# Thread: help in question using Intermidate Value Theorem

1. ## 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...

2. ## Re: help in question using Intermidate Value Theorem

Originally Posted by orir
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 $\displaystyle {\lim _{x \to {{\frac{{ - \pi }}{2}}^ + }}}f\;\& \;{\lim _{x \to {{\frac{\pi }{2}}^ - }}}f$

3. ## Re: help in question using Intermidate Value Theorem

what will help me consider those limits (infinity, and minus infinity)?

4. ## Re: help in question using Intermidate Value Theorem

Originally Posted by orir
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 $\displaystyle \left( { - \infty ,0} \right)$.