AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 5Lesson 9: Sketching curves of functions and their derivatives
Curve sketching with calculus: polynomial
Sal sketches a graph of f(x)=3x⁴-4x³+2 including extremum and inflection points. Created by Sal Khan.
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- How can the slope be zero if the graph is concave upwards? My brain hurts :((29 votes)
- Concave up means that the slope is increasing. If the slope was below zero and now it's zero, it has increased because 0 is a higher number than a negative number.(52 votes)
- How do I know if a function is differentiable? And does anyone know if there any videos on Asymptotes?
Thank youuu(52 votes)
- Asymptotes can be understood from the "conic sections:Hyperbola" videos in the algebra playlist(33 votes)
- what is the difference between a transition point and an inflection point?? I seem to have missed out on some of the things that Sal mentioned in the earlier lectures... Please help!(20 votes)
- There is no difference in this case. An inflection point (or point of inflection) is the point at which the concavity of the graph changes sign. In this case, the second derivative test is inconclusive, meaning that we must use a difference scheme to determine if x = 0 is in fact an inflection point.(29 votes)
- 6:40(I guess). Sal never tests 12 as an inflection point. He justifies it being a minimal point by saying its positive and thus concave upwards, however by the same logic 2/3 could be assumed a minimal point. I take it he doesn't test 12 as 12.0...1 and 11.0...1 plugged into 12x(3x-2) is always positive, thus a test was technically done and always should be?(7 votes)
- Are you thinking that 12 might be an inflection point because 12 was the coefficient on one of the factors? The only points that are critical points are those values, when substituted for x, cause the derivative to go to zero. 12 is a coefficient on one of the terms in the derivative, but setting x to 12 will NOT cause the second derivative to be zero.(11 votes)
- Why did Sal need to take the second derivative to find the inflection points? Couldnt he have used the first derivative? Im a bit confused.(6 votes)
- Inflection points are where the first derivative has relative max/mins (where the slope of the tangent line of the first derivative =0). He could have used the first derivative but not easily if he did it analytically. You can find points of inflection by looking at the graph of the first derivative, or by solving the 2nd derivative. (At least as far as I know...)(4 votes)
- how does a function look like (in graph) when the slope of a function is 0 and acceleration is 0?
I just cannot visualize it. ;o;(2 votes)
- A function that 0 slope and 0 concavity (acceleration) is just a horizontal line.(10 votes)
- Instead of the lengthy discussion starting at9:00, could the third derivative have been used at 2/3 to determine that the second derivative changes sign in the way discussed?(3 votes)
- I know this is a year late but, for future viewers, yes you could have. If the third derivative at 2/3 is not zero, you would know that it's an inflection point. Furthermore, if f'''(2/3) was positive (which it was) you would know that the slope is concave downwards below that point and concave upwards above that point. (You would know that the slope of the second derivative at that point is positive)(7 votes)
- How can I evaluate the derivative of a function at an indicated point?
Please help!(3 votes)
- You just take the derivative of that function and plug the x coordinate of the given point into the derivative.
So say we have f(x) = x^2 and we want to evaluate the derivative at point (2, 4). We take the derivative of f(x) to obtain f'(x) = 2x.
