롤의 정리와 평균값 정리

If $f(x)=\dfrac{x}{2}+\sin x$ is one–to–one over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if $f(a)=f(b)$ for distinct $a$ and $b$, then $f'(c)=0$ for some $c$ between $a$ and $b$. Since $f'(x)=\frac{1}{2}+\cos x$, $f'(x)=0$ when $x=2n\pi\pm\frac{2\pi}{3}$ for some integer $n$. So, in particular, if $a$ and $b$ are distinct numbers in the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$, then for every $c$ strictly between $a$ and $b$, $f'(c) \neq 0$, so by Rolle's Theorem $f(a) \neq f(b)$. Therefore $f(x)$ is one–to–one on the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$.

We should also show that the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$ cannot be extended to a larger interval over which $f(x)$ is still one–to–one. Consider the derivative $f'(x) = \frac{1}{2}+\cos x$. For all $-\frac{2\pi}{3} < x < \frac{2\pi}{3}$, we have $\cos x > -\frac{1}{2}$ (sketch the graph of $\cos x$ yourself) so that $f'(x)>0$ and $f(x)$ is increasing. But at $x=\frac{2\pi}{3}$, $f'(x)=0$, and then for $x$ a bit bigger than $\frac{2\pi}{3}$ we have $\cos x < -\frac{1}{2}$ so that $f'(x)<0$ and $f(x)$ is decreasing. So the graph “reverses direction", and $f(x)$ repeats values. (See the graph of $y=f(x)$ below.) The same is true for $x$ a little smaller than $-\frac{2\pi}{3}.$

When we define the inverse function $f^{-1}(x)$, first we restrict $f$ to $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. Then the range of $f^{-1}$ is also $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. The domain of $f^{-1}$ is the range of $f$ over this interval, so $\left[-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]$.