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Thread: help in question using Intermidate Value Theorem

  1. #1
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    help in question using Intermidate Value Theorem

    i need to prove that the function $\displaystyle f(x)= 2tanx-\frac{1}{cosx}$ gets every real value in the open range $\displaystyle (-\frac{\pi}{2},\frac{\pi}{2}) $
    what i was able to do so far is to translate the function to: $\displaystyle \frac{2sinx-1}{cosx}$ and say..
    let t>0 (at R); i know that $\displaystyle f(0)=-1 $, so what is left for me to do in order to use Intermidate Value Theorem is to find a point which is bigger than t. but i'm stuck...
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  2. #2
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    Re: help in question using Intermidate Value Theorem

    Quote Originally Posted by orir View Post
    i need to prove that the function $\displaystyle f(x)= 2tanx-\frac{1}{cosx}$ gets every real value in the open range $\displaystyle (-\frac{\pi}{2},\frac{\pi}{2}) $
    what i was able to do so far is to translate the function to: $\displaystyle \frac{2sinx-1}{cosx}$ and say..
    let t>0 (at R); i know that $\displaystyle f(0)=-1 $, so what is left for me to do in order to use Intermidate Value Theorem is to find a point which is bigger than t. but i'm stuck...
    First plot the graph.

    You can consider $\displaystyle {\lim _{x \to {{\frac{{ - \pi }}{2}}^ + }}}f\;\& \;{\lim _{x \to {{\frac{\pi }{2}}^ - }}}f$
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  3. #3
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    Re: help in question using Intermidate Value Theorem

    what will help me consider those limits (infinity, and minus infinity)?
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  4. #4
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    Re: help in question using Intermidate Value Theorem

    Quote Originally Posted by orir View Post
    what will help me consider those limits (infinity, and minus infinity)?

    Well for each $\displaystyle 0 < \varepsilon < 1$ the function $\displaystyle f(x)=\tan(x)-\sec(x)$ is continuous on $\displaystyle \left[ {0,\frac{\pi }{2} - \varepsilon } \right]$.

    As such it takes all values in $\displaystyle \left[ {f(0),f\left( {\frac{\pi }{2} - \varepsilon } \right)} \right] \to \left( {0,\infty } \right)$.

    You can make a similar argument for $\displaystyle \left( { - \infty ,0} \right)$.
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