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Limit at a point of discontinuity.mp3 | And just from an inspection, you can see that the function is not defined when x is equal to 3. You get 0 over 0. It's not defined. So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. ... |
Limit at a point of discontinuity.mp3 | So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. I'm going to think of the case when x is greater than 3 and when x is less than 3. So when x is, we'll do this in two different colo... |
Limit at a point of discontinuity.mp3 | So when x is, we'll do this in two different colors actually. When x, do it in green, that's not green. When x is greater than 3, what does this function simplify to? Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So... |
Limit at a point of discontinuity.mp3 | Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So for x is greater than 3, this is going to be the exact same thing as x minus 3 over x minus 3. Because if x is greater than 3, the numerator is going to be positive. ... |
Limit at a point of discontinuity.mp3 | Because if x is greater than 3, the numerator is going to be positive. You take the absolute value of that, you're not going to change its value. So you get this right over here. Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Si... |
Limit at a point of discontinuity.mp3 | Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Similarly, let's think about what happens when x is less than 3. When x is less than 3, well, x minus 3 is going to be a negative number. When you take the absolute value of that, y... |
Limit at a point of discontinuity.mp3 | When you take the absolute value of that, you're essentially negating it. So it's going to be the negative of x minus 3 over x minus 3. Or if you were to simplify these two things, for any value as long as x does not equal 3, this part right over here simplifies to 1, so you're left with a negative 1. Negative 1 for x ... |
Limit at a point of discontinuity.mp3 | Negative 1 for x is less than 3. I encourage you, if you don't believe what I just said, try it out with some numbers. Try out some numbers, 3.1, 3.001, 3.5, 4, 7, any number greater than 3, you're going to get 1. You're going to get the same thing divided by the same thing. And try values for x less than 3. You're goi... |
Limit at a point of discontinuity.mp3 | You're going to get the same thing divided by the same thing. And try values for x less than 3. You're going to get negative 1 no matter what you try. So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis. |
Limit at a point of discontinuity.mp3 | So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5. |
Limit at a point of discontinuity.mp3 | And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1. |
Limit at a point of discontinuity.mp3 | So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way. |
Limit at a point of discontinuity.mp3 | So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our f... |
Limit at a point of discontinuity.mp3 | We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our function is equal to 1. So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to n... |
Limit at a point of discontinuity.mp3 | So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this. |
Limit at a point of discontinuity.mp3 | And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question. |
Limit at a point of discontinuity.mp3 | It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at th... |
Limit at a point of discontinuity.mp3 | So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all thi... |
Limit at a point of discontinuity.mp3 | So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all this notation here, I wrote this negative as a superscript right after the 3 says. So let's think about the limit as we're approaching from the left. So in this case, if we start with val... |
Limit at a point of discontinuity.mp3 | So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1. We go to 1, f of x is equal to negative 1. We go to 2, f of x is equal to negative 1. If you go to 2.999999, f of x is equal to negative 1. So it looks like it is appro... |
Limit at a point of discontinuity.mp3 | If you go to 2.999999, f of x is equal to negative 1. So it looks like it is approaching negative 1 if you approach from the left-hand side. Now let's think about the limit of f of x as x approaches 3 from the positive direction, from values greater than 3. So here we see when x is equal to 5, f of x is equal to 1. Whe... |
Limit at a point of discontinuity.mp3 | So here we see when x is equal to 5, f of x is equal to 1. When x is equal to 4, f of x is equal to 1. When x is equal to 3.0000001, f of x is equal to 1. So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we a... |
Limit at a point of discontinuity.mp3 | So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we approach from the right. And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or anot... |
Limit at a point of discontinuity.mp3 | And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or another way of saying it, the limit of a function f of x as x approaches some value c is equal to L if and only if the limit of f of x as x approaches c from the negative direction is equal to ... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | I'm going to do it over various intervals. So first, let's think about the area under the curve between x is equal to 0 and x is equal to pi over 2. So we're talking about this area right over here. Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of rem... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. An... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | And that's kind of what this notation is trying to depict. But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluate... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0. This could be pi. This coul... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0. This could be pi. This could be 5. This could be a million. This could be a googol. This could be any crazy number. |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | This could be 5. This could be a million. This could be a googol. This could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, becaus... