# Use the intermediate value theorem to show that the equation, tan(x)=2x has an infinite amount of real solutions.

Mariah Sparks 2022-07-19 Answered
Use the intermediate value theorem to show that the equation,
$tan\left(x\right)=2x$
has an infinite amount of real solutions.
So far I have used the IVT to show that for $f\left(x\right)=tan\left(x\right)$ in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ there is a $L$ between $-\mathrm{\infty }$ and $\mathrm{\infty }$ such that there is a value $c$ where $f\left(c\right)=L$
I also have shown that for $g\left(x\right)=2x$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ there is a $K$ between $-\pi$ and $\pi$ such that there is a value $d$ where $f\left(d\right)=K$
Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f\left(x\right)$ consists of all reals that $f\left(x\right)$ must intersect with $g\left(x\right)$ or there is some value $x$ in the interval such that $2x=tan\left(x\right)$? If this is correct then how would I extend this result to all intervals $\left[\pi \left(n-\frac{1}{2}\right),\pi \left(n+\frac{1}{2}\right)\right]$?
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Consider $f\left(x\right)=\mathrm{tan}\left(x\right)-2x$. Then ${f}^{\prime }\left(x\right)=\frac{1}{{\mathrm{cos}}^{2}\left(x\right)}-2.$
On $\frac{\pi }{4},\frac{\pi }{2}$ and $-\frac{\pi }{2},-\frac{\pi }{4}$, ${f}^{\prime }\left(x\right)>0$.
Since $\underset{x\to \frac{\pi }{2}-}{lim}{f}^{\prime }\left(x\right)=+\mathrm{\infty },\underset{x\to \frac{\pi }{2}-}{lim}f\left(x\right)=+\mathrm{\infty }$.
Similarly since $\underset{x\to -\frac{\pi }{2}+}{lim}{f}^{\prime }\left(x\right)=-\mathrm{\infty },\underset{x\to -\frac{\pi }{2}+}{lim}f\left(x\right)=-\mathrm{\infty }$.
So $\mathrm{\exists }{x}_{1}\in \left(-\frac{\pi }{2},-\frac{\pi }{4}\right)$, such that, $f\left({x}_{1}\right)<0$, and $\mathrm{\exists }{x}_{2}\in \left(\frac{\pi }{4},\frac{\pi }{2}\right)$, such that $f\left({x}_{2}\right)>0$. By intermediate value theorem, $\mathrm{\exists }{x}_{0},{x}_{1}<{x}_{0}<{x}_{2},f\left({x}_{0}\right)=0$ since $f\left(x\right)$ is continuous.
Same reasoning can be applied to $-\frac{\pi }{2}+k\pi ,\frac{\pi }{2}+k\pi$, $k\in \mathbb{Z}$. So there are infinite real solution for $f\left(x\right)$.

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