anonymous
  • anonymous
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?
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jkristia
  • jkristia
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 a-b.
anonymous
  • anonymous
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, \[g(x) = f(a) + \frac{ f(b)-f(a) }{ b-a }(x-a)\] Then we can construct another function h(x) = f(x) - g(x) \[h(x) = f(x) - (f(a) + \frac{ f(b)-f(a) }{ b-a }(x-a))\] 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.