Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ a function twice differentiable such that $\lim\limits_{x \to \infty}f'(x) = \lim\limits_{x \to -\infty}f'(x) = A$. Prove that there exists $c \in \mathbb{R}$ such that $f''(c) = 0$.
I tried supose that $f''(c) \neq 0$ for all $c \in \mathbb{R}$, then $f'$ has no minimum and maximum. Can I say that $f'$ is increasing or decreasing? If yes, this is contrary to the hypothesis. If not, I would like suggestions.
On
If $f'$ is constant then we are done. Otherwise there is a value, say $B=f'(b) $, of $f'$ which is different from $A$ and let's assume that $B>A$ (the case $B<A$ is similar). Choose any number $C$ between $A$ and $B$ and then by the given limiting conditions we can ensure that there is a number $N>|b|$ such that $f'(x) < C$ whenever $x\leq -N$ or $x\leq N$.
Now $f'$ is continuous on $[-N, N] $ and hence attains its maximum value in this intervals. Since $b\in(-N, N) $ and $f'(b) =B>C$ it follows that this maximum value is attained at an interior point $c\in(-N, N) $. By usual principle of maxima/minima we have $f''(c) =0$.
Incomplete answer: I need the continuity of $f''$.
Consider the function $g(x)=f'(x)-f'(-x)$. This function is odd and $$\lim_{x\to\pm\infty}g(x)=0$$
Let $\epsilon >0$. Since $g(0)=0$ and $g$ is continuous, there exists some $\delta>0$ such that $|g(x)|<\epsilon$ if $|x|<\delta$. Let $\epsilon_1=|g(\delta/2)|$. There exists some $K>\delta$ such that $|g(x)|<\epsilon_1$ if $x>K$. Let $\epsilon_2=g(K+1)$. By continuity and the fact that $g$ is odd, there is some $\delta_1\in(0,\delta/2)$ such that $g(\delta_1)=g(K+1)$ or $g(-\delta_1)=g(K+1)$.
Now, by Rolle's theorem, there is some $c\in(-\delta_1,K+1)$ such that $g'(c)=0$. That is, $$f''(c)+f''(-c)=0$$
Again by continuity, there is some $c_1\in(-c,c)$ such that $f''(c_1)=0$.