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L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exists, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what? |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument, this right here is going to approach 0. |
L'Hôpital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3 | Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument, this right here is going to approach 0. So all you're left with is the 4 and the negative 3. So this is going to be equal to negative 4 over a negative 3 or negative 4 thirds. So you didn't have to use L'Hospital's Rule for this problem. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | For each point on the graph, is the object speeding up, slowing down, or neither? So pause this video and see if you can figure that out. All right, now let's do it together. And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? |
Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 | And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? Well, the speed here is two meters per second. Remember, it would be the absolute value of the velocity. And the absolute value is actually going down if we forward in time a little bit. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Let's say that we have the indefinite integral and the function is three x squared plus two x times e to the x to the third plus x squared dx. So how would we go about solving this? So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? |
_-substitution intro AP Calculus AB Khan Academy.mp3 | You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. U is x to the third plus x squared, x to the third plus x squared. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. U is x to the third plus x squared, x to the third plus x squared. And then we have our plus c. And we are done. We have found the antiderivative. And I encourage you to take the derivative of this, and I think you will find yourself using the chain rule and getting right back to what we had over here. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | So let's say we wanted to figure out the limit as x approaches zero of f of g of x, f of g of x. First of all, pause this video and think about whether this theorem even applies. Well, the first thing to think about is what is the limit as x approaches zero of g of x to see if we meet this first condition. So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two. So that's a check. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two. So that's a check. Now let's see the second condition. Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | So that's a check. Now let's see the second condition. Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero. So our right and left-hand limits are not the same here, so this thing does not exist. |
Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3 | Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero. So our right and left-hand limits are not the same here, so this thing does not exist. Does not exist. And so we don't meet this condition right over here, so we can't apply the theorem. But as we've already seen, just because you can't apply the theorem does not mean that the limit does not exist. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | So what I've drawn in red. At the point x equals a. And we've already seen this with the definition of the derivative. We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | Let's just take another arbitrary point, some here, some place. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here? |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing. We're finding, we're taking, we have an expression for the slope of a secant line, and then we're bringing those x values of those points closer and closer together, closer and closer together, so the slopes of those secant lines better and better and better approximate that slope of the tangent line, and at the limit, it does become the slope of the tangent line. That is the derivative of, or that's the definition of the derivative. |
Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3 | So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing. We're finding, we're taking, we have an expression for the slope of a secant line, and then we're bringing those x values of those points closer and closer together, closer and closer together, so the slopes of those secant lines better and better and better approximate that slope of the tangent line, and at the limit, it does become the slope of the tangent line. That is the derivative of, or that's the definition of the derivative. So this is the kind of the more standard definition of a derivative. It would give you your derivative as a function of x, and then you can then input your x that you want, the particular value of x, or you could use the alternate form of the derivative. If you know that, hey, look, I'm just looking to find the derivative exactly at a, I need a general function of f, then you could do this, but they're doing the same thing. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Now what I want to do in this video is introduce a third value, c, that is in between a and b. And it could be equal to a or it could be equal to b. So let me just introduce it. Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this. |
Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this. Well, you could say that the larger integral, which would be the area under the curve, it would be all of this. Let's just say it's a gap right there, or it jumps down there. So this entire area, you can break up into two smaller areas. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | What I encourage you in this video is to pause it and try to do the same type of proof for the derivative of the inverse cosine of x. So what is, so our goal here is to figure out, I want the derivative with respect to x of the inverse cosine of x. What is this going to be equal to? So assuming you've had a go at it, let's work through it. So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | So assuming you've had a go at it, let's work through it. So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | We want it in terms of x. And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y? |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y? Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | And what is cosine of y? Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that. |
Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3 | This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that. So copy and paste it. I'm gonna paste it down here. And now let's look at them side by side. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | So we'll now think about the extreme value theorem, which we'll see is a bit of common sense. But in all of these theorems, it's always fun to think about the edge cases. Why is it laid out the way it is? And that might give us a little bit of more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | And that might give us a little bit of more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here? |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis. That's my y-axis. And let's draw the interval. So the interval's from a to b. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | So this is my x-axis. That's my y-axis. And let's draw the interval. So the interval's from a to b. So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | So the interval's from a to b. So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see. Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a. |
Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3 | Well, let's see. Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a. That's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value. |
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