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Current time:0:00Total duration:4:41

so we've been given the value of H of X at a few values of X and then we're told James said that since H of 7 minus H of 3 over 7 minus 3 is equal to 1 so this is really the average rate of change between X is equal to 3 and X is equal to 7 between that point and that point right over there so since that is equal to 1 there must be a number C in the closed interval from 3 to 7 for which the derivative at that value C is equal to 1 so what James is trying to do is apply the mean value theorem which tells us if the conditions apply for the mean value theorem it tells us that if I'm going through to 2 values so let's say this is a let's say this is B and let's say the function does something like this if we meet the conditions for the mean value theorem it tells us that there's some C in the closed interval from A to B where the derivative of C is equal to the average rate of change between a and B so the average rate of change between a and B would be the slope of the secant line right over there and then we could just think about well looks like there are some points the way I've drawn it that point right over there seems to have the same slope this point right over there has seems to have the same slope and so that's all what the mean value theorem is claiming that there's going to be at least one C if we meet the conditions for the mean value theorem where the derivative at that point is the same as the average rate of change from the first endpoint to the second now what are the conditions for the mean value theorem to apply and we've reviewed this in multiple videos one way to think about it if we're talking about the closed interval from 3 to 7 one condition is that you have to be differentiable differentiable over the open interval from 3 to 7 so that's the interval but not including the endpoints and you have to be continuous continuous over the entire closed interval so including the endpoints and one interesting thing that we've mentioned before is that differentiability implies continuity so something is differentiable over the open interval it's also going to be continuous over this open interval and so the second condition would just say well then we also have to be continuous at the endpoints now let's look at the chancers so it says which condition makes it James's claim true so we have to feel good about these two things right over here in order to make jeans as claim choice a H is continuous over the closed interval from 3 to 7 so that does meet this second condition but continuity does not imply differentiability so that doesn't give us the confidence that we are differentiable over the open interval if you're differentiable you're continuous but if your continuous you're not necessarily differentiable classic example of that is if we have a sharp turn something like that we wouldn't be differentiable at that point even though we are continuous there so it's let me rule that one out so I'm gonna rule that one out choice B H is continuous and decreasing over the closed interval from 3 to 7 now that doesn't help us either because it still doesn't mean you're differentiable you could be continuous and then decreasing and still have one of these sharp turns where you're not differentiable so we will rule this one out H is differentiable over the closed interval from 3 to 7 this one feels good because if you're differentiable over the closed interval you're definitely gonna be differentiable over the open interval that does not include the endpoints this is a subset of this right over here and if your differentiable over a closed interval you're going to be continuous over differentiability implies continuity so I like this choice right over here now this last choice is the limit as X approaches 5 of H prime of X is equal to 1 so they're saying the limit of our derivative as we approach 5 is equal to 1 now that doesn't know we don't know for sure this limit might be true but that still does not necessarily imply that H prime of 5 is equal to 1 we still don't know that and you know five is in this interval but we still don't know just from this statement alone that there's definitely some C in the interval whose derivative is the same as the average rate of change over the interval so I would rule this one out as well