is, we're going to find all of the point at which this is true, and then see if we actually do have a sign change in the second derivative of 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 a second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go above or below that x, the second derivative has to actually change signs. Only then. So we can say, if f prime changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely going to be 0 right at x, but you have to actually see that if it's negative before x, has to be positive after x,or if 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 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So let's factor out a 4x squared. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x 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 21is 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 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. So let's solve this. This is going to be 0 when x is equal to 0, itself. This is going to be equal to 0 when x, let's see. If we add x to the fourth to 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 plus or minus three. So these are our candidate inflection points, x is equal to 0, x is equal to plus 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 label them inflection points. So what happens when x is slightly below 0? So let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them, necessarily, but if x is like 0.1. What is the second derivative going to be doing? If x is 0.1, 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 0, or just slightly less than 0, our second derivative is positive. Now what happens when x is slightly larger? When I write this notation, I want to be careful, I mean, really, just right below 0. Now when x is right above 0, what happens? Let's say x was 0.01, or 0.1, positive 0.1. Well, it's going to be the same thing. Because in both cases, we're squaring, and we're taking the fourth. So you're kind of losing your sign information. So if x is 0.1, 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 going to positive times a positive, so your second derivative is still going to be greater than 0. So something interesting here. f at your second derivative is 0 when x is equal to 0, but it is a not an inflection point. Because notice, the concavity did not change around 0. Our second derivative is positive as we approach 0 from the left, and it's positive as we approach 0 from the right. So in general, at 0, we're always, as we're near 0 from either direction, we're going to be concave upwards. So the fact that 0 is a critical point, and that we're always concave upward, as we approach 0 from either side, this tells us that this is a minimum point. Because we're concave upwards all around 0. So 0 is not an inflection point. Let's see if positive and negative 3 are inflection points. And if you study this equation, let me write our-- and actually, I just want to be clear. I've just been using the numerator of the second derivative. The whole second derivative 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 there 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 3. 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. That was our second derivative. Let's see if this changes signs around positive or negative 3. And actually, we should get the same answer, because regardless of whether we put positive or negative 3 here. you lose all your sign information because you're taking it to the fourth power, you're taking it to the second power. And obviously, anything to the fourth power is always going to be positive, anything to the second power is always going to be negative. So when we do our test, if it's true for positive 3, it's probably going to be true for negative 3 as well. But let's just try it out. So when x is just a little bit less than positive 3, what's the sign of f prime prime of x? So it's going to be 4 times 9, or it's going to be 4 times a positive number. It might be like 2.999, but this is still going to be positive. So this is going to be positive when x is approaching 3, and then this is going to be, well, if x is 3, this is 0, so x is a little bit less than 3. If x is a little bit less than 3, if it's like 2.9999, 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 3, is approaching from the left, we are concave upwards. This thing's going to be a positive. Then f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? Well, this first term is still going to be positive. But if x is just larger than 3, x to the fourth is going to be just larger than 81, and so this second term is going to be negative in that situation. Let me do it ina new color. It's going to be negative when x is larger than 3. 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 0, so we're going to be concave downwards. One last one. What happens when x is just a greater than minus 3? So just being greater than minus 3, that's like minus 2.99999. So when you take minus 2.99 square it, you're going to get a positive number, so this is going to be positive. And if you take minus 2.99 to the fourth, that's going to be a little bit less than 81, right? Because 2.99 to the fourth is a little bit less than 81, so this is still going to be positive. So you have a 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 0. Concave upwards. And then finally, when x is just, just less than negative 3, remember, when I write this down, I don't mean for all x's larger than negative 3,or all x's smaller than negative 3. There's actually no, well, I can't think of the notation that would say just, as we just approach three in this case, from the left, but what happens if we just go to minus 3.11? Or 3.01, I guess is a better one, or 3.1? Well, this term right here is going to be positive. But if we take minus 3.1 to the fourth, that's going to be larger than positive 81, right? The sign will become positive, it'll be larger than 81, so this'll 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. So I think we're ready to plot. so first of all, 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, the second derivative is 0. The second derivative's 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 change in the second derivative. So x is equal to 3. So 3 is definitely an 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. 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, 3.29, that this was a minimum. Because 0 was a critical point, the slope is 0 there, and because it's concave upwards all around 0. So 0 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-coordinates, we can just evaluate them. So they're actually going to have the same y-coordinates, because if you put a minus3 or positive 3 and take it to the fourth power, you're going to 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 want to take the natural log of it. Let's just say 4.7, just to get a rough idea. That's4.7. And that's true of whether 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. All right. Let me draw my axis, just like that. And this is my y-axis, this is my x-axis, this is y. You can even call it the f of x axis, if you like. This is x. And so the point 0, 3.29. Let's say this is 1, 2, 3, 4, 5, to the point 0, 3.29. That's 0, 1, 2, 3, a little bit above 3, it's right there. That's the minimum point. And then we're concave. The slope is 0 right there, we figured that out, because the first derivative was 0 there. So it's a critical point, and it's concave upwards around there. So that told us we arepoint at a minimum point, right there. And then at positive 3. So 1, 2, 3. At positive 3, 4.7. So 4.7 will look something like that. We have an inflection point. Before that, we're concave upwards, and then after that we're concave downwards. So it looks something like this. So we're concave upwards up with up to that point. Maybe, actually, you should, let me ignore that yellow thing I drew before. Let me get rid of that. Let me draw it like 1, 2, 3. 3, 4.7 looks like that, and minus 3, 4.7, 1, 2, 3, 4.7 looks like that. So we know at 0, we are slope of 0 and we're concave upwards, so we look like this. We're concave upwards, until x is equal to 3. And at x is equal to 3, we become concave downwards, and we go, let me try my best to draw it well, and we go off like that. And then we're concave upwards around 0, until we get, we're concave upwards as long as x is greater than minus 3, and then at minus 3 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 the red color-- this concave downwards right here, is this, right there. And then the concave upwards around 0 is right there. You could even imagine, this concave upwards that we measured, that's this, concave upwards, and then this concave upwards is that. And then around 0, we're always upwards. So this is my sense of what the graph will look like and maybe 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 grafted correctly using a graphing calculator. So let me get out 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 mathematics was correct. This was actually very satisfying. So hopefully you appreciate the usefulness of inflection points, and second derivative, and first derivative, in graphing some of these functions.