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OJFord
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Lecture 14: MVT proof is incomplete?
Moving parallel line up and down to find where it's a tangent to the curve is intuitive, sure, but that doesn't constitute a proof, does it? How do we *know* it will *always* become a tangent, for all possible curves? However intuitive it is, it isn't proven here.
Unless I'm missing something?
 one year ago
 one year ago
OJFord Group Title
Lecture 14: MVT proof is incomplete? Moving parallel line up and down to find where it's a tangent to the curve is intuitive, sure, but that doesn't constitute a proof, does it? How do we *know* it will *always* become a tangent, for all possible curves? However intuitive it is, it isn't proven here. Unless I'm missing something?
 one year ago
 one year ago

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jkristia Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure if this is correct (and I have not watched that lecture), but I think MVT is an extension to Rolle's theorem which states (something like), if you have 2 points on a curve a an b and f(a) = f(b), meaning a horizontal line from a to b. Then there must be at least one point 'c' on the curve where the tangent line is horizontal too, meaning the derivative is 0. Now if you slant the line between a and b, then the tangent line at 'c' will have the same slope as the slope ab.
 one year ago

Weidi Group TitleBest ResponseYou've already chosen the best response.0
It's the extension of the Rolle's theorem. Draw a curve,dw:1371142343381:dw A: (a,f(a)) B: (b,f(b)) g(x) is the line, \[g(x) = f(a) + \frac{ f(b)f(a) }{ ba }(xa)\] Then we can construct another function h(x) = f(x)  g(x) \[h(x) = f(x)  (f(a) + \frac{ f(b)f(a) }{ ba }(xa))\] Then use Rolle's theorem h(a) = h(b) = 0, so there must be a c h'(c) = 0 That's the proof, hope you like it
 one year ago
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