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so we have the graph of y is equal to H of X right over here and they ask us that the intermediate value theorem apply to H over the closed interval from negative 1 to 4 so the closed interval from negative 1 to 4 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 wait wait this function has a point of discontinuity but this is outside of that interval the interval we care about at negative 1 this is the value of H of X and then at 4 this is the value of H of X and 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 its 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 1 and H of 4 including those values and so H of negative 1 eyeballing it looks like it's negative 3 and H of 4 looks like it's zero so we know by the intermediate value theorem that we take on every value between negative 3 and zero and you can even see that right over here it actually does take on every one of those values in this part in this part of the curve right over there let's do another example so here we're sitting where asked does the extreme value theorem apply to G over the closed interval from 0 to 5 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 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 images in different colors say it was between negative two and zero then the extreme value theorem would apply where you 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 it does the extreme value theorem apply to s over the interval from negative 5 to negative two so once again let's just look at the interval we're going from negative five going from negative 5 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 where 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 be 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 it 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