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Thread: Proving infimum

  1. #1
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    Proving infimum

    $\displaystyle f(x)=e^{-x}+sinx $i need to prove that inf f([0,00))=-1
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  2. #2
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    Re: Proving infimum

    Isn't it because $\displaystyle \displaystyle e^{-x} > 0$ and $\displaystyle \displaystyle -1 \leq \sin{(x)} \leq 1$ for all x?
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  3. #3
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    Re: Proving infimum

    That is saying (-1) is a lowe bound of the function.
    It is something but it does not necesseraly say there is no bigger lower bound (inf).
    I need to prove that -1 is the biggest lower bound, and there is no else
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    Re: Proving infimum

    Quote Originally Posted by orir View Post
    $\displaystyle f(x)=e^{-x}+sinx $i need to prove that inf f([0,00))=-1
    Clearly $\displaystyle \forall x>0,~f(x)\ge -1$. $\displaystyle \forall n\in\mathbb{Z}^+$ define $\displaystyle a_n=\frac{(3n+4)\pi}{2}$.

    It should clear to you that $\displaystyle {\lim _{n \to \infty }}f({a_n}) = - 1$
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    Re: Proving infimum

    I need to solve that without sequences.. Havent learnt that yet..
    Is there another way?
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  6. #6
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    Re: Proving infimum

    Quote Originally Posted by orir View Post
    I need to solve that without sequences.. Havent learnt that yet..
    Is there another way?
    How are we expected to know what you have studied?

    Show that if $\displaystyle \varepsilon > 0$ show that $\displaystyle \exists x_{\varepsilon}$ such that $\displaystyle -1<f\left(x_{\varepsilon}\right)<-1+{\varepsilon}$.
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