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Current time:0:00Total duration:4:50

- [Narrator] We have the graph of y is equal to h of x right over here. They ask us does the
intermediate value theorem apply to h over the closed interval from negative one to four. The closed interval from
negative one to four right over here. We can first remind ourselves what is the intermediate value theorem. Although you don't have to even know what the theorem is actually saying, you just have to remember that the intermediate value theorem only applies over closed intervals where the function is continuous. You just have to say, you might be tempted to say wait, wait this function has a point of discontinuity but this is outside of that interval. The interval we care
about at negative one, this is the value of h of x and then at four this is the value of h of x and you can see over that entire interval right over here, the function is continuous. First, I would definitely say it meets the continuity requirement,
it's continuous over this closed interval. I would say yes, it does, the intermediate value theorem does apply. Since it does apply, I will be able to say that by the
intermediate value theorem because we are continuous
over this closed interval, the function in this case, h of x will take on every value between h of negative one and h of four, including those values. H of negative one, eyeballing it, looks like it's negative three. H of four looks like it's zero. We know by the intermediate value theorem that we take on every value between negative three and zero. You can even see that right over here. It actually does take on every one of those values in this part of the curve right over there. Let's do another example. Here we're asked does the
extreme value theorem apply to g over the closed interval from zero to five. Once again before we even think about what the extreme value theorem says, we have to remember that just like the intermediate value theorem it only applies over closed intervals where the function is continuous over the entire interval. Let's see, is our function continuous on the closed interval from zero to five? From zero to five. No we clearly have a discontinuity right over here at x is equal to three. First of all, we would say no. We can't say that over
this closed interval, that we have a well defined
maximum or minimum value. This is actually very clear over here that you don't have well
defined maximum value, we approach infinity over there and then we approach
negative infinity over there. Extreme value theorem does not apply. If instead it had offered
us a closed interval where we were continuous,
say they said between negative two and zero, let's
say it's that interval. It is a different color. Say it's between negative two and zero. Then the extreme value theorem would apply where you would say
okay, over this interval, the function has a well
defined minimum value and a maximum value. In this case it would actually be, the maximum would happen
at the right bound and the minimum value
happens at the left bound. But for this case, the closed interval, the one that we started with, definitely is not continuous so we do not, we would not say that the
extreme value theorem applies. Let's do one more example. Does the extreme value theorem apply to f over the interval from
negative five to negative two? Once again, let's just
look at the interval. We're going from negative
five to negative two. This is our function over that interval from x equals negative five
to x equals negative two. Even though there is a discontinuity in this graph that we
see, it's sitting outside of the interval. We only care about what's
happening inside of the interval. There we actually are continuous. We're continuous over
this closed interval. Because of that we know that
the extreme value theorem would apply, that there is a well defined minimum and maximum
value over the interval. You can see that the maximum value here happens when at f of negative two, it's looks like it's around negative 3.5 or so, or negative 3.6. Then the minimum value over this interval looks like it happens
at f of negative five. It looks like it's pretty
close to negative four because it looks like
we're just increasing over this entire interval. But the important thing here
is to recognize continuity. Continuity does not have to apply to the entire function or
over its entire domain, it just has to be true
over the closed interval that we are trying to
apply these theorems to.