Limits tricky problem Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is twice differentiable.
We know that:
$$\lim_{x\to-\infty}\ f(x) = 1$$
$$\lim_{x\to\infty}\ f(x) = 0$$
$$f(0) = \pi$$
We have to prove that there exist at least two points of the function in which $f''(x) = 0$.
How could we do it in a rigorous way? It is pretty intuitive, but in a rigorous way it isn't that simple for me...
 A: Let $x_0$ be a point where $f$ attains its maximum value (which is $\geq \pi$). At $x_0$ we have that $f''(x_0)\leq 0$. 
We claim that there is a point $x_1>x_0$ such that $f''(x_1)=0$. 
Assume that $x_1$ does not exist then $f''(x)\not=0$ for $x>x_0$. By Darboux theorem the sign of $f''(x)$ in  $(x_0,+\infty)$ should be always the same. We have two cases:
i) $f''(x)>0$ for all $x>x_0$ that is $f(x)$ is strictly convex in $(x_0,+\infty)$. Hence $f'$ is strictly increasing and $f'(x)>0$ for $x>x_0$ which implies that $f(x)>f(x_0)$ against the fact that $x_0$ is a maximum point.
ii) $f''(x)<0$ for all $x>x_0$ that is $f(x)$ is strictly concave in $(x_0,+\infty)$. Hence $f'$ is strictly decreasing and $f'(x)<0$ for $x>x_0$. Moreover the graph of $f$ remains under the tangent line at some $t>x_0$: for all $x\in (x_0,+\infty)$
$$f(x)\leq f'(t)(x-t)+f(t).$$
Now by taking the limit for $x\to +\infty$ to both sides, we get
$$0\leftarrow f(x)\leq f'(t)(x-t)+f(t)\to -\infty.$$ 
Contradiction!
By the same argument we get another zero of $f''(x)$ for $x<x_0$.
A: For the heck of it, let's do a proof that invokes only the Intermediate and Mean Value Theorems.  In particular, let's eschew any explicit use of the Extreme Value Theorem.
If $f$ ever takes the same value infinitely often, the Mean Value (or Rolle's) Theorem applied twice shows there are two values of $x$ for which $f''(x)=0$.  Indeed, it suffices for $f$ to take the same value four times:  If $f(a)=f(b)=f(c)=f(d)$ with $a\lt b\lt c\lt d$, then there are points $a'\in(a,b)$, $b'\in(b,c)$, and $c'\in(c,d)$ for which $f'(a')=f'(b')=f'(c')=0$, and this in turn implies there are points $a''\in(a',b')$ and $b''\in(b',c')$ for which $f''(a'')=f''(b'')=0$.
We henceforth assume $f$ never takes the same value infinitely often.  We now argue that there is a $c$ such that $f(c)\ge f(0)$ and $f'(c)=0$.  (If we allowed ourselves the Extreme Value Theorem, this conclusion would be almost automatic and in fact we could have skipped the previous paragraph as well.)
If $f'(0)=0$, we simply let $c=0$.  If $f'(0)\gt0$, then there exists $a\gt0$ for which $f(x)\gt f(0)$ for all $x\in(0,a)$.  Since $f(0)\gt\lim_{x\to\infty}f(x)$, the Intermediate Value Theorem gives us a $b\in[a,\infty)$ for which $f(b)=f(0)$.  Our new assumption on $f$ tells us there is a smallest such $b$, and the Intermediate Value Theorem guarantees $f(x)\gt f(b)=f(0)$ for all $x\in(0,b)$.  The Mean Value (or Rolle's) Theorem now says there is a $c\in(0,b)$ for which $f'(c)=0$.  A similar argument applies if $f'(0)\lt0$, with a largest $b\le a\lt0$ and $c\in(b,0)$.
