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• - [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

• 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.

- [Voiceover] Let F be a continuous function

# 15-2中值定理的示例（Intermediate value theorem example | Existence theorems | AP Calculus AB | Khan Academy）

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yukang920108 發佈於 2022 年 07 月 05 日