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | This could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate. So one way t... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate. So one way that we can denote this, the antiderivative of cosine of x, or an antiderivative of cosine of x, is sine of x. And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0. So this is ... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0. So this is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. No... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi o... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 p... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | What's sine of 3 pi over 2? If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle. So it's right over there. So the sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi ov... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | So the sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We've got a negative area. Now, how does that make sense? |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We've got a negative area. Now, how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Now, how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. ... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let'... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going ... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under al... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under all this orange area that I just outlined is clearly not negative, or any area isn't a negative number. And the area isn't even 1. But what just happened h... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | And the area isn't even 1. But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or anothe... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting t... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting the two below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get ... |
Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3 | And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis. If we get 0, then that means it all nets out. And if you want to see a case at 0, take the integral from 0 all the way to 2 pi. And this will evaluate to 0 because you have an area of 1 and another area ... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We are now going to cover the famous or perhaps infamous potato problem from the 2017 AP Calculus exam. At time t equals zero, a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius at time t equals zero, and the internal temperature of... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | I would guess that the ambient room temperature is 27 degrees Celsius, and so that's why the temperature would approach this, but it would always stay a little bit greater than that as t gets larger and larger. The internal temperature of the potato at time t minutes can be modeled by the function h that satisfies the ... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So before I even read part a, let's just make sense of what this differential equation is telling us, and let's see if it's consistent with our intuition. So let me draw some axes here. So this is my y-axis, so that is my y-axis, and this right over here is my t-axis. Now if the ambient room temperature is 27 degrees C... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Now if the ambient room temperature is 27 degrees Celsius, so I'll just draw there, that's what the room temperature is doing, and so we know at t equals zero, our potato is at 91 degrees, so let's see, that's 27, 91 might be right over there. 91, this is all in degrees Celsius, and what you would expect intuitively is... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So you would expect it to look something like this, and then asymptote towards a temperature of 27 degrees Celsius. So this is what you would expect to see, and this differential equation is consistent with that, where the rate of change, notice, this is for all t greater than zero, this is going to be a negative value... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | When there's a big difference, you expect a steeper rate of change, but then when there's less of a difference, the rate of change, you can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature. So with this out of the way, now let's tackle part A. Write an equation for the li... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Use this equation to approximate the internal temperature of the potato at time t equals three. So what are we going to do? Well, we're going to think about what's going on at time t equals zero, right over here. We want the equation of the tangent line, which might look something like this, at t equals zero. So this t... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We want the equation of the tangent line, which might look something like this, at t equals zero. So this thing would be of the form y is equal to the slope of the equation of the tangent line. Well, it would be the derivative of our function at that point, so dh dt, times t, plus our y-intercept. Where does it interse... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Where does it intersect the y-axis here? Well, when t is equal to zero, the value of this equation is going to be 91, because it intersects our graph right at that point, that point zero comma 91, plus 91. So what is our derivative of h with respect to t at time t equals zero, right at this point right over here? Well,... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Well, we just have to look at this. You could also write this as h prime of t right over here. So if we want to think about h prime of zero, that's going to be equal to negative 1 1⁴ times h of zero minus 27. What is our initial temperature minus 27? This is, of course, 91 degrees. They tell us that multiple times. We'... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | What is our initial temperature minus 27? This is, of course, 91 degrees. They tell us that multiple times. We've even drawn it a few times. 91 minus 27 is 64. 64 times negative 1⁴ is equal to negative 16. So this is negative 16 right over here. |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We've even drawn it a few times. 91 minus 27 is 64. 