# How do you use the intermediate value theorem to explain why f(x)=-4/x+tan((pix)/8) has a zero in the interval [1,3]?

##### 1 Answer
Nov 30, 2016

See explanation...

#### Explanation:

Given:

$f \left(x\right) = - \frac{4}{x} + \tan \left(\frac{\pi x}{8}\right)$

Note that:

• The function $- \frac{4}{x}$ has a vertical asymptote at $x = 0$, but is otherwise defined and continuous. So is defined and continuous on $\left[1 , 3\right]$

• The function $\tan \theta$ is defined for all $\theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, including all $\theta \in \left[\frac{\pi}{8} , \frac{3 \pi}{8}\right]$, and is continous in that interval.

Hence $f \left(x\right)$ is defined and continuous over the interval $x \in \left[1 , 3\right]$

We find:

$f \left(1\right) = - \frac{4}{1} + \tan \left(\frac{\pi}{8}\right) = - 4 + \left(\sqrt{2} - 1\right) = \sqrt{2} - 5 < 0$

$f \left(3\right) = - \frac{4}{3} + \tan \left(\frac{3 \pi}{8}\right) = - \frac{4}{3} + \left(\sqrt{2} + 1\right) = \sqrt{2} - \frac{1}{3} > 0$

In summary:

• $f \left(x\right)$ is defined and continuous on the interval $\left[1 , 3\right]$

• $f \left(1\right) < 0$ and $f \left(3\right) > 0$

So by the intermediate value theorem, $f \left(x\right)$ takes every value between $f \left(1\right) = \sqrt{2} - 5 < 0$ and $f \left(3\right) = \sqrt{2} - \frac{1}{3} > 0$ for some $x \in \left[1 , 3\right]$.

In particular, there is some value $a \in \left[1 , 3\right]$ such that $f \left(a\right) = 0$

graph{((x-1)^2+(y-sqrt(2)+5)^2-0.01)((x-3)^2+(y-sqrt(2)+1/3)^2-0.01)(y+4/x-tan((pix)/8)) = 0 [-8.94, 11.06, -6.24, 3.76]}