$\displaystyle f(x)=e^{-x}+sinx $i need to prove that inf f([0,00))=-1

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- Jun 16th 2013, 05:08 AMorirProving infimum
$\displaystyle f(x)=e^{-x}+sinx $i need to prove that inf f([0,00))=-1

- Jun 16th 2013, 05:19 AMProve ItRe: Proving infimum
Isn't it because $\displaystyle \displaystyle e^{-x} > 0$ and $\displaystyle \displaystyle -1 \leq \sin{(x)} \leq 1$ for all x?

- Jun 16th 2013, 05:52 AMorirRe: 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 - Jun 16th 2013, 06:03 AMPlatoRe: Proving infimum
- Jun 16th 2013, 08:00 AMorirRe: Proving infimum
I need to solve that without sequences.. Havent learnt that yet..

Is there another way? - Jun 16th 2013, 08:34 AMPlatoRe: Proving infimum