In lecture 14, Prof. Jerison says that one of the implications of the mean value theorem is that for an interval [a,b] of a function f that satisfies the hypotheses of the Mean Value Theorem, the average change on the interval is between the maximum and minimum values f'(x) reaches on the interval [a,b]. How does this conclusion follow from the MVT?

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- phi

the Mean Value Theorem
https://en.wikipedia.org/wiki/Mean_value_theorem
states that the the slope of the secant line (the line connecting the end points of the [a,b] interval) will equal the slope of a tangent line to the curve at least one point in that interval. (A picture might be clearer than the words, see link)
We can imagine that the curve is a straight line connecting a to b... in which case its slope is equal to the secant line. If the curve is not a straight line then it moves above the straight line (i.e. has higher slope), and (because it must come down at some point to hit the point b, will at some later point have a lesser slope than the straight line connecting a to b.
It should be clear that the slope of the secant line (i.e. straight line connecting a to b)
will be bracketed by the max and min slopes of the (tangents to the) curve.

- phi

to be complete, the curve could first dip below the secant line (i.e. have lesser slope) and at some point later rise up (i.e. have greater slope than the secant line). Just as above, the slope of the secant is bracketed by the min and max slopes of the curve.
And of course, the curve could cross above and below the secant line multiple times. This means there will be multiple points where the slope is steeper or shallower than that of the secant line. One of these slopes will be the max and one the min, and the secant's slope will be between them.

- anonymous

First of all what he said was that the mean value is not stricktly between maximum and minimum but it could also be the maximum or minimum. Which is the case if f'(x) is a constant then the mean value is also both maxima and minima.

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