Now let $B$ be a bound such that $f(x)\lt2$ for $|x|\ge B$, the existence of which is guaranteed by the limiting behavior of $f$. (Actually, $2$ is just a convenient number between between $\pi$ and $1$; any number between $f(c)$ and the larger of the two limiting values will do.)  Let $g_+(x)=m_+(x-c)+f(c)$ and $g_-(x)=m_-(x-c)+f(c)$ be linear functions with $g_+(B)=g_-(-B)=2$, and consider the functions $f_+(x)=f(x)-g_+(x)$ for $x\ge c$ and $f_-(x)=f(x)-g_-(x)$ for $x\le c$. Note that $f_+(c)=f_-(c)=0$ and observe that
$$m_+={g_+(B)-g_+(c)\over B-c}={2-f(c)\over B-c}\lt0\quad\text{and}\quad m_-={g_-(c)-g_-(-B)\over c+B}={f(c)-2\over c+B}\gt0$$
The sign of the slope $m_+$ implies $f_+(x)\gt0$ for $x\in(0,a)$ and also in the limit as $x\to\infty$.  But $f_+(B)=f(B)-g_+(B)\lt2-2=0$.  The Intermediate Value Theorem tells us that $f_+(c)=f_+(b)=f_+(b')=0$ for two values $b_1$ and $b_2$ with $c\lt b_1\lt B\lt b_2$.  Applying the Mean Value Theorem to $f_+(c)=f_+(b_1)$ and $f_+(b_1)=f_+(b_2)$ gives two values $b_1'$ and $b_2'$ with $c\lt b_1'\lt b_1\lt b_2'\lt b_2$ and $f_+'(b_1)=f_+'(b_2)=0$.  And applied to $f_+'(b_1)=f_+'(b_2)=0$, MVT coughs up a point $d\in(b_1,b_2)$ for which $f_+''(d)=0$.  But $g_+''(x)=0$ for all $x$, hence we have a point $d_+\gt c$ for which $f''(d_+)=0$.
The same argument, just with some of the signs reversed, shows there is a point $d_-\lt c$ for which $f''(d_-)=f_-''(d_-)=0$, and now we are truly done.  We have $f''(x)=0$ at two distinct points, $x=d_+$ and $x=d_-$.
A: since f(0) > 1 $\lim_\limits {x\to -\infty} f(x)$ and $\lim_\limits {x\to \infty} f(x)$ and $f(x)$ is not monotonic.
Since $f(x)$ is  continuous and differentiable, there it takes on a maximum value somewhere.  $f'(c) = 0$
There exists an $x$ on each tail such that  f'(x) = 0$
By the mean value theorem, there exist points in $(-\infty, c)$ and $(c,\infty)$ where $f''(x) = 0$ 
A: Since the limits of $f$, as $x$ tends to $\infty$ and $-\infty$ both exist, and 
$$
\lim_{x\to\infty}f(x),\,\,\lim_{x\to-\infty}f(x)<f(0)=\pi,
$$
then $f$ attains a total maximum, say at $x_0$, with $f(x_0)\ge\pi$, and thus $f'(x_0)=0$. 
The fact that $\,\lim_{x\to-\infty}f(x)<f(0)$, implies the existence of $x_1\in (-\infty,x_0)$, where $f'(x_1)>0$, and similarly, an  $x_2\in (x_0,\infty)$, where $f'(x_2)<0$. But the existence of the limits 
$\lim_{x\to\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$ also implies that
$$
\lim_{x\to-\infty} f(x+1)-f(x)=0\quad\text{and}\quad\lim_{x\to\infty} f(x+1)-f(x)=0,
$$ 
and hence, $f(x+1)-f(x)$ can become arbitrarily small both for $x$ or $-x$ sufficiently large. Due to Mean Value Theorem $f(x+1)-f(x)=f'(\xi)$, for some $\xi\in(x,x+1)$, and hence there exist $\xi_1<x_1$ and $\xi_2>x_2$, such that
$$
\xi_1<x_1<x_0<x_2<\xi_2
$$
and
$$
f'(\xi_1)<f'(x_1)>f'(x_0) \quad \text{and}\quad f'(x_0)>f'(x_2)<f'(\xi_2).
$$
Hence $f''$ vanishes at some point in each of the intervals
$$
(\xi_1,x_0) \quad\text{and}\quad (x_0,\xi_2).
$$
