# A problem on Rolle's theorem

So there's this question that asks us to prove that, between any two roots of $$\tan x=1$$ there exists at least one root of $$\tan x =-1$$. Suppose we assume that $$a,b$$ are two roots of $$\tan x-1=0$$, then $$f(a)=f(b)=0$$, where $$f(x)= \tan x-1$$. According to the theorem, $$f'(c)=0$$ where $$c \in (a,b)$$,i.e., $$\sec^2 c =0$$....and this is not defined. Either there's something wrong with my understanding or with the problem. Please help.

• The function is not continuous on some intervals. It has infinite discontinuities at $x=\frac{k\pi}{2}$ for some $k\in\mathbb{Z}-\{0\}$. So Rolle's theorem does not apply. – C Squared Aug 28 '20 at 8:31

## 2 Answers

The function $$f(x)=\tan(x)-1$$ is not continuous for every $$x\in[a,b]$$, ($$a,b$$ being consecutive roots of $$f$$), so Rolle's Theorem cannot be applied in such interval.

Let $$f(x)=\tan x -1$$, and $$g(x)=\tan x +1$$.

The roots of $$f(x)$$ occur in the interval $$I=[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$$ for some $$k\in\mathbb{Z}$$. The roots of $$f(x)$$ occur at the endpoints of interval $$I$$.

The roots of $$g(x)$$ occur in the interval $$J=[-\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$$ for some $$k\in\mathbb{Z}$$. The roots of $$g(x)$$ occur at the endpoints of interval $$J$$.

The interval $$(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$$ is contained in $$[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$$ and a root of $$g(x)$$ occurs at $$x=\frac{3\pi}{4}+\pi k$$. Roots of $$f(x)$$ occur at $$x=\frac{\pi}{4}+\pi k$$ and $$x=\frac{5\pi}{4}+\pi k$$.

$$\frac{\pi}{4}+\pi k<\frac{3\pi}{4}+\pi k<\frac{5\pi}{4}+\pi k$$.

So there is at least one root of $$g(x)$$ between any two roots of $$f(x)$$.