Afterwards, we just plug the x coordinate of (2,4) into f'(x). So, what we get is basically f'(x) = 2*2 which equals 4, and that's the slope of the original f(x) at the point (2, 4). The y coordinate, if they give you one, is just extraneous information, it's not needed in this problem.(6 votes)
- Guys I need some help with this: if you graph ((x+1)*lnx)/(x-1), the graph shows that when x=1, y=2, but why is it not undefined? This is so confusing for me.(2 votes)
- That's because the discontinuity is a "point discontinuity" or a "removable discontinuity". Graphical programs usually don't display this kind of discontinuities.(4 votes)
- how do you get the x and y intercept?(2 votes)
- To find the y-intercept, you make all x-values equal to 0 and solve for y. Inversely, to fine the x-intercept, you make all y-values equal to 0 and solve for x.(4 votes)
Let's see if we can use everything we know about differentiation and concativity, 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 4x to the third 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 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 here, critical points. And just as a refresher of what critical points means, it's the points 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. 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 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 of x, this is pretty straightforward. The derivative of 3x to the fourth, 4 times 3 is 12, 12x to the, we'll just decrement the 4 by 1, 3. Right? You just multiply times the exponent, and then decrease the new exponent by one, 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 could 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 0, or it's undefined. Now, I can 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 0. So f prime of x is equal to 0. So let's solve, which x is, let's solve-- I don't have to rewrite that, I just wrote that. Let's solve for when this is equal to 0. And I'll do it in the same color. So 12x to the third minus 12x squared is equal to 0. And so let's what we can do to solve this. We could factor out a 12x. So if we factor out a 12x, then this term becomes just x, and then-- actually, let's 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 distributed this 12x squared times this entire quantity, you would get my derivative right there. So the reason why did that is because, to solve for 0, or if I want all of the x's that make this equation equal to 0, I now have written it in a form where I'm multiplying one thing by another thing. And in order for this to be 0, one or both of these things must be equal to 0. So 12x squared are equal to 0, which means that x is equal to 0 will make this quantity equals 0. And 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 2 critical points. Our 2 critical points are 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 slope is 0. They might be maximum points, they might be minimum points, they might be inflection points, we don't know. They might be, you know, if this was a constant function, they could just be anything. So we really can't say a lot about them just yet, 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 concativity, and maybe we can get a better sense of this graph. So let's figure out the second derivative. I'll do that in this orange color. So the second derivative of my function f, let's see, 3 times 12 is 36x 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 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 is just equal to 0. We're neither 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, our second derivative, evaluated at 1 is. So that's 36 times 1, let me write it down, that's equal to 36 times 1 squared, which is 36, minus 24 times 1. So it's 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 derivative, our slope is increasing. The rate of change of our slope is positive here. So at this point right here, we are concave upwards. Which tells me that this is probably a minimum point, right? The slope is 0 here, but we are concave upwards at that point. So that's interesting. So let's see if there any other potential inflection points here. We already know that 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 any other inflection points, or potential inflection points. So let's see if this equals 0 anywhere else. So 36 x squared minus 24 x is equal to 0. Let's solve for x. Let us factor out, well, we can factor out 12x. 12x times 3x, right, 3x times 12x is 36x squared, minus 2, is equal to 0. So 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 if 12 x is equal to 0, so 12 x is equal to zero, that gives us x is equal to zero. So at x equals 0, this thing equals 0. So the second derivative is 0 there, and 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 zero. 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 is it might be, is because the second derivative is definitely 0 here. You put 2/3 here, you're going to get 0. So what we have to do, is see whether the second derivative is positive or negative on either side of 2/3. We already have a sense of that. I mean, we could try out a couple of numbers. We know that, you know, if we say that x is greater than 2/3. Let me scroll down a little bit, just so we have some space. So let's see what happens when x is greater than 2/3, 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. So let me rewrite it. f prime prime of x is equal to, let me write like this. I mean, I could write like that, but this might be easier to deal with. It's equal to 12x times 3x minus 2. So if x is greater than 2/3, this term right here is going to be positive. That's definitely, any positive number times 12 is going to be positive. But what about this term, right here? 3 times 2/3 minus 2 is exactly 0, right? That's 2 minus 2. But anything larger than that, 3 times, you know, if I had 2.1/3, this is going to be a positive quantity. Any value of x greater than 2/3 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 2/3, that tells us that the second derivative is positive. It is greater than 0. So in our domain, as long as x is larger than 2/3, we are concave upwards. And we saw that here, at x is equal to 1. We were concave upwards. But what about x being less than 2/3? 