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AP Calc: FUN‑1 (EU), FUN‑1.A (LO), FUN‑1.A.1 (EK)

- [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.

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