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Calle87 Group Title

Prove that sqrt(1+x)>1+(x/2) using MVT

  • one year ago
  • one year ago

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  1. Calle87 Group Title
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    should i find f(x) first?

    • one year ago
  2. quantum77 Group Title
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    This inequality isn't true, are you sure it's not sqrt(1+x)<1+(x/2)? For example if you try x=1 sqrt(1+1)>1+1/2 => sqrt(2)>1.5 is false.

    • one year ago
  3. Calle87 Group Title
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    yes i wrote it wrong, the sign is the wrong way

    • one year ago
  4. Calle87 Group Title
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    |dw:1354858780727:dw|

    • one year ago
  5. Calle87 Group Title
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    |dw:1354858978743:dw|

    • one year ago
  6. Calle87 Group Title
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    im just not sure how to apply the MVT

    • one year ago
  7. tamtoan Group Title
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    square both sides and see what you get .

    • one year ago
  8. quantum77 Group Title
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    You can square both sides but it says mean value theorem? Does the question give you some lower bound like for all x>0 or x>=-1? As the mean value theorem says: for a differentiable function f(x) on the closed interval (a,b) and is continuous on [a,b] then there exists a c where a<=c<=b where: \[f \prime(c)=\frac{ f(b)-f(a) }{ b-a }\] and you are correct that in this case f(x)=1+(x/2)-sqrt(1+x).

    • one year ago
  9. Calle87 Group Title
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    x>0

    • one year ago
  10. Calle87 Group Title
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    ya i cant square both sides

    • one year ago
  11. tamtoan Group Title
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    ty quantum, didn't even know what MVT were :), sorry calle87 :)

    • one year ago
  12. Calle87 Group Title
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    not sure how to do it with just {a,b} use to having an interval

    • one year ago
  13. Calle87 Group Title
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    s'ok :}

    • one year ago
  14. Calle87 Group Title
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    |dw:1354859852306:dw|

    • one year ago
  15. Calle87 Group Title
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    f'(c) is strange

    • one year ago
  16. quantum77 Group Title
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    Okay so that means that there exists a c where 0<c<=x: That: \[f \prime(c)=\frac{ f(x)-f(0) }{ x-0 }\] Then, \[1+c/2-1/(2*\sqrt{1+c})=\frac{ (1+x/2-\sqrt(1+x))-(1-1)}{ x }=\frac{ 1+x/2-\sqrt(1+x) }{ x }\] Notice that, if c>0: 1/sqrt(1+c)<1 therefore should be able to work out f'(c)>0 for all c>0. Then, \[\frac{ 1+x/2-\sqrt(1+x) }\] Since, x>0 the numerator is also greater than 0. Which is the original inequality :)

    • one year ago
  17. quantum77 Group Title
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    Sorry in the last equation it should have said:\[\frac{ 1+x/2-\sqrt(1+x) }{ x }>0\]

    • one year ago
  18. Calle87 Group Title
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    hmmm, f'(c) why is the one not zero?

    • one year ago
  19. Calle87 Group Title
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    |dw:1354860607643:dw|

    • one year ago
  20. Calle87 Group Title
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    this is were i get lost

    • one year ago
  21. quantum77 Group Title
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    Sorry (shoun't be a one). Let me work it out :p

    • one year ago
  22. Calle87 Group Title
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    |dw:1354860920108:dw|

    • one year ago
  23. mahmit2012 Group Title
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    |dw:1354861141850:dw|

    • one year ago
  24. quantum77 Group Title
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    Haha, we both messed up the derivative... it should be f(c)=1/2*(1-1/sqrt(1+c)). Which is not hard to show it is greater than 0

    • one year ago
  25. quantum77 Group Title
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    \[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(1-1/\sqrt(1+c))>=0\]

    • one year ago
  26. Calle87 Group Title
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    so c is a constant

    • one year ago
  27. quantum77 Group Title
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    c is some constant in the range (0,x)

    • one year ago
  28. mahmit2012 Group Title
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    |dw:1354861535246:dw|

    • one year ago
  29. Calle87 Group Title
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    mahmit how do you get that?

    • one year ago
  30. Calle87 Group Title
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    |dw:1354861444641:dw|

    • one year ago
  31. mahmit2012 Group Title
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    what are you doing? it is solved but with different inequality!!

    • one year ago
  32. mahmit2012 Group Title
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    MVT

    • one year ago
  33. Calle87 Group Title
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    so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly

    • one year ago
  34. mahmit2012 Group Title
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    |dw:1354861640565:dw|

    • one year ago
  35. mahmit2012 Group Title
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    yes.

    • one year ago
  36. Calle87 Group Title
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    wow omg, thank you so much!

    • one year ago
  37. mahmit2012 Group Title
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    but instead of > should be<

    • one year ago
  38. mahmit2012 Group Title
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    unless for x<0

    • one year ago
  39. Calle87 Group Title
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    well what do i do with this?

    • one year ago
  40. Calle87 Group Title
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    |dw:1354861684176:dw|

    • one year ago
  41. mahmit2012 Group Title
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    |dw:1354861815578:dw|

    • one year ago
  42. Calle87 Group Title
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    x>0

    • one year ago
  43. mahmit2012 Group Title
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    |dw:1354861953258:dw|

    • one year ago
  44. Calle87 Group Title
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    wait how do you just jump from here ?

    • one year ago
  45. Calle87 Group Title
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    |dw:1354861909373:dw|

    • one year ago
  46. Calle87 Group Title
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    @mahmit2012 do you know what happens to this?

    • one year ago
  47. Calle87 Group Title
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    |dw:1354862035437:dw|

    • one year ago
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