If we have a function $f(x)$ is differentiable in $\mathbb{R}$ and $\lim_{x\rightarrow \pm \infty} f(x) = 17 $. How can we show that $\exists \xi \in (-\infty, \infty) | f'(\xi) = 0$?
I'm wondering whether we can apply Rolles Theorem in this case, because now we've got an infinite interval for which our function is continous?
Thank you.
Best Answer
If $f(x)=17$ then $f'(x)=0$ and we are done. If not, let $x_0$ be a point such that $f(x_0)\neq 17$. Assume that $f(x_0)>17$ (the logic is the same if $f(x_0)<17$. By assumption, we know there exists $M>0$ such that $x\geq M$ implies $|f(x)-17|\leq \frac{f(x_0)-17}{2}$ and $x\leq -M$ implies $|f(x)-17|\leq \frac{f(x_0)-17}{2}$. Importantly, note that
$$f(M),f(-M)\in \left[17-\frac{f(x_0)-17}{2},17+\frac{f(x_0)-17}{2} \right]$$
By the mean value theorem, there exists $x_1\in (x_0,M]$ and $x_2\in [-M,x_0)$ such that
$$f(x_1)=17+\frac{f(x_0)-17}{2}$$
$$f(x_2)=17+\frac{f(x_0)-17}{2}$$
Then by Rolle's Theorem there exists $x_3\in (x_2,x_1)$ such that $f'(x_3)=0$.