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al245

  • one year ago

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

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  1. dan815
    • one year ago
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    wat is D

  2. dan815
    • one year ago
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    |dw:1372709144825:dw|

  3. dan815
    • one year ago
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    |dw:1372709185184:dw|

  4. dan815
    • one year ago
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    |dw:1372709228702:dw|

  5. dan815
    • one year ago
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    im assuming D is differential

  6. al245
    • one year ago
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    Thank you

  7. dan815
    • one year ago
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    hey i dont htink thats the way you want it proved tho, with the mean value theorm

  8. dan815
    • one year ago
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    go here and watch from time 28:04 mins http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-15-antiderivatives/

  9. dan815
    • one year ago
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    http://www.youtube.com/watch?v=-MI0b4h3rS0

  10. dan815
    • one year ago
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    or there

  11. dan815
    • one year ago
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    |dw:1372808395198:dw|

  12. Sohi
    • one year ago
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    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))/b-a = D(f(c)) and (g(b)-g(a))/b-a = 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))/x-a = D(f(x)) and (g(x)-g(a))/x-a = D(g(x)) And we are given D(f(x)) = D(g(x)), so (f(x)-f(a))/x-a = (g(x)-g(a))/x-a => 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.

  13. molonlabe
    • one year ago
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    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.

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