64 times negative 1⁴ is equal to negative 16. So this is negative 16 right over here. So just like that, we have the equation for the line tangent to the graph of h at t is equal to zero. I'll write it one more time. It is, we've got a mini drum roll here, y is equal ... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So this is negative 16 right over here. So just like that, we have the equation for the line tangent to the graph of h at t is equal to zero. I'll write it one more time. It is, we've got a mini drum roll here, y is equal to negative 16t plus 91. That's the equation of that tangent line right over there. And then they ... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | It is, we've got a mini drum roll here, y is equal to negative 16t plus 91. That's the equation of that tangent line right over there. And then they say, we want to use this equation to approximate the internal temperature of the potato at time t equals three. So let's say that this is time t equals three right over he... |
2017 AP Calculus AB BC 4a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So let's say that this is time t equals three right over here. We want to approximate the temperature that this model describes right over here, but we're gonna do it using the line. So we're gonna evaluate the line at t equals three. So then we would get, let's see, negative 16 times three plus 91 is equal to, this is... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | And so like always, pause this video and see if you can figure it out. All right, now the way I'm going to tackle it is I'm gonna look at each of these graphs and try to think what would their derivatives look like? So for this first one, we can see our derivative right over here, the slope of our tangent line. It woul... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | It would be a little bit negative, and then it gets more and more and more negative. And as we approach this vertical asymptote right over here, it looks like it's approaching negative infinity. So the derivative would actually, over here, it would be a little bit less than zero, but then it would get more and more and... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | So it would have a similar shape, general shape, to the graph itself, at least to the left of this vertical asymptote. Now what about to the right of the vertical asymptote? Right to the right of the vertical asymptote, it looks like the slope of the tangent line is very negative, it's very negative, but then it become... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | So on this side, the derivative starts out super negative, and then it looks like it is, the derivative is going to asymptote towards zero, something like that. So based on what we just, it actually looks like, so based on what I just sketched, it just looks like this right graph is a good candidate for the derivative ... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | Well, this blue graph out here, notice it's positive. So if this were the derivative of the left graph, then that means that the left graph would need a positive slope out here, but it doesn't have a positive slope. It's a slightly negative slope becoming super negative. And so right here, we're slightly negative, and ... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | And so right here, we're slightly negative, and then we become very negative. And so maybe this is, let's call, let's call this F, and maybe this is F prime. This is F prime right over here. And now let's look at this middle graph. What would its derivative do? So over here, our slope is slightly negative, and then it ... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | And now let's look at this middle graph. What would its derivative do? So over here, our slope is slightly negative, and then it becomes more and more and more and more and more negative. And so the derivative of this might look like it has to be slightly negative, but then it gets more and more and more and more negat... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | And so the derivative of this might look like it has to be slightly negative, but then it gets more and more and more and more negative as we approach that vertical asymptote. And in the right side of the vertical asymptote, our derivative is very positive here, and then it gets less and less and less and less and less... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | It looks like it might, the slope here might be asymptoting towards zero. So our graph might look something like that. Well, the left graph right here looks a lot like what I just sketched out as a candidate derivative for this blue graph, for this middle graph. And so I would say that this is f, then this is the deriv... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | And so I would say that this is f, then this is the derivative of that, which would make it f prime. And then we already established that this right graph is the derivative of the left one. So if it's the derivative of f prime, it's not f prime itself, it's the second derivative. So I feel pretty good about that. And j... |
Connecting f, f', and f'' graphically (another example) AP Calculus AB Khan Academy.mp3 | So I feel pretty good about that. And just for good measure, we could think about what the derivative of this graph would look like. Here the slope is slightly negative, but then it gets more and more and more and more and more negative. So the derivative would have a similar shape here. And then here, our derivative w... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | What we're going to do in this video is introduce ourselves to the notion of a definite integral. And with indefinite integrals and derivatives, this is really one of the pillars of calculus. And as we'll see, they are all related, and we'll see that more and more in future videos, and we'll also get a better appreciat... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So let me draw some functions here. And we're actually gonna start thinking about areas under curves. So let me draw coordinate axes here. So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases. So this is my y-axis. |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases. So this is my y-axis. This is my x-axis. And let's say I have some function here. So this is f of x, right over there. |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So this is my y-axis. This is my x-axis. And let's say I have some function here. So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that. And let's say that this is x equals b, just like that. |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that. And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis,... