Conditions for IVT and EVT: graph | Existence theorems | AP Calculus AB | Khan Academy

So we have the graph of ( y ) is equal to ( h ) of ( x ) right over here and 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. So let's we could 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.

So you just have to say you might be tempted to say, 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 ).

You can see over that entire interval right over here the function is continuous. So first I would definitely say it meets the continuity requirement. It's continuous over this closed interval and so I would say yes, it does. The intermediate value theorem does apply.

And 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.

And so ( h ) of negative one eyeballing it looks like it's negative three and ( h ) of four looks like it's zero. So we know by the intermediate value theorem that we take on every value between negative three and zero.

And 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. So here we say we're asked does the extreme value theorem apply to ( g ) over the closed interval from zero to five?

So once again, before we even think about what the extreme value theorem says, we just remember that just like the intermediate value theorem, it only applies over closed intervals where the function is continuous over the entire interval.

So let's see, is our function continuous on the closed interval from zero to five? So from zero to five. No, we clearly have a discontinuity right over here at ( x ) is equal to three.

So first of all, we would say no, and so the ( x ) we can't say that over this closed interval that we have a well-defined maximum or minimum value. And this is actually very clear over here that you don't have a well-defined maximum value; we approach infinity over there and we approach negative infinity over there.

And so the 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. Let me use a different color; say it was 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. And in this case, it would actually, 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?

So once again, let's just look at the interval we're going from negative five going from negative five to negative two. So this is our function over that interval from negative five from ( x ) equals negative five to ( x ) equals negative two.

And 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 and there we actually are continuous.

We're continuous over this closed interval and so 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.

And you can see that the maximum value here happens when at ( f ) of negative two looks like it's around negative three point five or so or negative three point six.

And 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.