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Math Help - Function with e problem

  1. #1
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    Function with e problem

    Prove that e^x >= 1+x for all x in [0,inf).

    My solution so far:

    I let f(x) = e^x - 1 - x, now if I can prove f is increasing on [0,inf), then it will work, but I have difficulties trying to do that...
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Prove that e^x >= 1+x for all x in [0,inf).

    My solution so far:

    I let f(x) = e^x - 1 - x, now if I can prove f is increasing on [0,inf), then it will work, but I have difficulties trying to do that...
    Let f(x)=e^x-x-1

    This function is differenciable apply the increasing/decreasing theorem,

    f'(x)=e^x-1
    Now if it true that e^x-1>0 on x>0.
    Yes, why?

    Because, it is equivalent to saying,
    e^x>1 and e^0=1 thus if x>0 then e^x > 1 because e^x is an increasing function. Q.E.D.
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  3. #3
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    oh, man, this is easy. I was trying to do this in a much harder way...

    Thanks!
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