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why does the MVT allow us to conclude that if D(f(x)) = D(g(x)) then f(x) = g(x) + C ???
 9 months ago
 9 months ago
why does the MVT allow us to conclude that if D(f(x)) = D(g(x)) then f(x) = g(x) + C ???
 9 months ago
 9 months ago

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dan815Best ResponseYou've already chosen the best response.0
im assuming D is differential
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
hey i dont htink thats the way you want it proved tho, with the mean value theorm
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
go here and watch from time 28:04 mins http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/videolectures/lecture15antiderivatives/
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
http://www.youtube.com/watch?v=MI0b4h3rS0
 9 months ago

SohiBest ResponseYou've already chosen the best response.0
I was also wondering the same thing when going through the course. I tried to make up the proof myself, I don't know if it is correct but here it is: MVT says (f(b)f(a))/ba = D(f(c)) and (g(b)g(a))/ba = D(g(c)) But if we take 'b' as variable(this is my major assumption) over the domain of the functions f and g, we get (f(x)f(a))/xa = D(f(x)) and (g(x)g(a))/xa = D(g(x)) And we are given D(f(x)) = D(g(x)), so (f(x)f(a))/xa = (g(x)g(a))/xa => f(x)f(a) = g(x)g(a) => f(x) = g(x) + (f(a)g(a)) => f(x) = g(x) + constant; as ((f(a)g(a)) is a constant Q.E.D.
 9 months ago

molonlabeBest ResponseYou've already chosen the best response.0
You can prove it without MVT: Proof by Contraposition: To prove that p implies q, you can instead attempt to prove NOT q implies NOT p, this is because the two statements are logically equivalent. Thus, if the "NOT" statement is true, then the other statement must be true as well. In this case, p is D(f(x))=D(g(x)) and q is f(x)=g(x)+c This means that NOT p is D(f(x)) =/= D(g(x)) and NOT q is f(x)=/=g(x)+c So lets try to prove NOT q implies NOT p: if f(x) =/= g(x) + C, then by the definition of the derivative, D(f(x)) =/= D(g(x)). Therefore, we have proven our contrapositive, and logically, proven the original.
 9 months ago
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