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Current time:0:00Total duration:6:32

Video transcript

we're told this table gives a few values of function s that tells us what f of X is equal to it 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 2 to 4 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 any time 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're 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 to 2 for the closed interval from 2 to 4 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 all right choice a says f is increasing in the open interval from 2 to 3 and decreasing on the open interval from 3 to 5 now you could completely this doesn't tell us that they were definitely continuous over that interval over over 2 to 4 so I would rule that out you could definitely still be discontinuous and statement a still being true so I'm going to rule that one out f is continuous over the closed interval from 2 to 5 well for continuous over the closed interval from 2 to 5 we're definitely continuous over a subset over that over the closed interval from 2 to 4 and so if we're able to make that if we know that that we can can we can use the extreme value theorem we've met the conditions 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 vegetable over the open interval from 2 to 4 and at X is equal to 2 so this is closed because differentiability does imply continuity but it isn't telling us what's happening at x equals 4 we could still be discontinuous at x equals 4 we're just differentiable for every point up to 4 because it's talking about it open interval so this is close if they said differentiable over the closed interval from 2 to 4 then this would allow us to conclude because differentiability over an interval implies continuity but I'm going to rule this one out because it does not ensure continuity at our right boundary at x equals 4 let's do another one of these H is a differentiable function and once again they've given us what are valid what our function H is equal to at certain sampled X points X values collide was asked whether there's a solution to H of X is equal to negative 2 on the interval from the closed interval from negative 1 to 3 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 try to pause this video and see if what Clyde is saying makes sense and if it if it doesn't make sense trying to pinpoint where he messed up what step did he make the wrong conclusion if he made any mistake at all all right so the first thing he did is he said look negative 2 we want to see is there an X value where H of X is equal to negative 2 so first she says what negative 2 is between H of negative 1 and H of 3 so let's see H of negative 1 is 2 and H of 3 is negative 5 and he is right that negative 3 or sorry negative 2 sits right in here so if we were to so negative 2 is right in between these two values it's right in between H of negative 1 and H of 3 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 1 to 3 and then he says the extreme value theorem guarantees a solution to H of x equals negative 2 on that closed interval so this feels a little bit fishy because the extreme value theorem says that it says well if we can 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 going to 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 and a value in between H of negative 1 and H of 3 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 it as I VT 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 1 and H of 3 and negative 2 is in is one of those values between H of negative 1 and H of 3 and so if he just wrote an intermediate value theorem instead he would have been correct so let's see which choice Quora is is consistent with what we just wrote he's definitely not correct step one wasn't correct wasn't incorrect two is between H negative two is between H of negative one and H of three step two was was actually correct we know that differentiability implies continuity step three is incorrect yes he applied the wrong theorem