Prove there exists a point $\xi$ so that $f'(\xi)=0$

Question

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.

Answer 1

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 $f'(x_3)=0$.

Answer 2

Okay: If $f$ is constant, this is trivial. Else your function needs to have at least one local extremum. That is clear: If for any interval $[a,b]$ $f$ takes it’s extreme values only in $a$ and $b$, then you have that either $f(a)< f(x) < f(b)$ or $f(a) > f(x) > f(b)$. For $x\in(a,b)$. If this holds for all $a,b$ then it is easy to see from continuity that in fact one of these equations holds for all $a,b$ and thus $f$ would be strictly monotonous or antitonous, which of course cannot happen.

So thus $f$ has at least one local extremum $x$, which implies $f'(x)=0$.

Answer 3

Using the IVP in the comments, it's pretty simple. If your function is constant, you are done. Assume $f'(x)>0$ and nonconstant, that means for some $f(c)$, $f(c)\neq 17$. But then it being strictly increasing means that if $f(c)>17$ then for all $x>c$, $f(x)>f(c)>17$, so you are bound away from 17 and thus contradicts the limit. Otherwise the same thing only to negative infinity. Reverse the two for strictly decreasing. Thus it can't be strictly increasing or decreasing, so you have a + and a - value, and the IVP gives you a 0