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A question in Calculus 1

  1. Apr 11, 2008 #1
    we have f(0)=1, f '(x)>=f(x)

    we shall prove f(x)>e^x for every x>=0

    thanx to the solvers
  2. jcsd
  3. Apr 11, 2008 #2
    what? that's not true f(0) = e^0
  4. Apr 11, 2008 #3
    sorry, i had a smal mistake:

    we have f(0)=1, f '(x)>=f(x)

    we shall prove f(x)>=e^x for every x>=0

    thanx to the solvers
  5. Apr 11, 2008 #4

    Gib Z

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    Ok. Welcome to Physicsforums shgidi, I hope you enjoy it here =]

    Please read our homework policy (theres a sticky at the top of this sections page), you must post your homework in the homework section! Also, we require you must show us any work that you have attempted (and yes you must attempt whatever you know) because otherwise its not helping you, its just doing your homework!

    It is only because I am so VERY nice that I will give you a hint :P
    In the interval we are interested in, [itex]x \geq 0[/itex], the function is always increasing because f(0) = 1, so f'(0) is great than 1, and they both just keep increasing. Since they are both positive, divide by f(x) on both sides. Now take an integral of that inequality from 0 to t. I basically did it for you :(
  6. Apr 11, 2008 #5
    OK, I promise for the next time to look and find the homework section.
    I've posted my question here, because I saw here many questions in the same style, so I thought it would fit.

    anyway, 10x for your solution, but it does'nt quiet help me, because Integrals is not part of the material of my current course.
  7. Apr 11, 2008 #6


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    And that's one of the reasons we ask that people show what they have tried: we have no idea what concepts you have available to do this.

    You might try the Mean Value Theorem. I hope you know what that is!
  8. Apr 11, 2008 #7
    Ok, Ill try again... this is a question from a test I had. The course is calculus 1, and I have for use all of the basic theorems, including mean value, lagrange etc..

    I tried some variations of those theorems, but I still have no solution.
    Ill be REALLY glad to see a solution.

  9. Apr 12, 2008 #8


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    Are you able to treat the case = instead of >= ?

    Do that and then try to think about > .
  10. Apr 12, 2008 #9
    you don't need to prove the >, if = is satisfied on the interval then the theorem is true. how is this not simply true definition since e^x's derivative is itself hence g(x)=e^x=g'(x)
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