Prove $f^\ {''}(c)=0$ 
Let $f:\mathbb{R} \rightarrow\mathbb{R}$ be twice differentiable and bounded, and suppose $f$ gets minimum at $x_0$. Prove there exists $c\in\mathbb{R}$ such that $f^\ {''}(c)=0$.

$f$ is bounded $\Rightarrow \exists M>0 \ s.t \ |f(x)|\leq M, \forall x\in\mathbb{R}$
So by MVT I got that the derivative is positive for all $x>x_0$ and negative for all $x<x_0$, and from that I concluded that $f(x)\rightarrow M$ when $x\rightarrow\pm\infty$.
Not sure how to continue from here.
Any help appreciated.
 A: We know that $f'(x_0) = 0$ and $f''(x_0) \geq 0$, since there's a minimum.  Suppose for contradiction that $f''$ is non-zero everywhere. Then, $f''(x_0) > 0$.  Now, we may state that $f'(x) > 0$ when $x > x_0$.
Pick an $x_1 > x_0$; note that $f'(x_1) > 0$.  Now, select an $R > 0$ large enough so that
$$
M - f(x_1) < f'(x_1)(R - x_1)
$$ 
We note that $f$ is bounded above by $M$ and increasing on $[x_1,R]$.  However, we have
$$
|f(R) - f(x_1)| = f(R) - f(x_1) \leq M - f(x_1) < f'(x_1)(R - x_1)
$$
by the mean value theorem, there exists an $x_2 \in [x_1,R]$ such that $f'(x_2) = \frac{f(R) - f(x_1)}{R - x_1} < f'(x_1)$.  By the MVT again, there exists an $x_3 \in [x_1,x_2]$ such that $f''(x_3) < 0$ (recall: $x_0 < x_1 < x_3 < x_2 < R$).  This contradicts our supposition.
A: Let us take the case when listed conditions are satisfied

*

*$f'(x)\gt 0,\space \forall x \in (x_0,\infty)$

*$f(x_n)\gt \cdots \gt f(x_2) \gt f(x_1), \space \forall x_1,x_2,\cdots,x_n \in (x_0,\infty), \space \exists f(x_k)\gt f(x_n) \space \forall k\gt n$

*$f'(x)\lt 0, \space \forall x \in (-\infty,x_0)$

*$f(x_n)\lt \cdots \lt f(x_2) \lt f(x_1), \space \forall x_1, x_2, \cdots,x_n \in (-\infty,x_0),\space \exists f(x_k)\lt f(x_n) \space \forall k\gt n$
These conditions indicate that the function is not bounded. Since we are said to take the function to be bounded, this situation is a contradiction. We can modify the conditions for the function to be bounded. If we say $f'(x)\gt 0, \space \forall x \in (x_0,x_1)$, we obtain a bounded function (we should set the similar conditions for $x\lt x_0$).$$f(x_0)=0, \space f(x_1)=0\Rightarrow \frac{f(x_1)-f(x_0)}{x_1-x_0}=f''(c)=0 $$
We can also assert that $f'(x)\gt 0,\space \forall x\in (x_0,\infty),\space \lim_{x\to \infty} f'(x)=0$. Since we know the function is concave up for $x \gt x_0$ and it has a horizontal asymptote, we can assert that the function is concave down for some interval $(c,\infty)$. Using intermediate value theorem we can deduce there is an inflection point at $c$, which means $f''(c)=0$.