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis, and between these two bounds, between x equals a and x equals b. So this area right over here. And you can already get an appre... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And you can already get an appreciation. We're not used to finding areas where one of the boundaries, or as we'll see in the future, many of the boundaries could actually be curves. But that's one of the powers of the definite integral, and one of the powers of integral calculus. And so the notation for this area right... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And so the notation for this area right over here would be the definite integral. And so we're gonna have our lower bound at x equals a, so we'll write it there. We'll have our upper bound at x equals b, right over there. We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're goi... |
Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus. This is known as the summa sym... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So it's a closed interval. It also includes 1 and it includes 4. You can view this as the domain of our function as we have defined it. So given this, given this information, this function definition, what I would like you to do is come up with the absolute maximum value of f. Where f is defined right over here, where ... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So given this, given this information, this function definition, what I would like you to do is come up with the absolute maximum value of f. Where f is defined right over here, where f is defined on this closed interval. I encourage you to pause this video and think about it on your own. So the extreme value theorem t... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | And let's say we have some function that's defined on a closed interval. We have a couple of different scenarios for what that function might look like on that closed interval. So we might hit a maximum point. We might hit a maximum point at the beginning of the interval, something like that. We might hit an absolute m... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | We might hit a maximum point at the beginning of the interval, something like that. We might hit an absolute maximum point at the end of the interval. So it might look something like this. So that's at the end of the interval. Or we might hit an absolute maximum point someplace in between. And that could look something... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So that's at the end of the interval. Or we might hit an absolute maximum point someplace in between. And that could look something like this. It could look like this. And at this maximum point, the slope of the tangent line is zero. So here, the derivative is zero. Or we could have a maximum point someplace in between... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | It could look like this. And at this maximum point, the slope of the tangent line is zero. So here, the derivative is zero. Or we could have a maximum point someplace in between that looks like this. And if it looks like this, then here the derivative would be undefined. So we have a couple of different tangent lines t... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | Or we could have a maximum point someplace in between that looks like this. And if it looks like this, then here the derivative would be undefined. So we have a couple of different tangent lines that you could place right over there. So what we need to do is let's test the different endpoints. Let's test the function a... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So what we need to do is let's test the different endpoints. Let's test the function at the beginning. Let's test the function at the end of the interval. And then let's see if there's any points where the derivative is either zero or the derivative is undefined. And these points where the derivative is either zero or ... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | And then let's see if there's any points where the derivative is either zero or the derivative is undefined. And these points where the derivative is either zero or it's undefined, we've seen them before, we call these, of course, critical numbers. And this would be either in either case, actually if we assume that tha... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So those are the different candidates. Now you could have a critical number in between where, say, the slope is zero, say something like this, but it isn't the maximum or minimum. But what we can do is if we can find all the critical numbers, we can then test the function evaluated at the critical numbers and the funct... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | All of those are the possible candidates for where f hits a maximum value. So first we could think about, well, actually, let's find the critical numbers first since we have to do it. So let's take the derivative of f. f prime of x is going to be equal to the derivative of natural log of x is 1 over x. So it's going to... |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | So it's going to be 8 over x minus 2x. And let's set that equal to zero. So if we focus on this part right over here, we could add 2x to both sides, and we would get 8 over x is equal to 2x. Multiply both sides by x. We get 8 is equal to 2x squared. Divide both sides by 2. You get 4 is equal to x squared. |
Finding absolute extrema on a closed interval AP Calculus AB Khan Academy.mp3 | Multiply both sides by x. We get 8 is equal to 2x squared. Divide both sides by 2. You get 4 is equal to x squared. And if we were just purely solving this equation, we would get x is equal to plus or minus 2. Now we're saying that the function is only defined over this interval, so negative 2 isn't part of its domain,... |
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