字幕列表 影片播放 列印英文字幕 - [Voiceover] Let F be a continuous function on the closed interval from negative two to one where F of negative two is equal to three and F of one is equal to six. Which of the following is guaranteed by the Intermediate Value Theorem? So before I even look at this, what do we know about the Intermediate Value Theorem? Well it applies here, it's a continuous function on this closed interval. We know what the value of the function is at negative two. It's three so let me write that. F of negative two is equal to three and F of one, they tell us right over here, is equal to six and all the Intermediate Value Theorem tells us and if this is completely unfamiliar to you, I encourage you to watch the video on the Intermediate Value Theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the endpoints of the interval or another way to say it is for any L between three and six, three and six, there is at least one C, there is at least one C, one C, between, or I could say once C in the interval from negative two to one, the closed interval, such that F of C is equal to L. This comes straight out of the Intermediate Value Theorem and just saying it in everyday language is look this is a continuous function. Actually I'll draw it visually in a few seconds. But it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well that it makes sense that I would have to take on every value between three and six or there's at least one point in this interval where I take on any given value between three and six. So let's see which of these answers are consistent with that and we only pick one. So F of C equals four. So that would be a case where L is equal to four. So there's at least one C in this interval such that F of C is equal to four. We could say that. But that's not exactly what they're saying here. F of C could be four for at least one C, not in this interval, remember the C is our X. This is our X right over here. So the C is going to be in this interval and I'll take a look at it visually in a second so that we can validate that. We're not saying for at least one C between three and six F of C is equal to four, we're saying for at least one C in this interval F of C is going to be equal to four. It's important that four is between three and six because that's the value of our function and the C needs to be in our closed interval along the x-axis. So I'm gonna rule this out. They're trying to confuse us. Alright. F of C equals zero for at least one C between negative two and one. Well here they got the interval along the x-axis right, that's where the C would be between, but it's not guaranteed by the Intermediate Value Theorem that F of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, it's saying F of C equals zero, and let's see, we're only left with this one so I hope it works. So F of C is equal to four, well that seems reasonable because four is between three and six, for at least one C between negative two and one. Well yeah because that's in this interval right over here. So I am feeling good about that and we could think about this visually as well. The Intermediate Value Theorem when you think about it visually makes a lot of sense. So let me draw the x-axis first actually and then let me draw my y-axis and I'm gonna draw them at different scales 'cause my y-axis, well let's see. If this is six, this is three. That's my y-axis. This is one, this is negative one, this is negative two and so we're continuous on the closed interval from negative two to one and F of negative two is equal to three. So let me plot that. F of negative two is equal to three. So that's right over there and F of one is equal to six. So that's right over there and so let's try to draw a continuous function. So a continuous function includes these points and it's continuous so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points. So I can't do that. That would be picking up my pencil. So it is a continuous function. So it takes on every value. As we can see, it definitely does that. It takes on every value between three and six. It might take on other values, but we know for sure it has to take on every value between three and six and so if we think about four, four is right over here. The way I drew it, it looks like it's almost taking on that value right at the y-axis. I forgot to label my x-axis here. But you can see it took on that value in for a C in this case between negative two and one and I could have drawn that graph multiple different ways. I could have drawn it something like I could have done it and actually it takes on, there's multiple times it takes on the value four here. So this could be our C, but once again it's between the interval negative two and one. This could be our C once again in the interval between negative two and one or this could be our C in between the interval of negative two and one and that's just the way I happen to draw it. I could have drawn this thing as just a straight line. I could have drawn it like this and then it looks like it's taking on for only one and it's doing it right around there. This isn't necessarily true that you take on, that you become four for at least one C between three and six. Three and six aren't even on our graph here. I would have to go all the way to two, three. There's not guarantee that our function takes on four for one C between three and six. We don't even know what the function does when X is between three and six.

A2 初級 美國腔 15-2中值定理的示例（Intermediate value theorem example | Existence theorems | AP Calculus AB | Khan Academy） 3 1 yukang920108 發佈於 2022 年 07 月 05 日 更多分享 分享 收藏 回報 影片單字