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- [Tutor] Let g of x equal one over x.

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Can we use the mean value theorem

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to say that the equation g[br]prime of x is equal to one half

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has a solution where negative one

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is less than x is less than two,

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if so, write a justification.

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Alright, pause this video

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and see if you can figure that out.

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So the key to using[br]the mean value theorem,

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even before you even think about using it,

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you have to make sure

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that you are continuous[br]over the closed interval

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and differentiable over the open interval,

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so this is the open interval here

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and then the closed interval[br]would include the end points.

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But you might immediately realize

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that both of these intervals[br]contain x equals zero

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and if x equals zero,[br]the function is undefined

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and if it's undefined there,

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well it's not going to continuous

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or differentiable at that point

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and so no,

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not continuous

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or differentiable,

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differentiable

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over the interval,

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over the interval.

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Alright, let's do the second part.

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Can we use the mean value theorem

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to say that there is a value[br]c such that g prime of c

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is equal to negative one half

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and one is less than c is less than two

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if so, write a justification.

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So pause the video again.

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Alright, so in this[br]situation, between one and two

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on both the open and the closed intervals,

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well, this is a rational function

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and a rational function[br]is going to be continuous

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and differentiable at[br]every point in its domain

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and its domain completely contains

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this open and closed interval

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or another way to think about it,

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every point on this open interval

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and on the closed[br]interval is in the domain,

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so we can write g of x

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is a rational

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function,

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which lets us know that it is continuous

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and differentiable at[br]every point in its domain,

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at every point

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in its domain,

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the closed interval from one to two

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is in domain

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and so now, let's see what[br]the average rate of change is

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from one to two

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and so we get g of two

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minus g of one

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over two minus one is equal to one half

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minus one over one, which is[br]equal to negative one half,

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therefore,

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therefore

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by the

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mean value theorem,

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there must be a c

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where one is less than c is less than two

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and g prime of c is equal to

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the average rate of change[br]between the end points,

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negative one half

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and we're done, so we can put[br]a big yes right over there

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and then this is our justification.