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Math Help - 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 f(x)= 2tanx-\frac{1}{cosx} gets every real value in the open range  (-\frac{\pi}{2},\frac{\pi}{2})
    what i was able to do so far is to translate the function to:  \frac{2sinx-1}{cosx} and say..
    let t>0 (at R); i know that  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 f(x)= 2tanx-\frac{1}{cosx} gets every real value in the open range  (-\frac{\pi}{2},\frac{\pi}{2})
    what i was able to do so far is to translate the function to:  \frac{2sinx-1}{cosx} and say..
    let t>0 (at R); i know that  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 {\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 0 < \varepsilon  < 1 the function f(x)=\tan(x)-\sec(x) is continuous on \left[ {0,\frac{\pi }{2} - \varepsilon } \right].

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

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