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 really far left, we're going to get a negative number here, and this might be negative. But if we just go right below 2/3, when we're still in the positive domain. So if this was like 1.9/3, which is a mix of a decimal and a fraction, or even 1/3, this thing is still going to be positive. Right below 2/3, this thing is still going to be positive. We're going to be multiplying 12 by a positive number. But what's going on right here? At 2/3, we're exactly 0. But as you go to anything less than 2/3, 3 times 1/3 is only 1. 1 minus 2, 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 2/3, the second derivative, right to the left, right when you go less than 2/3, the seconds derivative of x is less than 0. Now the fact that we have this transition, from when we're less than 2/3, we have a negative second derivative, and when we're greater than 2/3, we have a positive second derivative, that tells us that this, indeed, is an inflection point. That x is equal to 2/3 thirds is definitely an inflection point for our original function up here. Now, we have one more candidate inflection point, and then we're ready to graph. Then, you know, once you do all the inflection points and the max and the minimum, you are ready to graph the function. So let's see if x is equal to 0 is an inflection point. We know that the second derivative is 0 at 0. But what happens above and below the second derivative? So let me do our little test here. So when x is, let me draw a line so we don't get confused with all of the stuff that I wrote here. So when x is greater than 0, what's happening in the second derivative? Remember, the second derivative was equal to 12x times 3x minus 2. I like writing it this way, because you've kind of decomposed it into two linear expressions, and you could see whether each of them are positive or negative. So if x is greater than 0, 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 0, so we have to make sure to be very close to this number, right? So this number, let's say it's 0.1. You're right above 0. So this isn't going to be true for all of x greater than 0. We just want to test exactly what happens, right when we go right above 0. So this is 0.1. You would have 0.3, 0.3 minus 2, that would be a negative number, right? So right as x goes right above 0, this thing right here is negative. So at 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 at some point, we're going to be hitting a transition. Remember, we were concave downwards before we got to 2/3, right? So this is consistent. From 0 to 2/3, we are concave downwards, and then at 2/3, we become concave upwards. Now let's see what happens when x is right less than, when x is just barely, just barely less than 0. So once again, f prime, the second derivitive of x is equal to 12x times 3x minus 2. Well, right. If x was minus 0.1 or 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 just have some negative value here, times 12, is going to be negative. And then what's this going to be? Well, 3 times minus 0.1 is going to be minus 0.3, minus 2 is minus 2.3. 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 going to get a positive. So actually, right below x is less than 0, the second derivative is positive. Now, this all might have been a little bit confusing, but we should now have the payoff. We now have the payoff. We have all of the interesting things going on. We know that at x is equal to 1, we know that at x is equal to 1, let me write it over here. We've figured out at x is equal to 1, the slope is 0. So f prime prime is, sorry, let me write this way. I should have said, we know that the slope is 0. 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 function is concave upwards at this point. And that tells us that this is going to be a minimum point. And we should actually get the coordinates so we can actually graph it. That was the whole point of this video. So f of 1 is equal to what? f of 1, let's go back to our original function, is 3 times 1, right, 1 to the fourth is just 1, 3 times 1 minus 4 plus 2, right? So it's 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 one. And then we know at 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 concativity switches before and after. So this is an inflection point. And we are concave below 0, so when x is less than 0, we are upwards. Our second derivative is positive. And when x is it greater than 0, we are downwards. We're 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, because we want to graph that point? f of 0. See, f of 0, this is easy. 3 times 0 minus 4 times 0 plus 2, that's just 2. f of 0 is 2. And then finally we got the point, x is equal to 2/3. Let me do that in another color. We had the point x is equal to 2/3. We figured out that this was an inflection point. 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 less than 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. Now 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 out, 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 take aways. Let me do a rough graph. Let's see. So let me do my axes, just like that. So we're going to want to graph the point 0, 2. So let's say that the point, 0, 2. So this is x is equal to 0, and we go up, 1, 2. So this is the point 0, 2. 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 1, which is the point 1, 1, right? So this point right here. So that's the point 1, 1. This was the point 0, 2. And then we have 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. Maybe here someplace. Let's say f of 2/3 is right there. So that's the point, 2/3, and then wherever 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 thing. So we know that at x is equal to 1, the slope is 0. We know that the slope is 0. It's flat here. We know it's concave upwards. So we're dealing, it looks like this, 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? Let me do it in that color. We knew x equals 2/3 and on, we're concave upwards. And 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. We'll 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, 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. The graph looks like that. And we also know that x is equal to 0 was a critical point, the slope of 0. So the graph is actually flat right there, too. So this is an inflection point where the slope was also 0. 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 concativity, and transitions in concativity, to actually graph this fairly hairy-looking graph. But this should be kind of what it looks like, if you graph it on your calculator.