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## Intermediate and extreme value theorems

# Conditions for IVT and EVT: table

## Video transcript

- [Narrator] We're told this table gives a few values of function f. It tells us what f of x is equal to. That x is equal to two,
three, four and five. Which condition would
allow you to conclude that there exists a maximum value for f over the closed interval from two to four. So pause this video and see
if you can figure it out. So you might already remember, if we're trying to
conclude that there exists, anytime people are talking
about, there exists, something over an interval, we're probably dealing with some type of an existence theorem. And if we're talking about
the existence of a maximum or a minimum value for a
function over an interval. That means that we're talking about the extreme value theorem. Or we are likely talking about
the extreme value theorem and in order to apply the
extreme value theorem, sometimes abbreviated EVT, we need to know that our
function is continuous over this closed interval. So, in order to do that, we need to know that f is continuous over our
closed interval here, over two to four. The closed interval from two to four. So let's see which of these
will allow us to conclude this because if we're able to conclude this, then we're able to apply
the extreme value theorem, which tells us that there exists a maximum value for f over that interval. Alright, choice A says, f is increasing in the open
interval from two to three and decreasing on the open
interval from three to five. Now you can completely, this doesn't tell us that
we're definitely continuous over that interval, over two to four so I would rule that out. You could definitely
still be discontinuous and statement A still being true so I'm gonna rule that one out. F is continuous over the closed
interval from two to five. Well if we're continuous
over the closed interval from two to five, we're definitely continuous
over a sub-set over that, over the closed interval from two to four and so if we're able to
make that, if we know that, then we can use the extreme value theorem, we've met the condition for
the extreme value theorem. To say that there exists a maximum value for f over that interval. So, I like this one. Let's just look at this last one. F is differentiable over the open interval from two to four and at x is equal to two. So this is close because differentiability
does imply continuity. But, it isn't telling us what's happening at x equals four. We could still be
discontinuous at x equals four, we're just differentiable
for every point up to four because it's talking
about an open interval. So this is close. If they said differentiable over the closed interval from two to four, then this would allow us to conclude because differentiability
over an interval, implies continuity but I'm gonna rule this one out because it does not ensure continuity at our right boundary, at x equals four. Let's do another one of these. H is a differentiable function. And once again, they've given us what our
function h is equal to at certain sampled x points, x values. Clyde was asked whether
there's a solution to h of x is equal to negative two on the interval from the closed interval
from negative one to three. This is his solution and then we're asked, is Clyde's work correct? If not, what is his mistake? And then they give us some choices. But, before even looking at the choices, let us, pause this video and see if what Clyde
is saying makes sense and if it doesn't make sense. Try to pin point where he messed up. What step did he make
the wrong conclusion? If he made any mistake at all. Alright so, the first thing he did is he said look, negative two, we wanna
see, is there an x value where h of x is equal to negative two. So first he says look, negative two is between h of negative one and h of three, so let's see, h of negative one is two and h of three is negative five and he is right, that negative three,
or sorry negative two, sits right in here so negative two is right
between these two values. It's right in between h of negative one and h of three. Since h is differentiable, we know it is continuous
on that closed interval. That's right. We were just talking about that. Differentiability implies continuity. Continuity does not always
imply differentiability. You can't make that assumption but differentiability implies continuity. If we're differentiable over interval, we're definitely continuous
on that interval. And so, he's right here. That so far step one and step two, we can conclude that we are continuous
over the closed interval from negative one to three and then he says, the extreme value theorem
guarantees a solution to h of x equals negative
two on that closed interval. So this feels a little bit fishy. Because the extreme value theorem says, it says, if we meet our conditions and we did for even the
extreme value theorem, we're continuous on that closed interval. It says that we're gonna
have a well defined maximum and minimum value
on that closed interval but we don't care about minimum
and maximum values here, what we care is that we take on an intermediate value in that interval. Value in between h of negative one and h of three, possibly what each of
those boundaries as well. And so, he's applying the wrong theorem. It should be the
intermediate value theorem, I'll just abbreviate that as IVT, is what he should be saying. The intermediate value theorem, because we are continuous
on the closed interval. It says that we are going
to take on every value between h of negative one and h of three and negative two is one of those values between h of negative one and h of three and so if he just wrote
intermediate value theorem instead, he would've been correct. So let's see which choice is consistent with what we just wrote. He's definitely not correct. Step one wasn't incorrect. Negative two is between h of negative one and h of three. Step two was actually correct. We know that differentiability
implies continuity. Step three is incorrect, yes he applied the wrong theorem.